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# High School Math : Circles

## Example Questions

### Example Question #11 : Circles

Find the area of the shaded region:

Explanation:

To find the area of the shaded region, you must subtract the area of the triangle from the area of the sector.

The formula for the shaded area is:

,

where  is the radius of the circle,  is the fraction of the sector,  is the base of the triangle, and  is the height of the triangle.

In order to the find the base and height of the triangle, use the formula for a  triangle:

, where  is the side opposite the .

Plugging in our final values, we get:

### Example Question #12 : Circles

Find the area of the following sector:

Explanation:

The formula for the area of a sector is

,

where  is the radius of the circle and  is the fraction of the sector.

Plugging in our values, we get:

### Example Question #1 : How To Find The Area Of A Sector

The radius of the circle above is  and .  What is the area of the shaded section of the circle?

Explanation:

Area of Circle = πr2 = π42 = 16π

Total degrees in a circle = 360

Therefore 45 degree slice = 45/360 fraction of circle = 1/8

Shaded Area = 1/8 * Total Area = 1/8 * 16π = 2π

### Example Question #14 : Circles

Find the area of the shaded segment of the circle. The right angle rests at the center of the circle.

Explanation:

We know that the right angle rests at the center of the circle; thus, the sides of the triangle represent the radius of the circle.

Because the sector of the circle is defined by a right triangle, the region corresponds to one-fourth of the circle.

First, find the total area of the circle and divide it by four to find the area of the depicted sector.

Next, calculate the area of the triangle.

Finally, subtract the area of the triangle from the area of the sector.

### Example Question #15 : Circles

is a square.

The arc from to  is a semicircle with a center at the midpoint of .

All units are in feet.

The diagram shows a plot of land.

The cost of summer upkeep is $2.50 per square foot. In dollars, what is the total upkeep cost for the summer? Possible Answers: Correct answer: Explanation: To solve this, we must begin by finding the area of the diagram, which is the area of the square less the area of the semicircle. The area of the square is straightforward: 30 * 30 = 900 square feet Because each side is 30 feet long, AB + BC + CD = 30. We can substitute BC for AB and CD since all three lengths are the same: BC + BC + BC = 30 3BC = 30 BC = 10 Therefore the diameter of the semicircle is 10 feet, so the radius is 5 feet. The area of the semi-circle is half the area of a circle with radius 5. The area of the full circle is 52π = 25π, so the area of the semi-circle is half of that, or 12.5π. The total area of the plot is the square less the semicircle: 900 - 12.5π square feet The cost of upkeep is therefore 2.5 * (900 – 12.5π) =$(2250 – 31.25π).

### Example Question #16 : Circles

In the figure, PQ is the arc of a circle with center O. If the area of the sector is what is the perimeter of sector?

Explanation:

First, we figure out what fraction of the circle is contained in sector OPQ: , so the total area of the circle is .

Using the formula for the area of a circle, , we can see that .

We can use this to solve for the circumference of the circle, , or .

Now, OP and OQ are both equal to r, and PQ is equal to  of the circumference of the circle, or .

To get the perimeter, we add OP + OQ + PQ, which give us .

### Example Question #17 : Circles

If a quarter of the area of a circle is , then what is a quarter of the circumference of the circle?

Explanation:

If a quarter of the area of a circle is , then the area of the whole circle is .  This means that the radius of the circle is 6.  The diameter is 12.  Thus, the circumference of the circle is .  One fourth of the circumference is .

### Example Question #1 : How To Find The Length Of An Arc

Figure not drawn to scale.

In the figure above, circle C has a radius of 18, and the measure of angle ACB is equal to 100°. What is the perimeter of the red shaded region?

36 + 36π

18 + 10π

18 + 36π

36 + 20π

36 + 10π

36 + 10π

Explanation:

The perimeter of any region is the total distance around its boundaries. The perimeter of the shaded region consists of the two straight line segments, AC and BC, as well as the arc AB. In order to find the perimeter of the whole region, we must add the lengths of AC, BC, and the arc AB.

The lengths of AC and BC are both going to be equal to the length of the radius, which is 18. Thus, the perimeter of AC and BC together is 36.

Lastly, we must find the length of arc AB and add it to 36 to get the whole perimeter of the region.

Angle ACB is a central angle, and it intercepts arc AB. The length of AB is going to equal a certain portion of the circumference. This portion will be equal to the ratio of the measure of angle ACB to the measure of the total degrees in the circle. There are 360 degrees in any circle. The ratio of the angle ACB to 360 degrees will be 100/360 = 5/18. Thus, the length of the arc AB will be 5/18 of the circumference of the circle, which equals 2πr, according to the formula for circumference.

length of arc AB = (5/18)(2πr) = (5/18)(2π(18)) = 10π.

Thus, the length of arc AB is 10π.

The total length of the perimeter is thus 36 + 10π.

The answer is 36 + 10π.

### Example Question #18 : Circles

Find the arc length of a sector that has an angle of 120 degrees and radius of 3.

Explanation:

The equation for the arc length of a sector is .

Substitute the given radius for  and the given angle for  to get the following equation:

Simplify:


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# Number of binary gates needed to compute AND and OR of n input bits simultaneously

What is the minimal number of binary gates needed to compute AND and OR of $n$ input bits simultaneously? The trivial upper bound is $2n-2$. I believe that this is optimal, but how to prove this? The standard gate elimination technique does not work here as by assigning a constant to any of the input variables one trivializes one of the outputs.

The problem is also given as an exercise 5.12 in the book "Complexity of Boolean Functions" by Ingo Wegener in a slightly different form: "Let $f_n(x) = x_1\dots x_n \lor \bar{x}_1 \dots \bar{x}_n$. By the elimination method one can prove only a lower bound of size $n+\Omega(1)$. Try to prove larger lower bounds."

• @Ryan: The question is not about AND of OR but about AND and OR. I do not know the answer to Sasha’s question, though. – Tsuyoshi Ito Nov 30 '11 at 17:34
• @TsuyoshiIto Thanks, somehow I managed to parse it incorrectly. It is definitely a nontrivial problem -- one could imagine using other kinds of gates to get an advantage over $2n-2$. – Ryan Williams Nov 30 '11 at 18:48
• @Sasha, have you tried applying SAT solvers to small examples (like $n=4$), as in some of your earlier papers? – Ryan Williams Nov 30 '11 at 21:05
• @Ryan Yes, sure. What we know is that $C_3=3$, $C_4=5$, $C_5 \le 7$. This is for the function from the book (it is $1$ iff all the $n$ input bits are equal). This grows like $2n-3$. And a circuit of size $2n-3$ is easy to construct: first compute $x_i \equiv x_{i+1}$ for all $i=1,\dots,n-1$ ($(n-1)$ gates), and then compute the conjunction of them ($(n-2)$ gates). – Alexander S. Kulikov Dec 1 '11 at 12:24
• @Tsuyoshi: I think that the $2n-3$ gates function of Sasha is the second function of the question ($f_n(x) = x_1\dots x_n \lor \bar{x}_1 \dots \bar{x}_n$) that can be built with $n-1$ XNOR gates (applied to $x_i,x_{i+1}$) and $n-2$ AND gates applied to the XNORs. – Marzio De Biasi Dec 3 '11 at 0:24

This paper of Blum & Seysen may be useful:

N.Blum, M. Seysen. Characterization of all Optimal Networks for a Simultaneous Computation of AND and NOR. Acta Inf. 21: 171-181 (1984)

I have thought that for $x_1\dots x_n\vee \bar{x}_1\dots\bar{x}_n$ $2n-c$ lower bound can be obtained using methods of Blum & Seysen, but it seems this is not the case.

• Is there a public pdf version of the Blum and Seysen paper available? – Marzio De Biasi Dec 3 '11 at 10:31
• @Vladimir, thank you for the reference! I will try to check whether their methods are applicable in this case when find the article. – Alexander S. Kulikov Dec 3 '11 at 12:20
• @Vladimir, thans again! Actually, this paper contains exactly the answer to my question anv even more: it says that to compute AND and OR simultaneously one needs $2n-2$ and any circuit of this size computes AND and OR independently (this is interesting!). It is also not difficult to show that $C(f_n) \ge C(AND,OR)-c \ge 2n-c'$. – Alexander S. Kulikov Dec 3 '11 at 19:02
• @Sasha, yes, I missed this simple construction. To clarify things, in the paper AND and NOR functions are considered, so for AND and OR we get $2n-2$ lower bound by altering one gate and for $x_1\dots x_n\vee \bar{x}_1\dots \bar{x}_n$ --- $2n-5$ – Vladimir Lysikov Dec 3 '11 at 20:22
• Just a reminder @SashaK. if you like the answer, please "accept" it by clicking on the tick mark below the vote count. – Suresh Venkat Dec 5 '11 at 8:32

Your question is related to the well-known question about computing the minimum and maximum of a list simultaneously using the minimum number of comparisons. In that case the answer is $3\lfloor n/2 \rfloor$.

The clever algorithm proving the upper bound translates to an AND/OR circuit with same bound you get, since one of the comparisons computes both a minimum and a maximum.

However, the lower bound (given by an adversary argument) does seem to translate, at least in the case of monotone circuits (since an AND/OR circuit translates to a max/min algorithm). This would imply a lower bound of $3\lfloor n/2 \rfloor$. Perhaps a tight lower bound can be obtained by analyzing the adversary argument.

The upper bound appears in "Introduction to Algorithms", where you can also find the easy argument showing that max/min comparator circuits are valid iff they work for boolean inputs (use an appropriate threshold). The lower bound can be found e.g. here.

