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

Find the area of a circle with a diameter of 16.

$64\Pi$

$8\Pi$

$256\Pi$

$16\Pi$

Explanation

The formula for the area of a circle is as follows:

$A=\Pi\cdot r^{2}$

In this formula, A is area, and r is for radius. We know the diameter of the circle is 16, meaning we have to first find the radius. The diameter is twice the length of the radius of a circle, meaning to find radius divide the diameter by 2:

$\frac{16}{2}=8=r$

Now plug in the numbers to get the answer. Since pi is an irrational constant, it is okay to leave the answer in terms of pi.

$A=\Pi\cdot 8^{2}=64\Pi$

Find the area of a circle with a radius of 7.

$49\Pi$

$7\Pi$

$14\Pi$

$42\Pi$

Explanation

The formula for the area of a circle is as follows:

$A=\Pi\cdot r^{2}$

In this formula, A is area, and r is for radius. We know the radius of the circle is 7, so plug in the numbers to get the answer. Since pi is an irrational constant, it is okay to leave the answer in terms of pi.

$A=\Pi\cdot 7^{2}=49\Pi$

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

$4\pi$

$\frac{4}{\pi }$

$2\pi$

Explanation

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

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

Give the ratio of the area of a circle that circumscribes an equilateral triangle to that of a circle that is inscribed inside the same triangle.

$2\sqrt{2}:1$

$2\sqrt{3}:1$

$3\sqrt{2}:1$

Explanation

Examine the following diagram:

Thingy

If a (perpendicular) radius of the inscribed circle is constructed to the triangle, and a radius of the circumscribed circle is constructed to a neighboring vertex, a right triangle is formed. By symmetry, it can be shown that this is a 30-60-90 triangle, and, subsequently,

$BO = 2 \cdot AO$

If we let , the area of the inscribed circle is $A = \pi r^{2}$ .

Then , and the area of the circumscribed circle is $\pi\left ( 2r \right )^{2} = \pi \cdot 4r^{2} = 4 \pi r^{2} = 4A$

The ratio of the areas is therefore 4 to 1.

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

$\dpi{100} 64$

$\dpi{100} 254$

$\dpi{100} 255$

$\dpi{100} 81$

Explanation

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:

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

Give the area of a circle that is inscribed in an equilateral triangle with perimeter .

$48 \pi$

$24 \pi$

$72 \pi$

$96 \pi$

$144 \pi$

Explanation

An equilateral triangle of perimeter 72 has sidelength one-third of this, or 24.

Construct this triangle and its inscribed circle, as well as a radius to one side - which, by symmetry, is a perpendicular bisector - and a segment to one of that side's endpoints:

Thingy

Each side of the triangle has measure 24, so . Also, the triangle formed by the segments, by symmetry, is a 30-60-90 triangle. Therefore,

$BO = \frac{AB}{\sqrt{3}} = \frac{12}{\sqrt{3}} = \frac{12\cdot \sqrt{3}}{\sqrt{3}\cdot \sqrt{3}} = \frac{12 \sqrt{3}}{3} =4\sqrt{3}$

which is the radius of the circle. The area of this circle is

$A = \pi r^{2} = \pi (4 \sqrt{3})^{2} = \pi \cdot 4 ^{2} (\sqrt{3})^{2} = 16\cdot 3 \cdot \pi= 48\pi$

A $60 ^{\circ }$ central angle of a circle has a chord with length 7. Give the area of the circle.

$49 \pi$

$14 \pi$

$98 \pi$

$\frac{49 \pi}{4}$

The correct answer is not among the other responses.

Explanation

The figure below shows $\angle AOB$ , which matches this description, along with its chord $\overline{AB}$ :

Thingy

By way of the Isosceles Triangle Theorem, $\Delta AOB$ can be proved equilateral, so . This is the radius, so the area is

$A = \pi r^{2} = \pi \cdot 7^{2}= 49 \pi$

Give the area of a circle that circumscribes a triangle whose longer leg has length .

$\frac{121 \pi}{3}$

$\frac{121 \pi}{2}$

$121 \pi$

$242\pi$

$\frac{484\pi}{3}$

Explanation

If a right triangle is inscribed inside a circle, then the arc intercepted by the right angle is a semicircle, making the hypotenuse of the triangle a diameter.

By the 30-60-90 Theorem, the length of the shorter leg of a 30-60-90 triangle is that of the longer leg divided by $\sqrt{3}$ , so the shorter leg will have length $\frac{11}{\sqrt{3}}$ ; the hypotenuse will have length twice this length, or

$2 \cdot \frac{11}{\sqrt{3}} = \frac{22}{\sqrt{3}}$ .

The diameter of the circle is therefore $\frac{22}{\sqrt{3}}$ ; the radius is half this, or $\frac{11}{\sqrt{3}}$ . The area of the circle is therefore

$A = \pi r^{2} = \pi \cdot \left ( \frac{11}{\sqrt{3}} \right ) ^{2} = \frac{121}{3}\cdot \pi = \frac{121 \pi}{3}$

Find the area of a circle with a radius of 5.

$25\Pi$

$10\Pi$

$5\Pi$

Explanation

The formula for the area of a circle is as follows:

$A=\Pi\cdot r^{2}$

In this formula, A is area, and r is for radius. We know the radius of the circle is 5, so plug in the numbers to get the answer. Since pi is an irrational constant, it is okay to leave the answer in terms of pi.

$A=\Pi\cdot 5^{2}=25\Pi$