GMAT Quantitative - Calculating the area of a circle

Two circles are constructed; one is inscribed inside a given equilateral triangle, and the other is circumscribed about the same triangle.

The circumscribed circle has circumference $20 \pi$ . Give the area of the inscribed circle.

$25 \pi$

$400 \pi$

$50 \pi$

$100 \pi$

$200 \pi$

Explanation

Examine the diagram below, which shows the triangle, its three altitudes, and the two circles.

Thingy

The three altitudes of an equilateral triangle divide one another into two segments each, the longer of which is twice the length of the shorter. The length of each of the longer segments is the radius of the circumscribed circle, and the length of each of the shorter segments is the radius of the inscribed circle. Therefore, the inscribed circle has half the radius of the circumscribed circle.

The circumscribed circle has circumference $20 \pi$ , so its radius is

$20 \pi \div 2 \pi = 10$

The inscribed circle has radius half this, or 5, so its area is

$A= \pi r^{2} = \pi \cdot 5^{2}= 25 \pi$

Two circles are constructed; one is inscribed inside a given regular hexagon, and the other is circumscribed about the same hexagon.

The circumscribed circle has circumference $20 \pi$ . Give the area of the inscribed circle.

$75 \pi$

$50 \pi$

$200 \pi$

$\frac{400 \pi }{3}$

$\frac{200 \pi }{3}$

Explanation

Examine the diagram below, which shows the hexagon, segments from its center to a vertex and the midpoint of a side, and the two circles.

Thingy

Note that the segment from the center of the hexagon to the midpoint of a side is a radius of the inscribed circle, and the segment from the center to a vertex is a radius of the circumscribed circle. The two segments and half a side of the hexagon can be proved to form a 30-60-90 triangle.

The circumscribed circle has circumference $20 \pi$ , so its radius - and the length of the hypotenuse of the right triangle -

$20 \pi \div 2 \pi = 10$

By the 30-60-90 Theorem, the length of the shorter leg is half this, or 5. The length of the longer leg, which is the radius of the inscribed circle, is $\sqrt{3}$ times this, or $5 \sqrt{3}$ .

The area of the inscribed circle can now be calculated:

$A= \pi r^{2} = \pi \cdot \left ( 5 \sqrt{3 }\right ) ^{2}= \pi \cdot 25 \cdot 3 = 75 \pi$

Two circles are constructed; one is inscribed inside a given regular hexagon, and the other is circumscribed about the same hexagon.

The inscribed circle has circumference $20 \pi$ . Give the area of the circumscribed circle.

$\frac{400 \pi }{3}$

$200 \pi$

$50 \pi$

$75 \pi$

$\frac{200 \pi }{3}$

Explanation

Examine the diagram below, which shows the hexagon, segments from its center to a vertex and the midpoint of a side, and the two circles.

Thingy

The inscribed circle has circumference $20 \pi$ , so its radius - and the length of the longer leg of the right triangle - is

$20 \pi \div 2 \pi = 10$

By the 30-60-90 Theorem, the length of the shorter leg is this length divided by $\sqrt{3}$ , or $\frac{10}{\sqrt{3}}$ ; the length of the hypotenuse, which is the radius of the circumscribed circle, is twice this, or $\frac{20}{\sqrt{3}}$ .

The area of the circumscribed circle can now be calculated:

$A= \pi r^{2} = \pi \cdot \left (\frac{20}{\sqrt{3}}\right ) ^{2}= \pi \cdot \frac{400}{3} = \frac{400 \pi }{3}$

Two circles are constructed; one is inscribed inside a given square, and the other is circumscribed about the same square.

The circumscribed circle has circumference $20 \pi$ . Give the area of the inscribed circle.

$50 \pi$

$200 \pi$

$300 \pi$

$\frac{100 \pi}{3}$

The correct answer is not among the other responses.

Explanation

Examine the diagram below, which shows the square, segments from its center to a vertex and the midpoint of a side, and the two circles.

Thingy

Note that the segment from the center of the square to the midpoint of a side is a radius of the inscribed circle, and the segment from the center to a vertex is a radius of the circumscribed circle. The two radii and half a side of the square form a 45-45-90 Triangle, so by the 45-45-90 Theorem, the radius of the inscribed circle is equal to that of the circumscribed circle divided by $\sqrt{2}$ .