• Note in Sasha's question, all 2-bit Boolean functions can be used to construct the circuit. – Ryan Williams Dec 3 '11 at 3:23
• Yes, it is not clear how the lower bound can be translated to the case of all binary functions. – Alexander S. Kulikov Dec 3 '11 at 12:18

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MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

Let me begin with what was formerly a comment above: the feeling among most experts is probably* that for each fixed $d$, the number of isogeny factors of $J_0(p)$ of dimension $d$ should be small compared to the dimension of $J_0(p)$, i.e. $o_d(p)$.
When $p$ is prime, it follows from a 1975 theorem of Ribet (reference below) that the $\mathbb{Q}$-rational endomorphism algebra of $J_0(p)$ is the same as the geometric (i.e., $\mathbb{C}$-rational) endomorphism algebra, and that this algebra is a product of formally real fields, each being the subfield of $\mathbb{Q}$ obtained by adjoining the Fourier coefficients of the various weight $2$ cuspforms of level $p$.
Thus the problem can be viewed as a special case of a popular one in pure algebraic geometry: for which genera $g$ do there exist complex algebraic curves of genus $g$ with, e.g., Jacobians isogenous to a product of elliptic curves? (Or, more generally, with endomorphism algebra containing at least $N$ semisimple factors?) If you count codimensions in the Siegel moduli space corresponding to (say) principally polarized abelian varieties with certain nontrivial endomorphism algebras and the Torelli locus (i.e., of Jacobians), then you find that (at least in many special cases) the sum of these codimensions adds up to more than $\frac{g(g-1)}{2}$, \frac{g(g+1)}{2}$, the dimension of the Siegel moduli space. Thus unless there is excess intersection between these two loci, for sufficiently large$g$one expects Jacobians not to be highly decomposed. [This is a cue for Prof. JSE to weigh in on the matter.] For instance, I think it is at least conjectured that for sufficiently large$g$, no$g$-dimensional Jacobian splits completely into a product of elliptic curves. This is known to be true over finite fields, but maybe not over$\mathbb{C}$. Serre has worked on both the general geometric question and in on this special case. In his 1997 JAMS paper, Serre showed that the maximal dimension of a simple isogeny factor of$J_0(N)$approaches infinity with$N$. I think the result is quantitative, so if you look there you may get the most information currently known on the question you asked. Ribet, Kenneth A. Endomorphisms of semi-stable abelian varieties over number fields. Ann. Math. (2) 101 (1975), 555--562. Serre, Jean-Pierre Répartition asymptotique des valeurs propres de l'opérateur de Hecke$T_p$. (French) [Asymptotic distribution of the eigenvalues of the Hecke operator$T_p$] J. Amer. Math. Soc. 10 (1997), no. 1, 75--102. *: I think it's true, and the small number of people I've spoken to about this explicitly (several years ago) thought it was true. We'll see what others have to say. 1 When$p$is prime, it follows from a 1975 theorem of Ribet (reference below) that the$\mathbb{Q}$-rational endomorphism algebra of$J_0(p)$is the same as the geometric (i.e.,$\mathbb{C}$-rational) endomorphism algebra, and that this algebra is a product of formally real fields, each being the subfield of$\mathbb{Q}$obtained by adjoining the Fourier coefficients of the various weight$2$cuspforms of level$p$. Thus the problem can be viewed as a special case of a popular one in pure algebraic geometry: for which genera$g$do there exist complex algebraic curves of genus$g$with, e.g., Jacobians isogenous to a product of elliptic curves? (Or, more generally, with endomorphism algebra containing at least$N$semisimple factors?) If you count codimensions in the Siegel moduli space corresponding to (say) principally polarized abelian varieties with certain nontrivial endomorphism algebras and the Torelli locus (i.e., of Jacobians), then you find that (at least in many special cases) the sum of these codimensions adds up to more than$\frac{g(g-1)}{2}$, the dimension of the Siegel moduli space. Thus unless there is excess intersection between these two loci, for sufficiently large$g$one expects Jacobians not to be highly decomposed. For instance, I think it is at least conjectured that for sufficiently large$g$, no$g$-dimensional Jacobian splits completely into a product of elliptic curves. This is known to be true over finite fields, but maybe not over$\mathbb{C}$. Serre has worked on both the general geometric question and in this special case. In his 1997 JAMS paper, Serre showed that the maximal dimension of a simple isogeny factor of$J_0(N)$approaches infinity with$N$. I think the result is quantitative, so if you look there you may get the most information currently known on the question you asked. Ribet, Kenneth A. Endomorphisms of semi-stable abelian varieties over number fields. Ann. Math. (2) 101 (1975), 555--562. Serre, Jean-Pierre Répartition asymptotique des valeurs propres de l'opérateur de Hecke$T_p$. (French) [Asymptotic distribution of the eigenvalues of the Hecke operator$T_p\$] J. Amer. Math. Soc. 10 (1997), no. 1, 75--102.


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# glLogicOp

## Name

glLogicOp — specify a logical pixel operation for color index rendering

## C Specification

 void glLogicOp( GLenum opcode);

## Parameters

opcode

Specifies a symbolic constant that selects a logical operation. The following symbols are accepted: GL_CLEAR, GL_SET, GL_COPY, GL_COPY_INVERTED, GL_NOOP, GL_INVERT, GL_AND, GL_NAND, GL_OR, GL_NOR, GL_XOR, GL_EQUIV, GL_AND_REVERSE, GL_AND_INVERTED, GL_OR_REVERSE, and GL_OR_INVERTED. The initial value is GL_COPY.

## Description

glLogicOp specifies a logical operation that, when enabled, is applied between the incoming color index or RGBA color and the color index or RGBA color at the corresponding location in the frame buffer. To enable or disable the logical operation, call glEnable and glDisable using the symbolic constant GL_COLOR_LOGIC_OP for RGBA mode or GL_INDEX_LOGIC_OP for color index mode. The initial value is disabled for both operations.

Opcode Resulting Operation
GL_CLEAR 0
GL_SET 1
GL_COPY s
GL_COPY_INVERTED ~s
GL_NOOP d
GL_INVERT ~d
GL_AND s & d
GL_NAND ~(s & d)
GL_OR s | d
GL_NOR ~(s | d)
GL_XOR s ^ d
GL_EQUIV ~(s ^ d)
GL_AND_REVERSE s & ~d
GL_AND_INVERTED ~s & d
GL_OR_REVERSE s | ~d
GL_OR_INVERTED ~s | d

opcode is a symbolic constant chosen from the list above. In the explanation of the logical operations, s represents the incoming color index and d represents the index in the frame buffer. Standard C-language operators are used. As these bitwise operators suggest, the logical operation is applied independently to each bit pair of the source and destination indices or colors.

## Notes

Color index logical operations are always supported. RGBA logical operations are supported only if the GL version is 1.1 or greater.

When more than one RGBA color or index buffer is enabled for drawing, logical operations are performed separately for each enabled buffer, using for the destination value the contents of that buffer (see glDrawBuffer).

## Errors

GL_INVALID_ENUM is generated if opcode is not an accepted value.

GL_INVALID_OPERATION is generated if glLogicOp is executed between the execution of glBegin and the corresponding execution of glEnd.

## Associated Gets

glGet with argument GL_LOGIC_OP_MODE.

glIsEnabled with argument GL_COLOR_LOGIC_OP or GL_INDEX_LOGIC_OP.


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Kerodon

$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$ $\newcommand\empty{}$

Remark 4.6.1.18. Let $q: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be a morphism of simplicial sets, let $X$ and $Y$ be vertices of $\operatorname{\mathcal{C}}$, and let $e: q(X) \rightarrow q(Y)$ be an edge of $\operatorname{\mathcal{D}}$. Form a pullback diagram of simplicial sets

$\xymatrix@R =50pt@C=50pt{ \operatorname{\mathcal{C}}' \ar [r] \ar [d] & \operatorname{\mathcal{C}}\ar [d]^{q} \\ \Delta ^1 \ar [r]^-{e} & \operatorname{\mathcal{D}}, }$

so that $X$ lifts uniquely to a vertex $\widetilde{X} \in \operatorname{\mathcal{C}}'$ lying over the vertex $0 \in \Delta ^1$, and $Y$ lifts uniquely to a vertex $\widetilde{Y} \in \operatorname{\mathcal{C}}'$ lying over the vertex $1 \in \Delta ^1$. Remark 4.6.1.17 and Example 4.6.1.15 supply isomorphisms

$\operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,Y)_{e} \simeq \operatorname{Hom}_{\operatorname{\mathcal{C}}'}( \widetilde{X}, \widetilde{Y} )_{ \operatorname{id}_{\Delta ^1} } = \operatorname{Hom}_{\operatorname{\mathcal{C}}'}( \widetilde{X}, \widetilde{Y} ).$


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### Can we use quantum phase estimation to learn anything about the dynamics of puzzles like the Rubik's cube?

Introduction Consider a state $\vert\psi\rangle$ such as below, which is in a superposition of a difference between a Rubik's cube in a solved state and a Rubik's cube in the "superflip" ...
156 views

### How many qubits would be needed to do boson sampling in Qiskit?

In December 2020, there was this claim of quantum advantage/supremacy by a team of USTC using Gaussian boson sampling. Here is the paper and here is an explanatory news article in Nature. To get an ...
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### What is stopping FACTORING from being BQP-complete?

Classical complexity theory makes much of the study of so-called intermediate problems - that is, problems that are in $\mathsf{NP}$ but are nonetheless not known to be in $\mathsf{P}$ and further not ...
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### Is HHL still BQP-complete when the matrix entries are only in {0,1}?

I'm studying BQP-completeness proofs of a number of interesting problems of Janzing and Wocjan, and Wocjan and Zhang. Janzing and Wocjan show that estimating entries of matrix powers $(A^m)_{ij}$ with ...
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### What are the benefits of using quantum machine learning?

I have been investigating uses for quantum machine learning, and have made a few working examples (variations of variational quantum classifiers using PennyLane). However, my issue now is its ...
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### Strong vs weak simulations and the polynomial hierarchy collapse

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# Class 9 Maths Important Questions for Heron Formula Mensuration

Given below are the Class 9 Maths Important Questions for Heron Formula Mensuration
(a) Concepts questions
(b) Calculation problems
(c) Multiple choice questions
(e) Fill in the blank's

Question 1
Calculate the area in each case
1. Triangle have sides as a=5 cm ,b=4 cm,c=3 cm
2. Equilateral triangle having side a=2 cm
3. Right angle triangle have base=4 cm and Height =3 cm
4. Square whose diagonal is 10 cm
5. Rectangle whose length and breath are 6 and 4 cm
6. Parallelogram whose two sides are 10 cm and 16 cm and diagonal is 14 cm
7. Parallelogram whose base is 10 cm and height is 14 cm
8. Rhombus of diagonals to 10 and 24 cm
9. Two sides of trapezium are 36 and 24 cm and its altitude is 12 cm.
Solution
(a) $s=\frac{a+b+c}{2}$ =6
Area
$A=\sqrt{s(s-a)(s-b)(s-c)}$=6cm2
(b) Area of equilateral
$A=\frac{\sqrt{3}}{4}a^{2}=\sqrt{3}$
(c) Area of triangle
A=(1/2)BH=6cm2
(d) Area of square in terms diagonal
A=(1/2)d2=50cm2
(e) Rectangle area is given by
A=LXB=24cm2
(f) In parallelogram whose two sides and diagonal are given, Area is given by
$A=\sqrt{s(s-a)(s-b)(s-d)}$
Where $s=\frac{a+b+d}{2}$
So s=20cm
So A=80(3)1/2cm2
(g) Area is given by
A=Base X height =10X14=140cm2
(h) Area is given by
A=(1/2)d1d2=120cm2
(i) Area of trapezium is given by
A=(1/2)(Sum of parallel sides) Altitude
A=360cm2

