GMAT Math : Radius

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Example Questions

Example Question #1 : Radius

A circle is inscribed in a square with area 100. What is the area of the circle?

Possible Answers:

$25\pi$

$100\pi$

$5\pi$

Not enough information.

Correct answer:

$25\pi$

Explanation:

A square with area 100 would have a side length of 10, which is the diameter of the circle. The area of a circle is $\pi r^{2}$ , so the answer is $\pi * 5^{2} = 25\pi$ .

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Example Question #31 : Geometry

Thingy_2

The above figure shows a square inscribed inside a circle. What is the ratio of the area of the circle to that of the square?

Possible Answers:

$\pi : 2$

$\pi : \sqrt{2}$

$2\pi : 3$

$\sqrt{2}:1$

$4 : \pi$

Correct answer:

$\pi : 2$

Explanation:

Let be the radius of the circle. Its area is $A _{\bigcirc } = \pi r^{2}$

The diagonal of the square is equal to the diameter of the circle, or . The area of the square is half the product of its (congruent) diagonals: $A_{\square } = \frac{1}{2} \cdot2r \cdot 2r = 2r^{2}$

This makes the ratio of the area of the circle to that of the square $\frac{A _{\bigcirc }}{A_{\square }} = \frac{\pi r^{2} }{2r^{2} } = \frac{\pi }{2 }$ .

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Example Question #31 : Circles

Tom has a rope that is 60 feet long. Which of the following is closest to the largest area that Tom could enclose with this rope?

Possible Answers:

260

220

280

300

240

Correct answer:

280

Explanation:

The largest square you could make would be ${15}'\times {15}'$ with an area of 225 . However, the largest region that can be enclosed will be accomplished with a circle (so you don't lose distance creating the angles). This circle will have a circumference of 60 ft. This gives a radius of

$\frac{60}{2\pi} = \frac{30}{\pi}$

Then the area will be

$\pi \cdot r^2 = \pi \cdot (\frac{30}{\pi})^2 = \frac{900}{\pi} \approx 286$

This is closer to 280 than to 300

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Example Question #31 : Circles

What is the area of a circle which goes through the points (2,4), (14,4), (2,9) ?

Possible Answers:

$169\pi$

$144 \pi$

$36\pi$

$\frac{169\pi }{4}$

$72\pi$

Correct answer:

$\frac{169\pi }{4}$

Explanation:

As can be seen in this diagram, the three points form a right triangle with legs of length 5 and 12.

Right_triangle

A circle through these three points circumscribes this right triangle.

An inscribed right, or $90^{\circ }$ , angle intercepts a $2 \times 90 = 180 ^{\circ }$ arc, or a semicircle, making the hypotenuse a diameter of the circle. The diameter of the circle is therefore the hypotenuse of the right triangle, which we can find via the Pythagorean Theorem:

$d = \sqrt{ 5^{2} + 12 ^{2}} = \sqrt{ 25+ 144} = \sqrt{169} = 13$

The radius of the circle is half this, or $\frac{13}{2}$ .

The area of the circle is therefore:

$A = \pi r^{2} = \pi \cdot \left ( \frac{13}{2} \right) ^{2} =\frac{169\pi }{4}$

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Example Question #5 : Calculating The Area Of A Circle

A circle on the coordinate plane has equation

$x^{2} + y ^{2} = 50$

Which of the following represents its area?

Possible Answers:

$50 \pi$

$10 \pi \sqrt{2}$

$100 \pi$

$25 \pi$

$5 \pi \sqrt{2}$

Correct answer:

$50 \pi$

Explanation:

The equation of a circle centered at the origin is

$x^{2} + y ^{2} = r ^{2}$

where is the radius of the circle.

The area of a circle is $A = \pi r ^{2}$ ; since in the equation of the circle, $r ^{2} = 50$ , we can substitute to get the area $A = \pi \cdot 50 = 50 \pi$ .

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Example Question #2 : Calculating The Area Of A Circle

Refer to the above figure, in which the dot indicates the center of the larger circle. The larger circle is one yard in diameter. Give the area of the gray region in square inches.

Possible Answers:

$405 \pi \textrm{ in}^{2}$

$243 \pi \textrm{ in}^{2}$

$972 \pi \textrm{ in}^{2}$

$108 \pi \textrm{ in}^{2}$

$1,620 \pi \textrm{ in}^{2}$

Correct answer:

$243 \pi \textrm{ in}^{2}$

Explanation:

The larger circle has diameter one yard, or 36 inches; its radius is half that, or 18 inches, so its area is

$A_{1} = \pi r ^{2} = \pi \cdot 18 ^{2} = 324 \pi$ square inches.

The diameter of the smaller circle is equal to the radius of the larger, or 18 inches; its radius is half that, or 9 inches, so its area is:

$A_{2} = \pi r ^{2} = \pi \cdot 9 ^{2} = 81 \pi$ square inches.

