How to find the area of a circle - ISEE Upper Level Quantitative Reasoning

Card 0 of 128

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.

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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

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

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

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

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

Compare your answer with the correct one above

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

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

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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.

Compare your answer with the correct one above

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, $7 \pi$ .

Compare your answer with the correct one above

Question

What is the area of a circle with a diameter of , rounded to the nearest whole number?

Answer

The formula for the area of a circle is

$\dpi{100} \pi r^{2}$

Find the radius by dividing 9 by 2:

$\dpi{100} \frac{9}{2}=4.5$

So the formula for area would now be:

$\dpi{100} \pi r^{2}=\pi (4.5)^{2}=20.25\pi \approx 63.6= 64$

Compare your answer with the correct one above

Question

What is the area of a circle that has a diameter of inches?

Answer

The formula for finding the area of a circle is $\pi r^{2}$ . In this formula, represents the radius of the circle. Since the question only gives us the measurement of the diameter of the circle, we must calculate the radius. In order to do this, we divide the diameter by .

$\frac{15}{2}=7.5$

Now we use for in our equation.

$\pi (7.5)^{2}=176.7146 : in^{2}$

Compare your answer with the correct one above

Question

What is the area of a circle with a diameter equal to 6?

Answer

First, solve for radius:

$r=\frac{d}{2}=\frac{6}{2}=3$

Then, solve for area:

$A=r^2\pi=3^2\pi=9\pi$

Compare your answer with the correct one above

Question

The diameter of a circle is $4\ cm$ . Give the area of the circle.

Answer

The area of a circle can be calculated using the formula:

$Area=\frac{\pi d^2}{4}$ ,

where is the diameter of the circle, and $\pi$ is approximately .

$Area=\frac{\pi d^2}{4}=\frac{\pi\times 4^2}{4}=4\pi \Rightarrow Area\approx 4\times 3.14\Rightarrow Area\approx 12.56 \ cm^2$

Compare your answer with the correct one above

Question

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

Answer

The area of a circle can be calculated using the formula:

$Area=\frac{\pi d^2}{4}$ ,

where is the diameter of the circle and $\pi$ is approximately .

$Area=\frac{\pi (4t)^2}{4}=\frac{16\pi t^2}{4}=4\pi t^2 \Rightarrow Area\approx 4\times 3.14\times t^2$

$\Rightarrow Area\approx 12.56t^2$

Compare your answer with the correct one above

Question

The radius of a circle is $\frac{2}{\sqrt{\pi }}$ . Give the area of the circle.

Answer

The area of a circle can be calculated as $Area=\pi r^2$ , where is the radius of the circle, and $\pi$ is approximately .

$Area=\pi r^2=\pi\times (\frac{2}{\sqrt{\pi}})^2=\pi\times \frac{4}{\pi}\Rightarrow Area=4$

Compare your answer with the correct one above

Question

The circumference of a circle is inches. Find the area of the circle.

Let $\pi = 3.14$ .

Answer

First we need to find the radius of the circle. The circumference of a circle is $Circumference =2\pi r$ , where is the radius of the circle.

$12.56=2\times 3.14\times r\Rightarrow r=2\ in$

The area of a circle is $Area=\pi r^2$ where is the radius of the circle.

$Area=\pi r^2=3.14\times 2^2=12.56\ in^2$

Compare your answer with the correct one above

Question

The perpendicular distance from the chord to the center of a circle is $\sqrt{2}t$ , and the chord length is $2\sqrt{2}t$ . Give the area of the circle in terms of .

Answer

Chord length = $2\sqrt{r^2-d^2}$ , where is the radius of the circle and is the perpendicular distance from the chord to the circle center.

Chord length = $2\sqrt{r^2-d^2}\Rightarrow (2\sqrt{2})t=2\sqrt{r^2-(\sqrt{2}t)^2}$

$\Rightarrow 8t^2=4(r^2-2t^2)\Rightarrow 4r^2=16t^2\Rightarrow r^2=4t^2\Rightarrow r=2t$

$Area=\pi r^2$ , where is the radius of the circle and $\pi$ is approximately .

$Area=\pi r^2=3.14\times (2t)^2=12.56t^2$

Compare your answer with the correct one above

Question

In the above figure, .

What percent of the figure is shaded gray?

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

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

The inner gray ring is the region between the second-smallest and smallest circles, and has area

$A _{2} - A _{1} = 4 \pi - \pi = 3 \pi$

The total area of the gray regions is

$7 \pi + 3 \pi = 10 \pi$

Since $10 \pi$ out of total area $16 \pi$ is gray, the percent of the figure that is gray is

$\frac{10 \pi}{16 \pi } \times 100 % = \frac{5}{8} \times 100 % =62 \frac{1}{2} %$ .

Compare your answer with the correct one above

Question

In the above figure, .

Give the ratio of the area of the outer ring to that of the inner circle.

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 $A = \pi r^{2}$ .

The areas of the largest circle and the second-largest circle are, respectively,

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

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

The difference of their areas, which is the area of the outer ring, is

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

The inner circle has area

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

The ratio of these areas is therefore

$\frac{7 \pi}{\pi} = \frac{7}{1}$ , or 7 to 1.

Compare your answer with the correct one above

Question

The above figure depicts a dartboard, in which .

A blindfolded man throws a dart at the target. Disregarding any skill factor and assuming he hits the target, what are the odds against his hitting the white inner circle?

Answer

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

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

$A _{1}= \pi r^{2} = \pi \cdot 1 ^{2} = \pi$ - this is the white inner circle.

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

The area of the portion of the target outside the white inner circle is $16 \pi - \pi = 15 \pi$ , so the odds against hitting the inner circle are

$\frac{15 \pi}{\pi} = \frac{15}{1}$ - that is, 15 to 1 odds against.

Compare your answer with the correct one above

Tap the card to reveal the answer