Question 2
True or False statement
(a) Heron formula for area of triangle is not valid of all triangles
(b) If each side of the triangles is tripled, the area will becomes 9 times
(c) Base and corresponding altitude of the parallelogram are 8 and 5 cm respectively.Area of parallelogram is 40 cm2
(d) If each side of triangle is doubled, the perimeter will become 4 times
(e) If p is the perimeter of the triangle of sides a,b,c ,the area of triangle is
$A=\frac{1}{4}\sqrt{p(p-2a)(p-2b)(p-2c)}$
(f) When two triangles are congruent, there areas are same
(g) Heron’s belongs to America
(h) If the side of the equilateral triangle is a rational number, the area would always be irrational number
Solution
1. False
2. True
3. True
4. False
5. True
6. True
7. False
8. True

## Multiple choice Questions

Question 3
The difference between sides at right angles in a right angled triangle is 14 cm. The area of the triangle is 120 cm2 . The perimeter of the triangle is
(a) 80
(b) 45
(c) 60
(d) 64
Solution (c)
Let y be one of the at right angle ,then another side will be y-14
Now we know that
A=(1/2)BH
120=(1/2)y(y-14)
y2-14y-240
(y-24)(y+10)=0
y=24
So other side is 10
From pythogrous theorem
$\text{hyp}=\sqrt{10^{2}+24^{2}}$=26cm
So perimeter will be =10+24+26=60 cm

Question 4
Find the area of the equilateral triangle whose perimeter is 180 cm?
Which of the following is true?
(a)900(3)1/2 cm2
(b)300(3)1/2 cm2
(c)200(3)1/2 cm2
(d) None of these
Solution
P=3a => a=P/3=60 cm
Area of equilateral
$A=\frac{\sqrt{3}}{4}a^{2}$

Question 5
One diagonal and side of the rhombus are 24 and 13 cm respectively, Find the Area and other diagonal size?
(a) 160 cm2,10 cm
(b) 120 cm2,20 cm
(c) 120 cm2,10 cm
(d) None of the above
Solution (c)

Question 6
ABCD is a trapezium with AB =10cm, AD=5 cm, BC=4 cm and DC =7 cm?

Find the area of the ABCD
(a) 34 cm2
(b) 28cm2
(c) 20 cm2
(d) None of these
Solution a
BC is the altitude between the two parallel sides AB and DC
So Area of trapezium will be given by
$A=\frac{1}{2}BC(AB+DC)$=34cm2

Question 7
Find the area and perimeter of the right angle triangle whose hypotenuse is 5 cm and Base is 4 cm
(a) 6 cm2 ,12 cm
(b) 12 cm2 ,14 cm
(c) 4 cm2, 6 cm
(d) 12 cm2 ,6 cm
Solution (a)
By pythogorous theorem
$height=\sqrt{hyp^{2}-base^{2}}=\sqrt{25-19}=3$
So Area =(1/2) XBase X height=6 cm2
Perimeter = 5+4+3=12 cm

Question 8
In an isosceles triangle ?ABC with AB = AC=13 cm. D is mid point on BC. Also BC=10 cm
Which of the following is true?
(a) Area of Triangle ABD and ADC are equal
(b) Area of triangle ABD is 30 cm2
(c) Area of triangle ABC is 60 cm2
(d) All the above
Solution (d)
ABD an ADC are congruent triangle, So Area of Triangle ABD and ADC are equal
Also From pythogorous theorem, AD will be given as
$AD=\sqrt{AB^{2}-BD^{2}}=\sqrt{169-25}=12cm$
So Area of triangle ABC=(1/2)X base X height=60 cm2

Question 9
A triangle and a parallelogram have the same base and the same area. The sides of the triangle are 26 cm and 30 cm and parallelogram stands on the base 28 cm. calculate the height of the parallelogram
(a) 12 cm
(b) 14 cm
(c) 10cm
(d) 13 cm
Solution (a)
For triangle, all the sides are given, calculating the area using Heron formula
A=336 cm2
Now for parallelogram, Area is given by
A= Base X Altitude
336=28 X H
Or H=12 cm

Question 10
Find the percentage increase in size of the triangle if each side is doubled
(a)
(b) 300%
(c) 400%
(d) None of the above
Solution (a)


text:
Bulletin des Sciences Mathématiques ( IF 1.241 ) Pub Date : 2021-06-07 , DOI: 10.1016/j.bulsci.2021.103012
Thomas Kalmes

For a constant coefficient partial differential operator $P\left(D\right)$ with a single characteristic direction, such as the time-dependent free Schrödinger operator as well as non-degenerate parabolic differential operators like the heat operator, we characterize when open subsets ${X}_{1}\subseteq {X}_{2}$ of ${\mathbb{R}}^{d}$ form a P-Runge pair. The presented condition does not require any kind of regularity of the boundaries of ${X}_{1}$ nor ${X}_{2}$. As part of our result we prove that for a large class of non-elliptic operators $P\left(D\right)$ there are smooth solutions u of the equation $P\left(D\right)u=0$ on ${\mathbb{R}}^{d}$ with support contained in an arbitarily narrow slab bounded by two parallel characteristic hyperplanes for $P\left(D\right)$.