The inscribed circle has circumference $20 \pi$ , so its radius is

$20 \pi \div 2 \pi = 10$

Divide this by $\sqrt{2}$ to get the radius of the circumscribed circle:

$\frac{10}{\sqrt{2}}= \frac{10 \cdot \sqrt{2}}{\sqrt{2} \cdot \sqrt{2}} = \frac{10 \sqrt{2}}{2} = 5 \sqrt{2}$

The circumscribed circle has area

$A= \pi r^{2} = \pi \cdot \left (5 \sqrt{2} \right ) ^{2}= \pi \cdot 25 \cdot 2 = 50 \pi$

What is the area of a circle with a diameter of ?

$324\pi cm^2$

$81\pi cm^2$

$54\pi cm^2$

$36\pi cm^2$

$90\pi cm^2$

Explanation

The area of a circle is defined by $A=\pi r^{2}$ , where is the radius of the circle. We are provided with the diameter of the circle, which is twice the length of .

If , then $r=\frac{d}{2}=\frac{18cm}{2}=9cm$

Then, solving for :

$A=\pi r^{2}$

$A=\pi (9cm)^{2}$

$A=81\pi cm^2$

Two circles are constructed; one is inscribed inside a given square, and the other is circumscribed about the same square.

The inscribed circle has circumference $20 \pi$ . Give the area of the circumscribed circle.

$200 \pi$

$400 \pi$

$300 \pi$

$50 \pi$

$\frac{100 \pi}{3}$

Explanation

Examine the diagram below, which shows the square, segments from its center to a vertex and the midpoint of a side, and the two circles.

Thingy

Note that the segment from the center of the square to the midpoint of a side is a radius of the inscribed circle, and the segment from the center to a vertex is a radius of the circumscribed circle. The two segments and half a side of the square form a 45-45-90 triangle, so by the 45-45-90 Theorem, the radius of the circumscribed circle is $\sqrt{2}$ times that of the inscribed circle.

The inscribed circle has circumference $20 \pi$ , so its radius is

$20 \pi \div 2 \pi = 10$

The circumscribed circle has radius $\sqrt{2}$ times this, or $10 \sqrt{2}$ , so its area is

$A= \pi r^{2} = \pi \cdot \left (10 \sqrt{2} \right ) ^{2}= \pi \cdot 100 \cdot 2 = 200 \pi$

Two circles are constructed; one is inscribed inside a given equilateral triangle, and the other is circumscribed about the same triangle.

The inscribed circle has circumference $20 \pi$ . Give the area of the circumscribed circle.

$400 \pi$

$200 \pi$

$100 \pi$

$25 \pi$

$50 \pi$

Explanation

Examine the diagram below, which shows the triangle, its three altitudes, and the two circles.

Thingy

The three altitudes of an equilateral triangle divide one another into two segments each, the longer of which is twice the length of the shorter. The length of each of the longer segments is the radius of the circumscribed circle, and the length of each of the shorter segments is the radius of the inscribed circle. Therefore, the circumscribed circle has twice the radius of the inscribed circle.

The inscribed circle has circumference $20 \pi$ , so its radius is

$20 \pi \div 2 \pi = 10$

The circumscribed circle has radius twice this, or 20, so its area is

$A= \pi r^{2} = \pi \cdot 20^{2}= 400 \pi$

If the pitcher plant Sarracenia purpurea has a circular opening with a circumference of 6 inches, what is the area of the opening?

$A\approx 2.86 in^2$

$A\approx0.95 in^2$

$A\approx 11.45 in^2$

$A\approx 37.70 in^2$

Explanation

If the pitcher plant Sarracenia purpurea has a circular opening with a circumference of 6 inches, what is the area of the opening?

We need to work backward from circumference to find area.

Circumference can be found as follows:

$Circ=\pi 2 r$

Use this to find "r" which we will use to find the area:

$6=2 \pi r$

$r=\frac{3}{\pi}$

Next, find area using the following:

$A=\pi r^2$

$A=\pi (\frac{3}{\pi})^2$

$A\approx 2.86 in^2$

A circle on the coordinate plane is defined by the equation $x^{2}+y^{2}=25$ . What is the area of the circle?

$25\pi$

$50\pi$

$60\pi$

$5\pi$

Not enough information provided.

Explanation

The equation of a circle centered at the origin of the coordinate plane is $x^{2}+y^{2}=r^{2}$ , where is the radius of the circle.

The area of the circle, in turn, is defined by the equation $A=\pi r^{2}$ .

Since we are provided with the equation $x^{2}+y^{2}=25$ , we can deduce that $r^{2}=25$ and that $A=25\pi$ .

What is the area of a circle with a diameter of ?

$100\pi cm^2$

$400\pi cm^2$

$20\pi cm^2$

$40\pi cm^2$

Not enough information provided.

Explanation

The area of a circle is defined by $A=\pi r^{2}$ , where is the radius of the circle. We are provided with the diameter of the circle, which is twice the length of .