Circles

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GMAT Quantitative › Circles

Questions 1 - 10

The chord of a $60^{\circ }$ central angle of a circle with area $40 \pi$ has what length?

$2 \sqrt{10}$

$10\sqrt{2}$

$4 \sqrt{5}$

Explanation

The radius of a circle with area $40 \pi$ can be found as follows:

$\pi r^{2} = A$

$\pi r^{2} = 40 \pi$

$r^{2} = 40$

$r = \sqrt{40 } = \sqrt{4} \cdot \sqrt{10} = 2 \sqrt{10}$

The circle, the central angle, and the chord are shown below:

Chord

By way of the Isosceles Triangle Theorem, $\Delta AOB$ can be proved equilateral, so $AB = AO= 2 \sqrt{10}$ , the correct response.

The arc $\widehat{AB}$ of a circle measures $90^{\circ }$ and has length . Give the length of the chord $\overline{AB}$ .

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

$\frac{ \sqrt{2}}{\pi}x + \frac{ \sqrt{2}}{\pi}$

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

$\frac{ \pi \sqrt{2}}{2} x + \pi \sqrt{2}$

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

Explanation

The figure referenced is below.

Circle x

The arc is $\frac{90}{360} = \frac{1}{4}$ of the circle, so the circumference of the circle is

The radius is this circumference divided by $2 \pi$ , or

$\frac{ 4x+8}{2 \pi} = \frac{2}{\pi}x + \frac{4}{\pi}$ .

$\overline{AB}$ is, consequently, the hypotenuse of an isosceles right triangle with leg length $\frac{2}{\pi}x + \frac{4}{\pi}$ ; by the 45-45-90 Triangle Theorem, its length is $\sqrt{2}$ times this, or

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

Calculate the length of a chord in a circle with a radius of , given that the perpendicular distance from the center to the chord is .

Explanation

We are given the radius of the circle and the perpendicular distance from its center to the chord, which is all we need to calculate the length of the chord. Using the formula for chord length that involves these two quantities, we find the solution as follows, where is the chord length, is the perpendicular distance from the center of the circle to the chord, and is the radius:

$L=2\sqrt{r^2-d^2}=2\sqrt{5^2-3^3}=2\sqrt{25-9}=2\sqrt{16}=8$

A given circle has an area of $81\pi$ . What is the length of its diameter?

Not enough information provided

Explanation

The area of a circle is defined by the equation $A=\pi r^{2}$ , where is the length of the circle's radius. The radius, in turn, is defined by the equation $r=\frac{d}{2}$ , where is the length of the circle's diameter.

Given $A=81\pi$ , we can deduce that $r^{2}=81$ and therefore . Then, since $r=\frac{d}{2}$ , .

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:

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

What is the arc length for a sector with a central angle of $45^{\circ}$ if the radius of the circle is ?

$\frac{3\pi }{4}$

$\frac{\pi }{4}$

$\frac{3\pi }{2}$

$\frac{\pi }{2}$

$\pi$

Explanation

Using the formula for arc length, we can plug in the given angle and radius to calculate the length of the arc that subtends the central angle of the sector. The angle, however, must be in radians, so we make sure to convert degrees accordingly by multiplying the given angle by $\frac{\pi }{180}$ :

$S=r\theta$

$S=3(45^{\circ})(\frac{\pi }{180^{\circ}})=3(\frac{\pi }{4})=\frac{3\pi}{4}$

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 $90^{\circ }$ . The chord of the arc, $\overline{AB}$ , has length . Give the length of the arc $\widehat{AB}$ .

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

$\frac{ \pi \sqrt{2}}{2} x + \pi \sqrt{2}$

$\frac{ \sqrt{2}}{\pi}x + \frac{ \sqrt{2}}{\pi}$

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

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

Explanation

Examine the figure below, which shows the arc and chord in question.

Circle x

If we extend the figure to depict the circle as the composite of four quarter-circles, each a $90^{\circ }$ arc, we see that is also the side of an inscribed square. A diagonal of this square, which measures $\sqrt{2}$ times this sidelength, or

$(x+2) \sqrt{2} = x\sqrt{2} + 2\sqrt{2}$ ,

is a diameter of this circle. The circumference is $\pi$ times the diameter, or

$\pi ( x\sqrt{2} + 2\sqrt{2}) = x \pi \sqrt{2} + 2 \pi \sqrt{2}$ .

Since a $90^{\circ }$ arc is one fourth of a circle, the length of arc $\widehat{AB}$ is

$\frac{1}{4}( x \pi \sqrt{2} + 2 \pi \sqrt{2})$

$= \frac{ \pi }{4} x\sqrt{2} + \frac{ \pi }{2} \sqrt{2}$

$= \frac{ \pi \sqrt{2}}{4} x + \frac{ \pi \sqrt{2}}{2}$

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 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 $\sqrt{2}$ .