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text: Suppose has an incircle with radius r and center I. (i) C is the incircle of ΔPQR (see Fig.) answered Nov 11, 2020 by Maahi01 (24.4k points) selected Nov 13, 2020 by Taanaya . The radius of a circumcircle of an equilateral triangle is equal to (a / √3), where ‘a’ is the length of the side of equilateral triangle. Add your answer and earn points. Let D be the point where the incircle touches BC; the angles IDB, IDC are right angles. Every triangle has three distinct excircles, each tangent to one of the triangle's sides. ← Prev Question Next Question → 0 votes . Doubtnut is better on App. where A t is the area of the inscribed triangle.. Derivation: If you have some questions about the angle θ shown in the figure above, see the relationship between inscribed and central angles.. From triangle BDO $\sin \theta = \dfrac{a/2}{R}$ Now, radius of incircle of a triangle = where, s = semiperimeter. Calculates the radius and area of the circumcircle of a triangle given the three sides. Attention reader! By Euler's inequality, the equilateral triangle has the smallest ratio R/r of the circumradius to the inradius of any triangle: specifically, R/r = 2. Its radius, the inradius (usually denoted by r) is given by r = K/s, where K is the area of the triangle and s is the semiperimeter (a+b+c)/2 (a, b and c being the sides). Radius of incircle and circumcircle of an equilateral triangle. the point where the medians of the equilateral triangle intersect. Let the side be a . The area of incircle of an equilateral triangle of side 42 cm is : \begin{aligned} 462 cm^2 \end{aligned} \begin{aligned} 452 cm^2 \end{aligned} \begin{aligned} 442 cm^2 \end{aligned} \begin{aligned} 432 cm^2 \end{aligned} Answer: Option A. So I'm going to try my best to draw an equilateral triangle. Given below is the figure of Incircle of an Equilateral Triangle Also the radius of Incircle of an equilateral triangle = (side of the equilateral triangle)/ 3. Calculates the radius and area of the circumcircle of a triangle given the three sides. Physics. The Inradius of an Incircle of an equilateral triangle can be calculated using the formula: Below image shows an equilateral triangle with incircle: The formula used to calculate the area of Incircle using Inradius is: The formula used to calculate the perimeter of Incircle using Inradius is. The location of the center of the incircle. Thus, #A_"triangle"=1/2bh=1/2(2sqrt3)(3)=3sqrt3#. Thus, s=2sqrt3 and the height of the triangle can be found through a+h=2+1=3. A triangle can have three medians, all of which will intersect at the centroid (the arithmetic mean position of all the points in the triangle) of the triangle. The distances from the incenter to each side are equal to the inscribed circle's radius. Thus, #s=2sqrt3# and the height of the triangle can be found through #a+h=2+1=3#. What is the perimeter of a triangle with sides 1#3/5#, 3#1/5#, and 3#3/5#? Area of Incircle of a Right Angled Triangle - GeeksforGeeks. Options $22\sqrt{3} c m^2$ 231 cm 2. 000+ LIKES. The area of the incircle of an equilateral triangle of side 42 cm is. How do you find the area of the trapezoid below? The Area of the Incircle of an Equilateral Triangle of Side 42 Cm is - Mathematics. 1 Answer +1 vote . Also, let O be the centre and r be the radius of its incircle. The center of incircle is known as incenter and radius is known as inradius. The distances from the incenter to each side are equal to the inscribed circle's radius. Open App Continue with Mobile Browser. The radii of the incircles and excircles are closely related to the area of the triangle. Given the side lengths of the triangle, it is possible to determine the radius of the circle. Now, the incircle is tangent to AB at some point C′, and so \angle AC'I is right. To use this online calculator for Radius of the inscribed circle of an equilateral triangle, enter Side (s) and hit the calculate button. (a) 3 cm (b) 4 cm (c) 5 cm (d) 6 cm. Related Questions. Let R be the radius of cir-cumcircle of this triangle. So the area of the inscribed circle is ⅔π√3. Inscribed circle of an equilateral triangle is made through the midpoint of the edges of an equilateral triangle. An equilateral triangle is the most symmetrical triangle, having 3 lines of reflection and rotational symmetry of order 3 about its center. Relation to area of the triangle. Solution Show Solution. Incenter of a triangle - Mathematical Way . Using the properties of 30˚-60˚-90˚ triangles, it can be determined that h=1 and s/2=sqrt3. Equilateral triangles are the only triangles whose Steiner inellipse is a circle (specifically, it is the incircle). Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … close, link picture. The Incenter can be constructed by drawing the intersection of angle bisectors. Let ABC be the equilateral triangle such that AB = BC = CA = 42 cm. Let a be the length of BC, b the length of AC, and c the length of AB.. Now, the incircle is tangent to AB at some point C′, and so . How to check if a given point lies inside or outside a polygon? Area of triangle = IN RADIUS (r) × semi perimeter (s) use heron's formula for area of triangle.....hope you like it!!! where is the length of the side of equilateral triangle. So if this is side length a, then this is side length a, and that is also a side of length a. 8 years ago . If sides of a triangle are 13,14, 15 then it's incircle radius is ? and area = √3/4 × a². If you know all three sides If you know the length (a,b,c) of the three sides of a triangle, the radius of its circumcircle is given by the formula: obtaining an identity for the area of Pompeiu triangle. 4 views. where A t is the area of the inscribed triangle.. Derivation: If you have some questions about the angle θ shown in the figure above, see the relationship between inscribed and central angles.. From triangle BDO $\sin \theta = \dfrac{a/2}{R}$ How to check if two given line segments intersect? Options $22\sqrt{3} c m^2$ 231 cm 2. Let a be the length of the sides, A - the area of the triangle, p the perimeter, R - the radius of the circumscribed circle, r - the radius of the inscribed circle, h - the altitude (height) from any side.. Radius of Incircle. The radius of the inscribed circle and circumscribed circle in an equilateral triangle with side length 'a' Both circles have the same center. Find the ratio of the areas of the in-circle and circum-circle of an equilateral triangle? Radius of a circle inscribed. Books. In the example above, we know all three sides, so Heron's formula is used. Consider a circle incscrbed in a triangle ΔABC with centre O and radius r, the tangent function of one half of an angle of a triangle is equal to the ratio of the radius r over the sum of two sides adjacent to the angle. The segments from the incenter to each vertex bisects each angle. Another triangle calculator, which determines radius of incircle Well, having radius you can find out everything else about circle. the ratio: A_incircle / A_circumcircle = (1/12 * pi * x^2) / (1/3 * pi * x^2) = (1/12)/(1/3) = 3/12 = 1/4. The altitude of an equilateral triangle with side lengths of s units, ... What is the radius of the incircle of a triangle with sides 30, 28, and 26 cm? I think that's about as good as I'm going to be able to do. Note that the height can also be found through using #s# and #s/2# as a base and the hypotenuse of a right triangle where the other leg is #3#. p … Calculating the radius []. The center of incircle is known as incenter and radius is known as inradius. This is the scenario you've described, in which #a=2#. This is the same situation as Thales Theorem, where the diameter subtends a right angle to any point on a circle's circumference. You draw an equilateral triangle A, B, C. ℓ : length of a side. link brightness_4 code // C program to find the area of Inscribed circle // of equilateral triangle . The Pompeiu triangle PB C has the sides PA, PB, PC. Radius of the Circumcircle of a Triangle Brian Rogers August 11, 2003 The center of the circumcircle of a triangle is located at the intersection of the perpendicular bisectors of the triangle. So if this is side length a, then this is side length a, and that is also a side of length a. Geometry Perimeter, Area, and Volume Perimeter and Area of Triangle. Triangle; Equilateral triangle; Isosceles triangle; Right triangle; Square; Rhombus; Isosceles trapezoid ; Regular polygon; Regular hexagon ; All formulas for radius of a circle inscribed; Geometry theorems. Radius of incircle and circumcircle of an equilateral triangle. Chemistry . This online calculator determines the radius and area of the incircle of a triangle given the three sides. So, … Chemistry . diameter φ . Open App Continue with Mobile Browser. person_outlineTimurschedule 2011-06-24 21:08:38. Suppose triangle ABC is isosceles, with the two equal sides being 10 cm in length and the equal... What is the basic formula for finding the area of an isosceles triangle? Sum … Given the side lengths of the triangle, it is possible to determine the radius of the circle. It follows that R > 2r unless the two centres coincide (which only happens for an equilateral triangle). The radius of an incircle of a triangle (the inradius) with sides and area is ; The area of any triangle is where is the Semiperimeter of the triangle. So, circumradius = Now, inradius of the equilateral triangle = 924 cm 2. And let's say we know that the radius of this circle is 2. Inradius: The radius of the incircle. The radius of the circumcircle is also called the triangle's circumradius. Customer Voice. NCERT DC Pandey Sunil Batra HC Verma Pradeep Errorless. Find the lengths of the other two sides of the triangle. Please use ide.geeksforgeeks.org, Therefore, The formula used to calculate the area of Incircle using Inradius is: The formula used to calculate the perimeter of Incircle using Inradius is: C. filter_none. Incircle is the circle that lies inside the triangle which means the center of circle is same as of triangle as shown in the figure below. In particular, a new proof of (21) will be given. NCERT DC Pandey Sunil Batra HC Verma Pradeep Errorless. Program to calculate area of Circumcircle of an Equilateral ... picture. Equilateral Triangle: All three sides have equal length All three angles are equal to 60 degrees. Books. What is the radius of a circle inscribed in a triangle having side lengths 35 cm, 44 cm and 75 cm? 0 0. la console. For an independent proof see [12, p.52]. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … The center of the incircle, ca Given with the side of an equilateral triangle the task is to find the area and perimeter of an incircle inside it where area is the space occupied by the shape and volume is the space that a shape can contain. MCQ. In an equilateral triangle, the altitudes are the angle bisector. The point where the angle bisectors meet. The radius of a circumcircle of an equilateral triangle is equal to (a / √3), where ‘a’ is the length of the side of equilateral triangle. Let I be the incentre. Perimeter: Semiperimeter: Area: Altitude: Median: Angle Bisector : Circumscribed Circle Radius: Inscribed Circle Radius: Right Triangle: One angle is equal to 90 degrees. Let P be a point inside the equilateral triangle ABC (see Figure 3). Thus the radius C'I is an altitude of \triangle IAB. Please give the answer in details. 4.1. The radius of this Apollonius circle is {\displaystyle {\frac {r^ {2}+s^ {2}} {4r}}} where r is the incircle radius and s is the semiperimeter of the triangle. Equilateral triangle formulas. The location of the center of the incircle. The square of the radius of the incircle is the square of the length of the shortest side of these triangles which derives from the area of one smaller triangle: ½r²√3 = 1. The area of the incircle of an equilateral triangle of side 42 cm is. The circular hull of the excircles is internally tangent to each of the excircles, and thus is an Apollonius circle. The length of a leg of an isosceles right triangle is #5sqrt2# units. What is the area of an equilateral triangle inside a circle ... picture. Hence, its semiperimeter = 3a/2 . side a: side b: side c : circumradius r . Doubtnut is better on App. 3sqrt3 This is the scenario you've described, in which a=2. The point where the angle bisectors meet. Radius of the Incircle of a Triangle Brian Rogers August 4, 2003 The center of the incircle of a triangle is located at the intersection of the angle bisectors of the triangle. All formulas for radius of a circumscribed circle. Angle IBD = … 462 cm 2. area ratio Sc/St . Note that the height can also be found through using s and s/2 as a base and the hypotenuse of a right triangle where the other leg is 3. Another triangle calculator, which determines radius of incircle Well, having radius you can find out everything else about circle. Also the radius of Incircle of an equilateral triangle = (side of the equilateral triangle)/ 3. Now, radius of incircle of a triangle = where, s = semiperimeter. The formula above can be simplified with Heron's Formula, yielding The radius of an incircle of a right triangle (the inradius) with legs and hypotenuse is. Therefore, edit Its symmetry group is the dihedral group of order 6 D 3. Area of circle = and perimeter of circle =, where r is the radius of given circle. In #Delta OBD, angleOBD=30^@, angle ODB=90^@ => … circumcircle area Sc . And when I say equilateral that means all of these sides are the same length. This is based on the following idea: let M1 be … Find the length of one side of an equilateral triangle inscribed in a circle of the measure of a radius is 10 radical 3? Paiye sabhi sawalon ka Video solution sirf photo khinch kar. picture. Experience. First, draw three radius segments, originating from each triangle vertex (A, B, C). Now for an equilateral triangle, sides are equal. For right triangles In the case of a right triangle, the hypotenuse is a diameter of the circumcircle, and its center is exactly at the midpoint of the hypotenuse. 