The area of the gray region is the difference of the two:

$A= 324 \pi - 81 \pi = 243 \pi$ square inches.

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Example Question #1 : Calculating The Area Of A Circle

For $10, Brandon can order either a 12"-diameter pizza, two 6"-diameter pizzas, or three 4"-diameter pizzas. Which option is the best value, assuming all pizzas are the same thickness?

Possible Answers:

Cannot be determined

All three pizzas have the same value.

one 12" pizza

three 4" pizzas

two 6" pizzas

Correct answer:

one 12" pizza

Explanation:

For a circle, $Area = \pi r ^{2}$ .

Therefore, the area of the 12" pizza $=6^{2}\pi =36\pi$ .

The area of the two 6" pizzas $=2\times 3^{2}\pi=2\times 9\pi =18\pi$ .

The area of the three 4" pizzas = $3\times 2^{2}\pi=3\times 4\pi =12\pi$ .

The 12" pizza is the best option.

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Example Question #3 : Calculating The Area Of A Circle

Refer to the above circle. The smaller circle has diameter feet. In terms of , give the area of the gray region in square inches.

Possible Answers:

$108 \pi k \textrm{ in}^{2}$

$432 \pi k ^{2} \textrm{ in}^{2}$

$216 \pi k \textrm{ in}^{2}$

$108\pi k ^{2} \textrm{ in}^{2}$

$432 \pi k \textrm{ in}^{2}$

Correct answer:

$108\pi k ^{2} \textrm{ in}^{2}$

Explanation:

Multiply the diameter of the smaller circle by 12 to convert it to inches - this will be 12k , which is also the radius of the larger circle. The radius of the smaller circle will be half this, or .

Using the area formula for a circle $A = \pi r^{2}$ , we can substitute these quantities for and subtract the area of the smaller circle from that of the larger:

$A_{1} = \pi r^{2} = \pi \left (12k \right ) ^{2} = 144 \pi k^{2}$ square inches

$A_{2} = \pi r^{2} = \pi \left (6k \right ) ^{2} = 36 \pi k^{2}$ square inches

$A = A_{1} - A_{2} = 144 \pi k ^{2}- 36 \pi k ^{2} = 108 \pi k ^{2}$ square inches

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Example Question #4 : Calculating The Area Of A Circle

A circle on the coordinate plane has equation

$x^{2} + 4x + y ^{2} + 6y = 12$

What is its area?

Possible Answers:

$36 \pi$

$25 \pi$

$12 \pi$

$5 \pi$

$10 \pi$

Correct answer:

$25 \pi$

Explanation:

The area of a circle is equal to $A = \pi r ^{2}$ , where is the radius.

The standard form of the area of a circle with radius and center (h,k ) is

$(x-h)^{2} + (y-k)^{2} = r^{2}$

Once we get the equation in standard form, we know $r ^{2}$ , which can be multiplied by $\pi$ .

Complete the squares:

$\left (\frac{4}{2} \right )^{2} = 4 ;\left (\frac{6}{2} \right )^{2} = 9$

so $x^{2} + 4x + y ^{2} + 6y = 12$ can be rewritten as follows:

$x^{2} + 4x + 4 + y ^{2} + 6y + 9 = 12 + 4 + 9$

$\left (x + 2 \right ) ^{2} +\left ( y + 3 \right ) ^{2} = 25$

Therefore, $r^{2} = 25$ and $A = \pi r ^{2} = 25 \pi$ .

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Example Question #5 : Calculating The Area Of A Circle

A square has the same area as a circle with a radius of 12 inches. What is the sidelength of that square, in terms of $\pi$ ?

Possible Answers:

$12\sqrt{\pi} \textrm{ in }$

$144\pi\textrm{ in }$

$12\pi\textrm{ in }$

$12\pi^{2}\textrm{ in }$

$\sqrt{12\pi} \textrm{ in }$

Correct answer:

$12\sqrt{\pi} \textrm{ in }$

Explanation:

The area of a circle with radius 12 is

$A = \pi r^{2} = \pi \cdot 12^{2} = 144\pi$ .

This is also the area of the square, so the sidelength of that square is the square root of the area:

$s = \sqrt{A} = \sqrt{144\pi} = \sqrt{144} \cdot \sqrt{\pi}= 12\sqrt{\pi}$

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GMAT Math : Radius

Study concepts, example questions & explanations for GMAT Math

All GMAT Math Resources

Example Questions

Example Question #1 : Radius

Example Question #31 : Geometry

Example Question #31 : Circles

Example Question #31 : Circles

Example Question #5 : Calculating The Area Of A Circle

Example Question #2 : Calculating The Area Of A Circle

Example Question #1 : Calculating The Area Of A Circle

Example Question #3 : Calculating The Area Of A Circle

Example Question #4 : Calculating The Area Of A Circle

Example Question #5 : Calculating The Area Of A Circle

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