Area and Circumference of a Circle - SSAT Upper Level Quantitative

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Question

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

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Answer

The formula for the area of a circle is

pi $r^{2}$

Find the radius by dividing 9 by 2:

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

So the formula for area would now be:

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

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Question

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

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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}$$

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Question

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

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Answer

First, solve for radius:

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

Then, solve for area:

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Question

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

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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$$

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Question

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

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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$$

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Question

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

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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$

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Question

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

Let $\pi = 3.14$ .

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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$$

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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 .

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Answer

Chord length = , where is the radius of the circle and is the perpendicular distance from the chord to the circle center.

Chord length =

$\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$$=12.56t^2$$

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Question

A circle on the coordinate plane has equation

.

Which of the following gives the area of the circle?

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Answer

The equation of a circle on the coordinate plane is

,

where is the radius. Therefore, in this equation,

.

The area of a circle is found using the formula

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

so we substitute 66 for , yielding

$A = 66 \pi$ .

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Question

Give the area of the above figure.

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Answer

The figure is a semicircle - one-half of a circle - with radius 5.5, or $\frac{11}{2}$ . Its area is one-half of the square of the radius multiplied by $\pi$ - that is,

$A = $\frac{1}{2}$ \pi $r^{2}$$

$A = $\frac{1}{2}$ \pi \left ($\frac{11}{2}$ \right $)^{2}$$

$A = $\frac{1}{2}$ \pi \cdot $\frac{121}{4}$$

$A = $\frac{121\pi}{8 }$$

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Question

Give the area of the figure in the above diagram.

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Answer

The figure is a sector of a circle with radius 8; the sector has degree measure $180 ^{\circ } - 45 ^{\circ }= 135 ^{\circ }$ . The area of the sector is

$A = $\frac{ $135^{\circ }$$}{360 ^{\circ }} \cdot \pi $r^{2}$$

$A = $\frac{ $135^{\circ }$$}{360 ^{\circ }} \cdot \pi \cdot $8^{2}$$

$A = $\frac{ 3}{8}$ \cdot 64 \pi$

$A = 24 \pi$

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Question

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

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Answer

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}$$

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Question

Give the area of a circle that circumscribes an equilateral triangle with perimeter 54.

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Answer

An equilateral triangle of perimeter 54 has sidelength one-third of this, or 18.

Construct this triangle and its circumscribed circle, as well as a perpendicular bisector to one side and a radius to one of that side's endpoints:

Each side of the triangle has measure 18, so . Also, the triangle formed by the segments, by symmetry, is a 30-60-90 triangle. By the 30-60-90 Theorem,

$BO = $\frac{AB}{\sqrt{3}$} = $\frac{9}{\sqrt{3}$} = $\frac{9\cdot \sqrt{3}$}{$\sqrt{3}$\cdot $\sqrt{3}$} = $\frac{9 \sqrt{3}$}{3} = 3 $\sqrt{3}$$

and $AO = 2 \cdot BO = 2 \cdot 3$\sqrt{3}$ = 6 $\sqrt{3}$$ .

The latter is the radius, so the area of this circle is

$A = \pi $r^{2}$ = \pi (6 $$\sqrt{3}$)^{2}$ = \pi \cdot 6 ^{2} $($\sqrt{3}$)^{2}$ = 36 \cdot 3 \cdot \pi = 108 \pi$

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Question

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

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Answer

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:

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$

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Question

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.

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Answer

Examine the following diagram:

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.

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Question

Give the area of a circle that circumscribes a right triangle with legs of length and .

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Answer

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

The length of the hypotenuse of this triangle can be calculated using the Pythagorean Theorem:

$d = \sqrt ${10^{2}$$+24^{2}$} = $\sqrt{100+576}$ = $\sqrt{676}$ = 26$

The radius is half this, or 13, so the area is

$A = \pi $r^{2}$ = \pi \cdot 13 ^{2} = 169 \pi$

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Question

Give the area of a circle that circumscribes a 30-60-90 triangle whose shorter leg has length 11.

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Answer

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

The length of a hypotenuse of a 30-60-90 triangle is twice that of its short leg, so the hypotenuse of this triangle will be twice 11, or 22. The diameter of the circle is therefore 22, and the radius is half this, or 11. The area of the circle is therefore

$A = \pi $r^{2}$ = \pi \cdot 11 ^{2} = 121 \pi$

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Question

A $90 ^{\circ }$ central angle of a circle has a chord with length . Give the area of the circle.

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Answer

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

By way of the Isoscelese Triangle Theorem, $\Delta AOB$ can be proved a 45-45-90 triangle with hypotenuse 15. By the 45-45-90 Theorem, its legs, each a radius, have length that can be determined by dividing this by $\sqrt{2}$ :

$A O = $\frac{AB }{\sqrt{2}$} = $\frac{15}{\sqrt{2}$}$

The area is therefore

$A = \pi $r^{2}$ = \pi \cdot \left ( $\frac{15}{\sqrt{2}$} \right $)^{2}$ = \pi \cdot $\frac{225}{2}$ =\frac{225 \pi}{2}$$

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Question

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

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Answer

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

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$

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Question

A $120 ^{\circ }$ central angle of a circle has a chord with length . Give the area of the circle.

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Answer

The figure below shows $\angle AOB$ , which matches this description, along with its chord $\overline{AB}$ and triangle bisector $\overline{OM}$ .

We will concentrate on $\Delta AOM$ , which is a 30-60-90 triangle.

Chord $\overline{AB}$ has length 15, so $AM = $\frac{1}{2}$ AB = $\frac{1}{2}$ \cdot 15 = $\frac{15}{2}$$

By the 30-60-90 Theorem,

$OM = AM \div $\sqrt{3}$= $\frac{15}{2}$ \div $\sqrt{3}$= $\frac{15}{2\sqrt{3}$}$

and

$AO = 2 \cdot OM = 2 \cdot $\frac{15}{2\sqrt{3}$} = $\frac{15}{ \sqrt{3}$}$

This is the radius, so the area is

$A = \pi $r^{2}$ = \pi \cdot \left ( $\frac{15}{ \sqrt{3}$} \right $)^{2}$= $\frac{225}{3}$ \pi = 75 \pi$

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