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Schreiber (1935), see [7], [13]). Radius of circumcircle of a triangle = Where, a, b and c are sides of the triangle. p is the perimeter of the triangle… Let a be the length of BC, b the length of AC, and c the length of AB. Right Triangle Equations. Inradius and circumradius of equilateral triangle formula ... picture. FAQ. asked 1 day ago in Perimeter and Area of Plane Figures by Harithik (12.7k points) What is the radius of a circle inscribed in a triangle having side lengths 35 cm, 44 cm and 75 cm? In geometry, the incircle or inscribed circle of a triangle is the largest circle contained in the triangle; it touches the three sides. And let's say we know that the radius of this circle is 2. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready. Let the side be a . around the world. generate link and share the link here. Here is how the Radius of the inscribed circle of an equilateral triangle calculation can be explained with given input values -> 2.598076 = (sqrt(3)*9)/6. A_incircle = 1/12 * pi * x^2. Geometry Perimeter, Area, and Volume Perimeter and Area of Triangle. I think that's about as good as I'm going to be able to do. Therefore \triangle IAB has base length c and height r, and so has area \tfrac{1}{2}cr. Writing code in comment? Hence, y 1 = 2 ⇒ x 1 = 2√3 and y 1 = 1 ⇒ x 1 = √3 . 462 cm 2. For a triangle, the center of the incircle is the Incenter, where the incircle is the largest circle that can be inscribed in the polygon. The length of the base of an isosceles triangle is 4 inches less than the length of one of the... What is the value of the hypotenuse of an isosceles triangle with a perimeter equal to #16 + 16sqrt2#? The center of the incircle is a triangle center called the triangle's incenter. Please give the answer in details. Radius of circumcircle of a triangle = Where, a, b and c are sides of the triangle. Lv 7. The radius of an incircle of a triangle (the inradius) with sides and area is The area of any triangle is where is the Semiperimeter of the triangle. Find the ratio of the areas of the in-circle and circum-circle of an equilateral triangle? Now for an equilateral triangle, sides are equal. Area of circle = and perimeter of circle = , where r is the radius of given circle. The equilateral triangle is comprised of six 30-60-90 triangles, each of area 1. play_arrow. 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Physics. The circum-radius of an equilateral triangle is x cm. The radius of incircle is given by the formula r = A t s where A t = area of the triangle and s = semi-perimeter. edit close. The segments into which one side is divided by the points of contact are 36 cm and 48 cm. Another formula for the radius . 924 cm 2. Below image shows an equilateral triangle with circumcircle: The formula used to calculate the area of circumscribed circle is: (π*a 2)/3. where its incentre is I and its radius is 1 (by hypothesis). By using our site, you Inradius, Semiperimeter, and Area - Expii. What is the length of the ... See all questions in Perimeter and Area of Triangle. and area = √3/4 × a². Radius of incircle and circumcircle of an equilateral triangle. Problem. Suppose triangle ABC is isosceles, with the two equal sides being 10 cm in length and the equal... What is the basic formula for finding the area of an isosceles triangle? 1 See answer ADITHYANARAYANAN is waiting for your help. Example Problems. Suppose \triangle ABC has an incircle with radius r and center I. code. Let the radius of the incircle be r. ⇒ Area of this circle = πr 2 = 154 (22/7) × r 2 = 154 ⇒ r 2 = 154 × (7/22) = 49 ∴ r = 7 cm Recall that incentre of a circle is the point of intersection of the angular bisectors. Using the properties of #30˚-60˚-90˚# triangles, it can be determined that #h=1# and #s/2=sqrt3#. Equilateral Triangle Equations. Inradius: The radius of the incircle. The product of the incircle radius r and the circumcircle radius R of a triangle with sides a, b, and c is [7]:p. 189, #298(d) rR=\frac{abc}{2(a+b+c)}. Don’t stop learning now. For equilateral triangles In the case of an equilateral triangle, where all three sides (a,b,c) are have the same length, the radius of the circumcircle is given by the formula: where s is the length of a side of the triangle. And when I say equilateral that means all of these sides are the same length. abhisek26 abhisek26 BY FORMULA. The radius is given by the formula: where: a is the area of the triangle. Consider the triangle BIC. The center of the Incircle is same as the center of the triangle i.e. Best answer. Radius of incircle and circumcircle of an equilateral triangle. The area of the triangle is equal to \frac {1} {2}\times r\times (\text {the triangle's perimeter}), 21 The area of the triangle is equal to 1 2 × r × (the triangle’s perimeter), \frac{1}{2}\times r\times(\text{the triangle's perimeter}), 2 1 × r × (the triangle’s perimeter), where r r r is the inscribed circle's radius. The Pythagorean Theorem; The law of Sines; The law of Cosines; Theorems; Trigonometric identities. An excircle or escribed circle of the triangle is a circle lying outside the triangle, tangent to one of its sides and tangent to the extensions of the other two. Question By default show hide Solutions. Let ABC be the equilateral triangle such that AB = BC = CA = 42 cm. Also the radius of Incircle of an equilateral triangle = (side of the equilateral triangle)/ 3. circles; class-10; Share It On Facebook Twitter Email. The radius of the incircle of a triangle is 24 cm. Solution Show Solution. Given the length of sides of an equilateral triangle, the task is to find the area and perimeter of Incircle of the given equilateral triangle. Now, in our case 6r = l √ 3, (5) gives PA≥ l √ 3 (6) for any point P in the plane of equilateral triangle ABC. picture. Calculate the radius of a inscribed circle of an equilateral triangle if given side ( r ) : radius of a circle inscribed in an equilateral triangle : = Digit 2 1 2 4 6 10 F Into which one side of length a of BC, b and c the of. The triangle can be determined that h=1 and s/2=sqrt3 triangle formula... picture in-circle and of! Triangle such that AB = BC = CA = 42 cm, s semiperimeter... Through # a+h=2+1=3 # from the incenter to each vertex bisects each angle midpoint of incircle. C. ℓ: length of AB the only triangles whose Steiner inellipse is a radius of incircle of equilateral triangle = the of... Which determines radius of the edges of an isosceles triangle whose base is 16 cm and whose height 15. ) 4 cm ( D ) 6 cm ncert DC Pandey Sunil Batra HC Verma Pradeep.. Base length c and height r, and so \angle AC ' is... Of an isosceles right triangle is x cm else about circle 30˚-60˚-90˚ triangles, of... # 1/5 #, and thus is an altitude of \triangle IAB has base length c and height,! Is waiting for your help { 3 } c m^2\ ] 231 cm 2 incenter and radius is given the! The lengths of the triangle in the example above, we know that the radius of incircle and of! Triangle and AD = h be the altitude DR whose equation is BC, b length. From the incenter can be found through # a+h=2+1=3 # Pompeiu triangle geometry Perimeter, area, and that also! R > 2r unless the two centres coincide ( which only happens for an equilateral triangle length a b! See all questions in Perimeter and area of the circle ABC is an equilateral triangle ) /.... The Pythagorean Theorem ; the law of Sines ; the law of Cosines ; Theorems ; Trigonometric identities inside equilateral... That AB = BC = CA = 42 cm is khinch kar therefore \triangle.! What is the length of the other two sides of a leg of an equilateral )... The important DSA concepts with the DSA Self Paced Course at a price! Δpqr ( see Fig. isosceles right triangle is the dihedral group of 3. [ 12, p.52 ] from the incenter can be determined that h=1 s/2=sqrt3! Of ΔPQR ( see Fig. link brightness_4 code // c program to find the ratio of areas!, so Heron 's formula is used circumcircle is also called the triangle in the example above, we that... Of contact are 36 cm and 48 cm, generate link and the., b the length of BC, b, C. ℓ: length of the excircles, and Volume and! 2√3 and y 1 = √3 that AB = BC = CA = 42.! C is the incircle is known as inradius an Apollonius circle of in... 5Sqrt2 # units # a+h=2+1=3 # circle // of equilateral triangle so Heron 's formula is used of ( ). Segments into which one side is divided by the formula: where: a is the Perimeter of a.! Right Angled triangle - GeeksforGeeks Nov 13, 2020 by Maahi01 ( 24.4k points selected... Selected Nov 13, 2020 by Taanaya as Thales Theorem, where incircle... Batra HC Verma Pradeep Errorless equilateral that means all of these sides are equal terms of x the altitude whose... Right angle to any point on a circle ( specifically, it is possible to determine the radius of of. In particular, a new proof of ( 21 ) will be given Well, having you. This triangle ) ( 3 ) =3sqrt3 # D be the altitude DR whose equation is IAB has length... I say equilateral that means all of these sides are the only triangles whose Steiner inellipse is a circle specifically. Its symmetry group is the area of the circle is used # #! # 5sqrt2 # units is side length a, b, c ) h... So the area of inscribed circle of an equilateral triangle is made through the midpoint of...... Subtends a right Angled triangle - GeeksforGeeks equation is two given line segments intersect with a hypotenuse of in... Dr whose equation is I say equilateral that means all of these are...... see all questions in Perimeter and area of triangle and height r, and so AC. The incenter can be determined that h=1 and s/2=sqrt3 of Cosines ; Theorems Trigonometric... Right angle to any point on a circle of the triangle IDB, IDC are right angles and radius! # A_ '' triangle '' =1/2bh=1/2 ( 2sqrt3 ) ( 3 ) =3sqrt3 # the two... See Fig. same length find out everything else about circle in length Theorem ; the angles IDB, are... The angles IDB, IDC are right angles is internally tangent to AB at some point C′ and., PB, PC 3/5 #, 3 # 3/5 #, and is... 2Sqrt3 ) ( 3 ) =3sqrt3 # radius of incircle of equilateral triangle Figure of incircle and circumcircle of an equilateral triangle ABC ( Fig. Pb c has the sides PA, PB, PC and circumradius equilateral! Subtends a right angle to any point on a circle 's circumference determines radius of given.. 'S about as good as I 'm going to be able to do 8mm. Circumcircle of a triangle with sides 1 # 3/5 # with radius r and center.. Say we know that the radius of the triangle i.e or outside a polygon I. A point inside the equilateral triangle is comprised of six 30-60-90 triangles, it can be constructed drawing... Pradeep Errorless the excircles, each tangent to one of the equilateral triangle = side! ) 3 cm ( b ) 4 cm ( c ) is divided by formula. Circle // of equilateral triangle such that AB = BC = CA = 42 cm sides have length! That 's about as good as I 'm going to be able to do, PC if given. 'M going to try my best to draw an equilateral triangle then it 's incircle radius 10... # A_ '' triangle '' =1/2bh=1/2 ( 2sqrt3 ) ( 3 ) = now, inradius of triangle! As inradius circumradius of equilateral triangle = where, s = semiperimeter find the area of the areas the!, s = semiperimeter p.52 ] of ΔPQR ( see Figure 3 ) such that AB = BC CA... Determines radius of incircle Well, having radius you can find out radius of incircle of equilateral triangle... Therefore \triangle IAB PA, PB, PC Heron 's formula is used a: side b: c. Segments from the incenter to each side are equal to the inscribed circle 's circumference let be. Pa, PB, PC circumradius of equilateral triangle of side 42 is. The properties of 30˚-60˚-90˚ triangles, it can be constructed by drawing the of... Three sides have equal length all three sides have equal length all three sides, so 's. Radical 3 can find out everything else about circle = 2√3 and y =! Calculate area of an equilateral triangle intersect incentre is I and its radius 1... D be the altitude of equilateral triangle 3 # 1/5 #, 3 # 1/5,... The properties of 30˚-60˚-90˚ triangles, each of the trapezoid below, s = semiperimeter reflection and rotational symmetry order! To check if two given line segments intersect divided by the formula::... 13, 2020 by Taanaya of AB circular hull of the equilateral triangle,. Incentre l lies on the altitude DR whose equation is 16 cm and whose height is cm! 3 cm ( c ) 5 cm ( c ) 5 cm ( c ) an independent proof see 12! Tangent to one of the excircles, each tangent to one of the equilateral triangle of side cm... Are 13,14, 15 then it 's incircle radius is 1 ( by hypothesis ) the most symmetrical,! The sides PA, PB, PC determined that # h=1 # and s/2=sqrt3! A ) 3 cm ( b ) 4 cm ( D ) 6 cm a+h=2+1=3... By the points of contact are 36 cm and 48 cm to check if given. 24.4K points ) selected Nov 13, 2020 by Maahi01 ( 24.4k points selected. For your help h=1 # and the height of the excircles is internally tangent to AB at some C′..., so Heron 's formula is used so \angle AC ' I right... The example above, we know that the radius of incircle of isosceles! Then this is side length a, then this is side length a, then this is side a. Solution sirf photo khinch kar is possible to determine the radius of the incircle ), … find lengths... Diameter subtends a right Angled triangle - GeeksforGeeks > 2r unless the two centres coincide ( which only happens an!... see all questions in Perimeter and area of circumcircle of radius of incircle of equilateral triangle triangle center called triangle! Answered Nov 11, 2020 by Maahi01 ( 24.4k points ) selected Nov,! Can find out everything else about circle 3 cm ( c ) cm! Circle // of equilateral triangle = now, inradius of the equilateral ). Hc Verma Pradeep Errorless 13,14, 15 then it 's incircle radius is given by the points of contact 36. Of AC, and 3 # 1/5 #, and thus is an circle... 13, 2020 by Maahi01 ( 24.4k points ) selected Nov 13, by... Intersection of angle bisectors the circle now for an equilateral triangle ABC ( see Figure 3 ) by... Each vertex bisects each angle 24 cm 's formula is used a 45-45-90 triangle, radius... Geometry Perimeter, area, and Volume Perimeter and area of circle =, where incircle!


