How to find the area of a circle - ISEE Upper Level (grades 9-12) Quantitative Reasoning

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Question

Circle B has a radius $\frac{2}{3}$ as long as that of Circle A.

Which is the greater quantity?

(a) The area of Circle A

(b) Twice the area of Circle B

Answer

If we call the radius of Circle A , then the radius of Circle B is $\frac{2}{3} r$ .

The areas of the circles are:

(a) $\pi r^{2}$

(b) $\pi \left( \frac{2}{3} r \right ) ^{2} = \frac{4}{9} \pi r^{2}$

Twice the area of Circle B is

$2 \cdot \pi \left( \frac{2}{3} r \right ) ^{2} =2 \cdot \frac{4}{9} \pi r^{2} = \frac{8}{9} \pi r^{2} < \pi r^{2}$ ,

making (a) the greater number.

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Question

Circle 1 is inscribed inside a square. The square is inscribed inside Circle 2.

Which is the greater quantity?

(a) Twice the area of Circle 1

(b) The area of Circle 2

Answer

If the radius of Circle 1 is , then the square will have sidelength equal to the diameter of the circle, or . Circle 2 will have as its diameter the length of a diagonal of the square, which by the $45 ^{\circ }-45 ^{\circ }-90^{\circ }$ Theorem is $\sqrt{2}$ times that, or $2r \sqrt{2}$ . The radius of Circle 2 will therefore be half that, or $r \sqrt{2}$ .

The area of Circle 1 will be $A = \pi r^{2}$ . The area of Circle 2 will be $A = \pi \left ( r\sqrt{2} \right ) ^{2} = \pi \left ( r^{2} \cdot 2 \right ) = 2 \pi r^{2}$ , twice that of Circle 1.

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Question

Compare the two quantities:

Quantity A: The area of a circle with radius

Quantity B: The circumference of a circle with radius

Answer

Recall for this question that the formulae for the area and circumference of a circle are, respectively:

$A=\pi * r^2$

$C=2\pir$

For our two quantities, we have:

Quantity A: $\pi * 13^2 = 169\pi$

Quantity B: $2 * \pi * 63 = 126\pi$

Therefore, quantity A is greater.

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Question

The radius of a circle is . Give the area of the circle in terms of .

Answer

The area of a circle with radius can be found using the formula

$A = \pi r^{2}$

Since , the area is

$A = \pi (x- 5)^{2}$

$A = \pi (x^{2} - 2 \cdot x \cdot 5 + 5^{2})$

$A = \pi (x^{2} - 10 x + 25)$

$A = \pi x^{2} - 10 \pi x +25 \pi$

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Question

The radius of a circle is . Give the circumference of the circle in terms of .

Answer

The circumference of a circle is $2 \pi$ times its radius. Therefore, since the radius is , the circumference is

$2 \pi \left (x - 5 \right ) = 2 \pi x - 2 \pi \cdot 5 = 2 \pi x - 10 \pi$

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Question

The radii of six circles form an arithmetic sequence. The radius of the second-smallest circle is twice that of the smallest circle. Which of the following, if either, is the greater quantity?

(a) The area of the largest circle

(b) Twice the area of the third-smallest circle

Answer

Call the radius of the smallest circle . The radius of the second-smallest circle is then , and the common difference of the radii is .

The radii of the six circles are, from least to greatest:

The largest circle has area

$A_{6} = \pi (6r)^{2} = \pi \cdot 36 r^{2} = 36 \pi r^{2}$

The third-smallest circle has area:

$A_{3} = \pi (3r)^{2} = \pi \cdot 9 r^{2} = 9 \pi r^{2}$

Twice this is

$2 A_{3} = 2 \cdot 9 \pi r^{2} = 18 \pi r^{2}$

The area of the sixth circle is greater than twice that of the third-smallest circle, so the correct choice is that (a) is greater.

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Question

The areas of six circles form an arithmetic sequence. The second-smallest circle has a radius twice that of the smallest circle.

Which is the greater quantity?

(a) The area of the largest circle.

(b) Twice the area of the third-largest circle.

Answer

Let be the radius of the smallest circle. Then the second-smallest circle has radius . Their areas, respectively, are

$A_{1} = \pi r ^{2}$

and

$A_{2} = \pi \left (2r \right )^{2} = 4 \pi r^{2}$

The areas form an arithmetic sequence, so their common difference is

$4 \pi r^{2} - \pi r^{2} = 3 \pi r^{2}$ .

The six areas are

$\pi r ^{2}, 4 \pi r ^{2}, 7 \pi r ^{2}, 10 \pi r ^{2}, 13 \pi r ^{2}, 17 \pi r ^{2}$

The third-largest circle has area $10 \pi r ^{2}$ ; twice this is $20 \pi r ^{2}$ . This is greater than the area of the largest circle, which is $17\pi r ^{2}$ . (b) is the greater quantity.

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Question

In the above figure, .

Which is the greater quantity?

(a) Twice the area of inner gray ring

(b) The area of the white ring

Answer

For the sake of simplicity, we will assume that ; this reasoning is independent of the actual length.

The four concentric circles have radii 1, 2, 3, and 4, respectively, and their areas can be found by substituting each radius for in the formula $A = \pi r^{2}$ :

$A _{1}= \pi r^{2} = \pi \cdot 1 ^{2} = \pi$

$A _{2}= \pi r^{2} = \pi \cdot 2^{2} =4 \pi$

$A _{3}= \pi r^{2} = \pi \cdot 3^{2} =9 \pi$

$A _{4}= \pi r^{2} = \pi \cdot 4^{2} =16 \pi$

The white ring has as its area the difference of the areas of the second-largest and third-largest circles:

$A_{3} - A_{2} = 9 \pi - 4 \pi = 5 \pi$

The inner gray ring has as its area the difference of the areas of the third-largest and smallest circles:

$A_{2} - A_{1} = 4 \pi - \pi = 3 \pi$ .

Twice this is $2 \cdot 3 \pi = 6 \pi$ , which is greater than the area of the white ring.

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Question

In the above figure, .

Which is the greater quantity?

(a) Six times the area of the white circle

(b) The area of the outer ring

Answer

For the sake of simplicity, we will assume that ; this reasoning is independent of the actual length.

The four concentric circles have radii 1, 2, 3, and 4, respectively, and their areas can be found by substituting each radius for in the formula $A = \pi r^{2}$ :

$A _{1}= \pi r^{2} = \pi \cdot 1 ^{2} = \pi$

$A _{2}= \pi r^{2} = \pi \cdot 2^{2} =4 \pi$

$A _{3}= \pi r^{2} = \pi \cdot 3^{2} =9 \pi$

$A _{4}= \pi r^{2} = \pi \cdot 4^{2} =16 \pi$

The outer gray ring is the region between the largest and second-largest circles, and has area

$A _{4} - A _{3} = 16 \pi - 9 \pi = 7 \pi$

Six times the area of the white (inner) circle is $6 A_{1} = 6 \pi$ , which is less than the area of the outer ring, $7 \pi$ .

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