Radius

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GMAT Quantitative › Radius

Questions 1 - 10

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

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 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

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

The arc $\widehat{AB}$ of a circle measures $60^{\circ }$ . The chord of the arc, $\overline{AB}$ , has length . Give the length of the radius of the circle.

$2 \pi x + 4 \pi$

$\pi x +2 \pi$

$\textup{None of the other responses gives a correct answer.}$

Explanation

A circle can be divided into congruent arcs that measure

$\frac{1}{6} \cdot 360 ^{\circ } = 60 ^{\circ }$ .

If the (congruent) chords are constructed, the figure will be a regular hexagon. The radius of this hexagon will be equal to the length of one side - one $60 ^{\circ }$ chord of the circle; this radius will coincide with the radius of the circle. Therefore, the radius of the circle is the length of chord $\overline{AB}$ , or .

The arc $\widehat{AB}$ of a circle measures $120^{\circ }$ . The chord of the arc, $\overline{AB}$ , has length . Give the length of the radius of the circle.

$\frac{\sqrt{3}}{6} x+ \frac{ \sqrt{3}}{3}$

$\frac{\sqrt{3}}{12} x+ \frac{ \sqrt{3}}{6}$

$\frac{\sqrt{3}}{4} x+ \frac{ \sqrt{3}}{2}$

$x \sqrt{3} + 2 \sqrt{3}$

Explanation

A circle can be divided into three congruent arcs that measure

$\frac{1}{3} \cdot 360 ^{\circ } = 120 ^{\circ }$ .

If the three (congruent) chords are constructed, the figure will be an equilateral triangle. The figure is below, along with the altitudes of the triangle:

Circle and triangle

Since , it follows by way of the 30-60-90 Triangle Theorem that

$BM = \frac{1}{2} AB = \frac{1}{2} (x+2) = \frac{1}{2} x+1$

and

$AM = \frac{ \sqrt{3}}{2} \cdot BM = \frac{ \sqrt{3}}{2} \cdot \left ( \frac{1}{2} x+1 \right ) = \frac{\sqrt{3}}{4} x+ \frac{ \sqrt{3}}{2}$

The three altitudes of an equilateral triangle split each other into segments that have ratio 2:1. Therefore,

$OA = \frac{2}{3} \cdot AM$

$=\frac{2}{3} \cdot \left ( \frac{\sqrt{3}}{4} x+ \frac{ \sqrt{3}}{2} \right )$

$= \frac{\sqrt{3}}{6} x+ \frac{ \sqrt{3}}{3}$

A $60^{\circ }$ arc of a circle measures . Give the radius of this circle.

$\frac{3}{\pi}x+\frac{12}{\pi}$

$\frac{6}{\pi}x+\frac{24}{\pi}$

$6 \pi x+24 \pi$

$9 \pi x^{2}+ 72 \pi x+ 144 \pi$

$3\pi x+12\pi$

Explanation

A $60^{\circ }$ arc of a circle is $\frac{60}{360 } = \frac{1}{6}$ of the circle. Since the length of this arc is , the circumference is $\textup{six times }$ this, or

The radius of a circle is its circumference divided by $2 \pi$ ; therefore, the radius is

$r = \frac{C}{2 \pi} = \frac{6x+24}{2\pi} = \frac{3}{\pi}x+\frac{12}{\pi}$

The points and form a line which passes through the center of circle Q. Both points are on circle Q.

To the nearest hundreth, what is the length of the radius of circle Q?

Explanation

To begin this problem, we need to recognize that the distance between points L and K is our diameter. Segment LK passes from one point on circle Q through the center, to another point on circle Q. Sounds like a diameter to me! Use distance formula to find the length of LK.

$d=\sqrt{(x-x')^2+(y-y')^2}$

Plug in our points and simplify:

$d=\sqrt{(9--4)^2+(3-5)^2}=\sqrt{13^2+2^2}=\sqrt{173}\approx13.15$

Now, don't be fooled into choosing 13.15. That is our diameter, so our radius will be half of 13.15, or 6.575. This rounds to 6.58

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

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

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