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Ticket change [52ac211279] - New ticket [aed11eb1c8168ce6|aed11eb1c8] <i>LIMIT with OFFSET in select statement is not working properly in android platform</i>. by anonymous 2018-05-15 16:59:11.
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text:
# Error arising when using \typeout & \meaning with \midrule

In order to facilitate debugging some experimental code, I created a version of \typeout that will break its content into multiple lines:

%% dummy macro for simulating newlines in output to terminal
\def\my@linefeed{}
\def\mytypeout#1\my@linefeed#2\@nil{%%
\typeout{\detokenize{#1}}%%
% \typeout{\meaning#1}%%
\expandafter\ifx\expandafter\relax\detokenize{#2}\relax\else
\mytypeout#2\@nil
\fi}

I would use it as follows:

\mytypeout ==> A testing mytypeout on its own\my@linefeed
==> B testing mytypeout on its own\my@linefeed
==> C testing mytypeout on its own\my@linefeed
==> D testing mytypeout on its own\my@linefeed
==> E testing mytypeout on its own\my@linefeed
==> F testing mytypeout on its own\my@linefeed\@nil

my \mytypeout{...} seemed to work fine until I started using it in conjunction with booktabs.

The following MWE compiles:

\documentclass{article}
\usepackage{booktabs}
\usepackage{etoolbox}
\usepackage{pgffor}

\makeatletter
%%\let\my@linefeed\relax%% --> raises errors!
%% dummy macro for simulating newlines in output to terminal
\def\my@linefeed{}
\def\mytypeout#1\my@linefeed#2\@nil{%%
\typeout{\detokenize{#1}}%%
% \typeout{\meaning#1}%%
\expandafter\ifx\expandafter\relax\detokenize{#2}\relax\else
\mytypeout#2\@nil
\fi}

%% \spaces added to make \typeout more readable
\def\my@row@of@table{%%
First column matter \space
&
\expandonce\x \space
\noexpand\\
%% Neither \meaning nor \detokenize raise an error
%% if this next line is commented out
% \noexpand\midrule
\noexpand\my@linefeed
}

\def\full@agenda{}
\def\build@agenda{%%
\foreach \x in \list@of@agenda@items {%%
\xdef\full@agenda{%%
\expandonce\full@agenda
\my@row@of@table}%%
\expandafter\mytypeout\full@agenda\my@linefeed\@nil
\typeout{---DONE----}%%
}}

\def\setagenda#1{%%
\bgroup
\def\list@of@agenda@items{#1}%%
\build@agenda
\begin{tabular}{cl}
\toprule
\full@agenda
\bottomrule
\end{tabular}%%
\egroup
\def\full@agenda{}}

\makeatother
\begin{document}

\setagenda{
{A Line of random content},
{B Not too involved math: $x\cdot y$ },
{C Some more random text.},
{D Random },
{E Hello world},
{F no more}
}

\end{document}

If I uncomment the line

  % \noexpand\midrule

then in order to get the above MWE to compile, I had to change the definition of \mytypeout. Namely, I had to change

\typeout{\meaning#1}

to

\typeout{\detokenize{#1}}

I can't quite figure out what's going on here. I tried isolating the error by testing the following code

%% No problems with \typeout + \meaning in either of the next two examples:
%% example #1
\def\aetemp{\midrule}
\typeout{TEMP(A):\meaning\aetemp}
%% example #2
\def\aetemp{\noexpand\midrule}
\typeout{TEMP(B):\meaning\aetemp}

But I didn't get any errors.

What's happening here? What am I missing?

• I think I get what's going on here. I think the problem is that \meaning works on one token. Instead I'm applying \meaning to #1 which is a sequence of tokens. – A.Ellett Oct 27 '13 at 6:25
• Moderators, should I keep this question posted? Or is this too localized? – A.Ellett Oct 27 '13 at 6:31
• I am not a moderator, but vote for keep it open – I have learned something from it. I'd suggest to add an answer. – Daniel Oct 27 '13 at 7:43
• If you want sensible answers, you should make an example that shows the error, or give precise instructions for modifying the presented code. If I leave \noexpand\midrule commented and uncomment the \typeout{\meaning#1} line I get several errors. – egreg Oct 27 '13 at 10:13
• @egreg I see what you mean. I had tried improving the answer and in the process didn't notice that I broke the code either way you went with \typeout{\meaning#1) and \noexpand\midrule. – A.Ellett Oct 27 '13 at 14:58

If I add \showtokens{#1} in the definition of \mytypeout, just before the \typeout, I get, for the first call,

> First column matter  & A Line of random content \\ .
\mytypeout ...@linefeed #2\@nil ->\showtokens {#1}
\typeout {\detokenize {#1}...
l.62     }

This clearly means that \meaning#1 is completely out of place, because it would just act on F, with an expansion such as

The character F

So \typeout{\meaning#1} is definitely not what you want. You can use \meaning, but like

  \begingroup\def\next{#1}%
\typeout{\expandafter\strip@prefix\meaning\next}%%
\endgroup

This will give as output

First column matter  & A Line of random content \\
First column matter  & B Not too involved math: $x\cdot y$  \\
First column matter  & C Some more random text. \\
First column matter  & D Random  \\
First column matter  & E Hello world \\
First column matter  & {F no more}  \\

which of course is no different from what you get with the \detokenize version.


text:
# Calculating the relevant variables for a rotating habitat

Bizmuth

1. Homework Statement

I want to calculate the rotational period and surface velocity of a Bishop Ring with a radius of 80 km and a surface 'gravity' of 0.86 G

I should note that this isn't for homework. It's for personal interest. I'm looking for a sanity check on this. The homework forum just seemed the best place for this type of question.

## Homework Equations

From https://en.wikipedia.org/wiki/Artificial_gravity the formula to calculate the rotational period for a given radius and desired gravity is

t = 2 * pi * (r/a)^.5

where t is time, r is radius of the habitat, and a is desired acceleration (gravity)

## The Attempt at a Solution

[/B]
0.86 G = 8.428 m/s^2

2 * pi * (80,000/8.428)^.5 = 612 seconds

circumference = 2 * pi * 80 km = 502.65 km

linear velocity = 502.65 / 612 = 0.82 km/s = 2957 km/hr

So the habitat will rotate once every 612 seconds, and the velocity of the rim will be 2957 km/hr

Did I brain-fart anywhere in here?

Homework Helper
Gold Member

## Homework Statement

I want to calculate the rotational period and surface velocity of a Bishop Ring with a radius of 80 km and a surface 'gravity' of 0.86 G

I should note that this isn't for homework. It's for personal interest. I'm looking for a sanity check on this. The homework forum just seemed the best place for this type of question.

## Homework Equations

From https://en.wikipedia.org/wiki/Artificial_gravity the formula to calculate the rotational period for a given radius and desired gravity is

t = 2 * pi * (r/a)^.5

where t is time, r is radius of the habitat, and a is desired acceleration (gravity)

## The Attempt at a Solution

[/B]
0.86 G = 8.428 m/s^2

2 * pi * (80,000/8.428)^.5 = 612 seconds

circumference = 2 * pi * 80 km = 502.65 km

linear velocity = 502.65 / 612 = 0.82 km/s = 2957 km/hr

So the habitat will rotate once every 612 seconds, and the velocity of the rim will be 2957 km/hr

Did I brain-fart anywhere in here?
It looks good to me.

There is the possibility of a very minor rounding error in your work. But given your choice of g = 9.8 m/s2, which has only two significant figures, any such rounding errors would be pretty meaningless anyway.

Oh, and one last thing, the period -- the amount of time for a single rotation -- is typically denoted as capital T rather than lowercase t.

Btw, nice work!

Last edited:
Bizmuth
Thank you!

collinsmark


text:
## Vickrey Auctions for Railway Tracks

Please always quote using this URN: urn:nbn:de:0297-zib-10855
• We consider an auction of slots to run trains through a railway network. In contrast to the classical setting for combinatorial auctions, there is not only competition for slots, but slots can mutually exclude each other, such that general conflict constraints on bids arise. This turns the winner determination problem associated with such an auction into a complex combinatorial optimization problem. It also raises a number of auction design questions, in particular, on incentive compatibilty. We propose a single-shot second price auction for railway slots, the Vickrey Track Auction (VTA). We show that this auction is incentive compatible, i.e., rational bidders are always motivated to bid their true valuation, and that it produces efficient allocations, even in the presence of constraints on allocations. These properties are, however, lost when rules on the submission of bids such as, e.g., lowest bids, are imposed. Our results carry over to generalized" Vickrey auctions with combinatorial constraints.

$Rev: 13581$


text:
Confused about set-theoretic definition of a function

1. Sep 15, 2011

poochie_d

I have read that a function f: A -> B can be defined as an ordered triple of sets (A,B,X), where X is the set of all ordered pairs $X = \{(a,f(a)) \in A \times B\}.$ But ordered tuples are really functions from $\{1, ..., n\}$ to (whatever set under consideration), right? So isn't this a circular definition? Or is there a more basic definition of functions that does not involve tuples? Any help would be greatly appreciated.

2. Sep 15, 2011

micromass

No, this is not true. The ordered tuple (a,b) is defined as {{a},{a,b}}. It's not defined as a function.

3. Sep 15, 2011

poochie_d

But aren't tuples other than the ordered pair defined as functions, so that the definition of functions as triples would still be circular?

4. Sep 15, 2011

micromass

No, triples can be defined as

$$(a,b,c)=((a,b),c)$$

And the definition of a function only uses ordered pairs and triples. So there is nothing circular.

5. Sep 15, 2011

poochie_d

Oh I think I get it now. Thanks micromass!


text:
## Friday, April 28, 2017

### Faster Homomorphic Function Evaluation using Non-Integral Base Encoding.

Homomorphic evaluation of arithmetic circuits is exactly what FHE and SHE schemes were created for. In practice, those circuits imply the use of real numbers which seems inconsistent with the fact that the plaintext space of the most efficient and commonly used FHE/SHE schemes is of the form
$$R_t = \mathbb{Z}_t[X]/(X^d+1).$$
Thus, basically we must encode real values into polynomials, which is done by expansion with respect to a base $b$ and then replacing all occurrences of $b$ by $X$. But then the problem arises of what SHE parameters to choose. Indeed, each homomorphic operation increases the encoding coefficients, so after a few consecutive multiplications this may lead to wrapping around modulo $t$ and invalid decoding results. To solve this problem $t$ must be chosen big enough. However, at the same time it should be small to suppress the growth of the inherent noise inside ciphertexts, in order to decrypt correctly. In practice, this problem is solved by the Chinese Remainder theorem (CRT) that splits a homomorphic algorithm into a number of threads with different plaintext spaces each modulo a factor of $t$. Nevertheless, to reduce the number of threads it is desirable to be able to take $t$ as small as possible.
To decrease the plaintext modulus one should choose an encoding scheme such that the encoding coefficients are as small as possible. The first such algorithm was presented by Dowlin et al. and studied by our colleagues Costache et al. This approach uses a balanced base $b=3$ representation where each coefficient belongs to the set $\{-1,0,1\}$. However, this choice is not optimal, because the degrees of the outcomes of the homomorphic computations are much smaller than required by the security recommendations of the underlying SHE scheme.

On the picture above the grey zone denotes the available plaintext space while the dark one is what actually was used. The lion's share of the plaintext space stays untouched. To use that space more efficiently, one might reduce the degree $d$ but it induces a significant decrease in the security level. Hence, informally, the optimisation problem consists in "squashing" the grey zone such that it almost coincides with the dark one as shown on the picture below.
Our idea is to encode real numbers with long but very sparse polynomials. In this setting the non-zero coefficients group together to a lesser extent, thereby hopefully decreasing the required plaintext modulus. One such representation is already well-known, namely, w-NAF. It is a base $b=2$ representation with coefficients from the set $\{0, \pm1, \pm3, \dots, \pm2^{w-1}-1\}$ and the property that any $w$ consecutive coefficients have at most one non-zero. Bigger $w$'s reduce the average number of non-zero digits but increase their size, which is problematic for our needs as it imposes the coefficients to grow exponentially. To address this issue we switch from the base $b=2$ to a smaller non-integral base $b>1$. Then it will be possible to construct a w-NAF-like encoding but only with the smallest possible set of digits $\{-1,0,1\}$. We called this representation "w-NIBNAF" which can be read as "non-integral base w-NAF".
For any real number $a$ the w-NIBNAF algorithm generates a ternary sequence $a_0, \dots, a_n$ in the following way:

1. Define the base $b_w$ as the unique positive root of the polynomial $F_w(x) = x^{w+1} - x^w - x - 1.$
2. Find a power $r$ such that $t = -b_w^r$ or $b_w^r$ is the closest number to $a$.
3. $a_r =$ sgn$(t)$
4. Go to step 2 with $a \leftarrow a - t$.

The algorithm stops when $t < \varepsilon$ for some real $\varepsilon$ which determines the precision of the encoding.

Our w-NIBNAF representation drastically reduces the average number of non-zero coefficients per representation. We encoded 10 000 random integers in the range from $-2^{40}$ to $2^{40}$ with the precision $\varepsilon = 1$ and found that even for 1-NIBNAF the percentage of non-zeros is around 50 percent while it is 67 percent for balanced ternary expansions. It means that the w-NIBNAF encoding leads to smaller coefficients in the worst case where the input values of an arithmetic circuit are chosen such that the circuit output contains the maximal possible coefficient.

To illustrate the practical impact of the new encoding scheme we applied 950-NIBNAF to a non-trivial arithmetic circuit containing multiplications and additions. The choice of the circuit was motivated by the load forecasting use-case of the HEAT discussed in our previous paper. In a nutshell, the prediction was performed there by a neural-network-like structure that can be expressed by a polynomial with real coefficients and degree 16. To preserve correct evaluation of the algorithm with balanced ternary expansion, it required to take the plaintext modulus around 104 bits long. This huge modulus imposes to use a CRT decomposition of the plaintext space that boosts up the execution time of the homomorphic algorithm to 32 seconds per run. Evaluated with 950-NIBNAF encodings the same circuit needs only 6 bits for a plaintext modulus, and as a consequence does not require to split the plaintext space. Hence, one run takes only 2.5 seconds that is a speed-up by a factor 13.

A more thorough explanation with sharp bounds on the maximal coefficients can be found via ePrint IACR archive.

The KU Leuven and NXP teams of the HEAT project.
SaveSaveSaveSave
SaveSave


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# Concatenation Property?

$14, \ \ \ 21, \ \ \ 28, \ \ \ 35, \ \ \ 42, \ \ \ 49, \ \ \ 56$

The above shows a sequence of multiples of 7. Is it true that if I combine these numbers together by their numerals in that order, then it is also divisible by 7? In other words, is the large number

$14 21 28 35 42 49 56$

divisible by 7?

×


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## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

(a) $\phi = \frac{5\pi}{3}$ (b) $v = 0.136~m/s$ (c) $v_{max} = 0.157~m/s$
(a) The general equation for the motion is: $x(t) = A~cos(\omega~t+\phi)$ At $t = 0$: $x = A~cos(\phi) = \frac{A}{2}$ $cos(\phi) = \frac{1}{2}$ $\phi = arccos(\frac{1}{2})$ $\phi = \frac{\pi}{3}, \frac{5\pi}{3}$ Since the basic cos-curve is shifted to the left $\frac{5\pi}{3}$, the phase constant is $\phi = \frac{5\pi}{3}$ (b) From the graph, we can see that one cycle is completed in 4 seconds. Therefore, the period $T = 4~s$. We can find the angular frequency. $T = \frac{2\pi}{\omega}$ $\omega = \frac{2\pi}{T}$ $\omega = \frac{2\pi}{4~s}$ $\omega = 1.57~rad/s$ From the graph, we can see that the amplitude is 0.10 meters. We can find the velocity at t = 0: $v(t) = -A~\omega~sin(\omega~t+\phi)$ $v = -A~\omega~sin(\phi)$ $v = -(0.10~m)(1.57~rad/s)~sin(\frac{5\pi}{3})$ $v = 0.136~m/s$ (c) We can find the maximum speed as: $v_{max} = A~\omega$ $v_{max} = (0.10~m)(1.57~rad/s)$ $v_{max} = 0.157~m/s$


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# open3d.registration.CorrespondenceChecker¶

class open3d.registration.CorrespondenceChecker

Base class that checks if two (small) point clouds can be aligned. This class is used in feature based matching algorithms (such as RANSAC and FastGlobalRegistration) to prune out outlier correspondences. The virtual function Check() must be implemented in subclasses.

Check(self, source, target, corres, transformation)

Function to check if two points can be aligned. The two input point clouds must have exact the same number of points.

Parameters
Returns

bool

__init__()

Initialize self. See help(type(self)) for accurate signature.


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⚠ Forum Maintenance: Feb 6th, 8am - 14pm (UTC+2)

# Created .so file in Eclipse for raspberry pi and it works with Eclipse C++ but not inside QT

• Hi everyone,
I wrote some c++ code to produce a shared object .so for the raspberry pi using eclipse. I tested have it in eclipse and the test program can run the .so file perfectly. I had placed the library in 3 places so that Eclipse and QT have their best chance to be able to find it somewhere. These places are as follows:
/lib /usr/lib /usr/local/lib
Now I am trying to use the same .so file inside of QT creator. So I find I also needed to include any header files I want to use. I grabbed the header I wanted to test and put it in 2 places.
/usr/include /usr/local/include
This allowed QT to recognise my class members that I defined, however when I test them inside the qt code by running the program , the computer locks up.
I have added the .so library to the .pro file by right clicking the project name and clicking ** Add Library**

I added in the needed header file to the project directory and removed it from both include dirs above
and the .so NOW WORKS... eek heh I joined up and solved my problem all at the same time ... Anyhow , Hello to all of you other souls who are also hooked on programming away on their computers !!!


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lilypond-user
[Top][All Lists]

 From: eagle136 Subject: Re: Problem with tuplets ... please help Date: Wed, 27 Feb 2008 13:07:58 -0800 (PST)

```

James E. Bailey-2 wrote:
>
>
> On 27.02.2008, at 18:44, Graham Percival wrote:
>
>> On Wed, 27 Feb 2008 18:29:32 +0100
>> "James E. Bailey" <address@hidden> wrote:
>>
>>> Another way of answering that is, lilypond uses "english.ly" by
>>> default, so unless you specify \include "deutsch.ly", you'll get
>>> errors like that (I forget to put that in my files all the time).
>>
>> False.  Look at the docs -- we use gis all the time, and never
>> gsharp.  "english.ly" is definitely *not* the default.
>>
>> Cheers,
>> - Graham
>
>
> Ah, true enough. I just assumed because it always errored on "h" that
> it was english.
>
>
> _______________________________________________
> lilypond-user mailing list
> http://lists.gnu.org/mailman/listinfo/lilypond-user
>
>

Thank you soooo much!  You were quite right I was including 'english.ly', or
rather the template I was using was using this and as it was the first time
I had required an accidental the error cropped up.  I am very grateful.
Such lists are a wonderful thing!  I hope I will be able to help someone
else out when I have got to grips with this program ....
--
View this message in context:


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# How to find the number of ordered pairs (A,B), where A and B are subsets of S and A is a proper subset of B.

Question-> Let $S$ be a set of $n$ consecutive natural numbers. How to find the number of ordered pairs $(A,B)$, where $A$ and $B$ are subsets of $S$ and $A$ is a proper subset of $B$.

Number of subsets -> $2^n$ --> here $2^8$ number of proper subsets -> $(2^n)-1$

then how the answer is $3^n-2^n$ ?

-
Wow, I have a deja vu. I feel like this exact question was posted a year or two ago. – Asaf Karagila Jun 26 '13 at 6:00

Consider the complement.

Hint: There are $2^n$ pairs of subsets where $A = B$, which is a subset of $S$.

This follows from the rule of product, because there are 2 possibilities for each element.

Hint: There are $3^n$ pairs of subsets where $A$ is a subset of $B$, which is a subset of $S$.

This follows from the rule of product, because there are 3 possibilities for each element.

-

Let $X$, $Y$, and $Z$ be pairwise disjoint subsets of $S$ whose union is $S$. Set $A=X$, $B=X\cup Y$. Then the pair $(A,B)$ has the right property, except that if $B=A$ then $A$ is not a proper subset of $B$.

Conversely, any pair $(A,B)$ such that $A$ is a subset of $B$ uniquely determines a partition of $S$ into pairwise disjoint subsets $X$, $Y$, and $Z$ by the rule $X=A$, $Y=B\setminus A$, and $Z$ equal to the rest.

We first count the number of ways to partition $S$ into three pairwise disjoint subsets. This is just the number of functions from $S$ to the set $\{1,2,3\}$. So there are $3^n$ ways to do the partioning

Now we throw away the number of "bad" partitions where $Y=\emptyset$. This is just the number of ways to divide $S$ into $2$ sets. There are $2^n$ ways to do that.

-

When you find sets $A$ and $B$ such that $A\subsetneqq B\subseteq S$, you’re really dividing $S$ into three parts: things that are in $A$ (and of course also in $B$), things that are in $B$ but not $A$, and things that are not in $B$. In other words, evern element of $S$ belongs to exactly one of the sets $A$, $B\setminus A$, and $S\setminus B$. Conversely, if you divide $S$ into parts $X,Y$, and $Z$, you can let $A=X$ and $B=X\cup Y$ to get a pair of subsets $A$ and $B$ of $S$ such that $A\subseteq B$.

There’s nothing in the statement of the problem that prevents $A$ from being empty, so there’s no harm in letting $X=\varnothing$. There’s nothing that prevents $B$ from being all of $S$, so there’s no harm in letting $Z=\varnothing$. But we’re told that $A\subsetneqq B$, so $B\setminus A\ne\varnothing$. Thus, we must ensure that the part $Y$ of our $3$-way split of $S$ gets at least one member.

In order to build such a division of $S$ into $3$ parts, you can go through $S$ one element at a time, assigning each element in turn to $X$, $Y$, or $Z$. That’s a sequence of $n$ $3$-way choices, so by the multiplication principle it can be made in $3^n$ different ways. That is, there are $3^n$ ways to divide up $S$ into the $3$ parts $X$, $Y$, and $Z$, from which, as noted above, we get $A$ and $B$ by setting $A=X$ and $B=X\cup Y$.

However, that calculation failed to take into account the requirement that $Y$ be non-empty. How many of these unwanted divisions are there? They are essentially just divisions of $S$ into two parts, $X$ and $Z$, and the same reasoning that we used in the previous paragraph shows that there must be $2^n$ of them. Subtracting them, we find that there are $3^n-2^n$ divisions of $S$ into $3$ parts $X,Y$, and $Z$ with $Y\ne\varnothing$, so that when we set $A=X$ and $B=X\cup Y$, $A$ really is a proper subset of $B$.

-


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# Chapter 22 - The Transition Elements and coordination Compounds - Questions and Problems - Page 970: 22.10

#### Work Step by Step

Werner showed that electrical coductance of a solution of $[Pt(NH_3)_4Cl_2]$ refers to that of three ions in solution and that two of the chloride ions can be precipitated as $AgCl$.

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


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# Tag Info

## Hot answers tagged coherence

6

Is the coherence length consistently defined as the length at which these two waves achieve a phase difference of 1 radian? No Coherence length is the maximal difference in way traveled by the to rays without losing the phase relation which allows interference. This lenghts can be some µmeters (eg for white light from a glowing body) or several ...

5

There is a phenomenon called decoherence in quantum mechanics which is largely responsible for this. Basically (the following is a simplification), all the strange behavior that occurs in QM tends to happen when the wavefunctions of different particles are in phase. Decoherence occurs when the phases are randomized, so there's no special correlation between ...

5

The coherence length is just the coherence time multiplied by the propagation speed. To understand the coherence time, say you have a wave described, in complex notation, by $$E(t) = A(t) e^{i \omega t}$$ where $A(t)$ is a slowly varying complex amplitude. You make this wave interfere with a delayed version of itself and collect the intensity  |E(t) ...

2

Environmentally induced decoherence makes wave function collapse unnecessary. As far as I know, that's still a conjecture. I might be wrong about that, though. But the environment, usually taken to be some heat bath, introduces a preferred frame. (That in which the total (spatial) momentum vanishes.) It's the rest frame of the system + reservoir, ...

2

This is just the comments formalised into a partial answer. First, it's important to realise that entanglement is a type of quantum correlation between two distinct systems. So, it's not useful to consider a single two-level system, and there is no such thing as entanglement between states. As Lubos Motl points out, you need to consider a system that has a ...

2

No. Coherence means "fixed, definite phase relation", or in a polychromatic context it can mean "definite phase relation at any given frequency". What would be the phase relation of the horizontally- and vertically-polarized waves? $0^\circ$? $90^\circ$? There's no reason for any definite phase relation. Here's a more specific way to think about it. ...

2

At the time of Young there were no lasers to provide coherent light. Incandescent light is incoherent because it comes from many sources and the same is true for sunlight. By passing the light through the one slit he created a single coherent wave front . It is worth reading his description "on the nature of light and colors" in the link above. Edit: I ...

2

Yes, coherent radiation means that the phases of two ( or more ) waves representing the radiation differ by a known constant. Incoherence means that the phase differences are unknown/random. Laser radiation is coherent because stimulated emission assures phase differences are constant . Radiation from an incandescent lamp is incoherent because the ...

1

I believe, that the derivation is wrong... If you assume a translationally invariant state, such that $G^1(r, r') = G^1(r - r')$ then you can get the result. Rewrite the exponential as $p r - p' r' = p( r- r') + r'(p - p')$. Since, in this case, the left-hand-side of Eq. (2.27) can only depend on $r - r'$ it must be such that $p = p'$ from the second term. ...

1

Brandon, the simple truth is that you have just asked one of the hardest and least understood questions in all of physics. So, don't feel bad if you don't understand it very well, because, er... no one else really does either? It's not that we can't model this stuff mathematically. Shoot, Richard Feynman's version of something called Quantum Electrodynamics ...

1

They are two different but closely related concepts Given two sine waves of equal frequency once can ask what is their relative phase, and are they in-phase or out of phase. So $\sin(\omega t)$ and $sin(\omega t+\pi)$ have a relative phase $\pi$ or 180 degrees and so we would say they are out of phase. The question of coherence is: how stable is the ...

1

For two waves to be 'perfectly' coherent (I'm assuming electromagnetic, transverse waves) they have to have the same wavelength, polarization and phase. To go a bit deeper, the coherence can never be perfect. Only an infinitely long wave has a wavelength that is 'fully defined', ie. has no uncertainty (or width) associated with it. One therefore defines a ...

1

The beam of white light is refracted by the prism, but the different wavelengths in the light are refracted by different angles. That means when the light falls on the screen the different wavelengths are spread out over a region of the screen. This is what is meant by dispersion. The dispersion is linear if the different wavelengths of light are spread ...

1

If I understand your question correctly, you wish to use quantum interference to observe phenomena that otherwise have a low probability of occurring? If so, exploiting this is well known and well established in quantum optics. For example, take phenomena that occur with coherently prepared atomic systems, such as Lasing without inversion, ...

Only top voted, non community-wiki answers of a minimum length are eligible


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# Hong Kong Nov 2009 - Day 2

27 Nov 2009

On vacation but still have to wake up at 6.10am. Going to catch the 8.30am ferry to Macau. from Gravity

And that was the only tweet for the day, because as I headed to Macau, I do not have pre-paid SIM GPRS service.

This was our breakfast, taken at a nearby restaurant just a minute's walk from our hotel.

We took a cab to China Hong Kong Harbour Ferry where we boarded the First Ferry Macau ship. We bought the tickets in Singapore via the Internet and so we just went to the e-ticket counter to get our tickets printed.

We chose First Ferry instead of TurboJet because
1. First Ferry offers a free upgrade to Deluxe Class for 8.30am timing.
2. First Ferry is slightly cheaper.
3. First Ferry is nearer to our hotel.

Deluxe Class serves free food and refreshments, but Angie felt seasick and I wanted to save my tummy for Macanese food.

Halfway the voyage, my ST550 battery indicated "low". My fault for not charging overnight, but the battery indicator was full. I should not have trusted the indicator. Afterall, the ST550 is good for only 180 shots.

Once we arrive at Macau, we were advised by the tourist office to hop onto Lisboa Hotel's free shuttle bus that will bring us into the tourist belt.

Above: Grand Lisboa

Armed with the tourist map, we just walked down the main road and traced across several tourist attractions.

Below: this shop looks so rustic that the shop owner placed a "no photo" sign. Obviously his shop is a frequent target for avid photographers.

Finally approaching the main draw of Macau: The Ruins of St. Paul's.

Below, shot with Nokia E72. Looks fabulous!

From here, we walked back to Lisboa Hotel and checked out the casino. Angie played jackpot and initially lost almost half the capital. Then when she won back, I told her to let go, and so she won HK$9.50 (less than S$2). Wahahah. Then she went to another table to play some "wheel of fortune" thingie and lost HK$100. By then it's already 1pm but we weren't hungry. I remembered there was this "Fisherman's Wharf" mentioned in the tourist map that there are some entertainment and shopping and I recalled \it was near the Ferry, so we tried to head back to the Ferry by hopping onto the Hotel Lisboa's shuttle bus. But neither did we realise that you can only take the bus if you had a ticket, and the ticket was given free inside the casino. We went back to the casion, found the staff and requested for tickets. She thought we were free-loaders, but we indeed played in the casino. Anyway, we got it and so headed back to Ferry. Once we reached there, we found that the Fisherman's Wharf was actually quite a distance from the Ferry and didn't seem walkable, so we took a public bus and got off the next stop. And once we got off, reality set in: it's a freaking theme park! I felt kindda cheated and embarrassed to bring my wife all the way there for nothing. So we decided to just grab a cab to The Venetian Macao, where it's guaranteed to be happening. With no cab in sight, we walked to Sands Casino and before we hopped on a cab, decided to visit the casino first. And wow it was much better than Hotel Lisboa. They had live shows featuring a guy performer with many other cabaret girls, with cheezy songs and dance movements. And so we left the casino and walking towards the taxi stand when we saw The Venetian shuttle bus at the Sands Casino lobby area. Thinking that we might be able to hitch a ride, we walked further, and lo and behold! There really is a shuttle bus from Sands to The Venetian! We saved our cab fare! So travel tip: if the Venetian shuttle bus at the Ferry is too crowded (and usually is), just hop on the Sands shuttle bus and transfer to The Venetian one. Needless to say, The Venetian was breakthtaking. And by then, my ST550 totally died, so I used my E72 to shoot. I had a great time shopping here. Bought some shirts, pants, a watch, some souvenirs. You can use HK$ in Macau cos it's about 3% higher value than MOP$. Technically, you lose when you use HK$ to pay for MOP$priced items. Interestingly, some merchandise are quoted in HK$, like the G2000 I went. But when I paid by credit card, they will use the same amount and charge in MOP\$. So you gain that 3%, but not withstanding the banks exchange rate.

We took the 8.30pm shuttle bus back to the Ferry terminal. Our e-ticket was pre-bought for 9.30pm trip, but it was just 8.50pm and we were already at the Ferry gate. We asked the gate keeper if we could just hop in the 9pm trip and he said no problem. Interesting, we thought. What if the Ferry was full? Perhaps they knew it won't be.

Back to HK! Waaaa! What an adventure in Macau, feels constipated not able to go online to share my experience. Now i can! So we saved half an hour's wait and reached HK around 10pm. We grabbed a cab back to hotel, but instead of heading back to our room, we walked to the nearest Guilinggao 龜苓膏 (http://en.wikipedia.org/wiki/Guilinggao) stall, the famous "Hoi Tin Tong" for our nightly bowl. And also got some BBQ sticks for supper.

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