Calculating the length of a chord - GMAT Quantitative

Card 0 of 72

Question

The chord of a $60^{\circ }$ central angle of a circle with circumference $90 \pi$ has what length?

Answer

A circle with circumference $90 \pi$ has as its radius

$90\pi \div 2\pi = 45$ .

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

By way of the Isosceles Triangle Theorem, $\Delta AOB$ can be proved equilateral, so , the correct response.

Compare your answer with the correct one above

Question

The chord of a $90^{\circ }$ central angle of a circle with circumference $60 \pi$ has what length?

Answer

A circle with circumference $60 \pi$ has as its radius

$60\pi \div 2\pi = 30$ .

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

By way of the Isosceles Triangle Theorem, $\Delta AOB$ can be proved a 45-45-90 triangle with legs of length 30. By the 45-45-90 Theorem, its hypotenuse - the chord of the central angle - has length $\sqrt{ 2}$ times this, or $30 \sqrt{2}$ . This is the correct response.

Compare your answer with the correct one above

Question

The chord of a $90^{\circ }$ central angle of a circle with area $50 \pi$ has what length?

Answer

The radius of a circle with area $50 \pi$ can be found as follows:

$\pi r^{2} = A$

$\pi r^{2} = 50 \pi$

$r^{2} = 50$

$r = \sqrt{50 } = \sqrt{25} \cdot \sqrt{2} = 5 \sqrt{2}$

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

By way of the Isosceles Triangle Theorem, $\Delta AOB$ can be proved a 45-45-90 triangle with legs of length $5 \sqrt{2}$ . Its hypotenuse has length $\sqrt{ 2}$ times this, or

$5 \sqrt{2} \times \sqrt{2}= 5 \times 2 = 10$

This is the correct response.

Compare your answer with the correct one above

Question

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

Answer

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:

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

Compare your answer with the correct one above

Question

The chord of a $120^{\circ }$ central angle of a circle with area $32 \pi$ has what length?

Answer

The radius of a circle with area $32 \pi$ can be found as follows:

$\pi r^{2} = A$

$\pi r^{2} = 32 \pi$

$r^{2} = 32$

$r = \sqrt{32} = \sqrt{16} \cdot \sqrt{2} = 4 \sqrt{2}$

The circle, the central angle, and the chord are shown below, along with $\overline{AM}$ , which bisects isosceles $\Delta AOB$

We concentrate on $\Delta AOM$ , a 30-60-90 triangle. By the 30-60-90 Theorem,

$OM = \frac{1}{2}AO = \frac{1}{2} \cdot 4 \sqrt{2} = 2\sqrt{2}$

and

$AM = OM \sqrt{3} = 2 \sqrt{2} \cdot \sqrt{3} = 2 \sqrt{6}$

The chord $\overline{AB}$ has length twice this, or

$2 \cdot 2 \sqrt{6} = 4 \sqrt{6}$

Compare your answer with the correct one above

Question

The chord of a $120^{\circ }$ central angle of a circle with circumference $40 \pi$ has what length?

Answer

A circle with circumference $40 \pi$ has as its radius

$40 \div 2\pi = 20$ .

The circle, the central angle, and the chord are shown below, along with $\overline{AM}$ , which bisects isosceles $\Delta AOB$ :

We concentrate on $\Delta AOM$ , a 30-60-90 triangle. By the 30-60-90 Theorem,

$\overline{OM}$ has half the length of $\overline{AO}$ , so

and

$AM = OM \sqrt{3} =10 \sqrt{3}$

The chord $\overline{AB}$ has length twice this, or

$2 \cdot 10 \sqrt{3}= 20 \sqrt{3}$

Compare your answer with the correct one above

Question

Consider the Circle :

(Figure not drawn to scale.)

If $\angle BOX$ is a $90^{\circ}$ angle, what is the measure of segment ?

Answer

This is a triangle question in disguise. We have a ninety-degree triangle with two sides made up of the radii of the circle. This means the other two angles ( $\angle OXB$ and $\angle OBX$ ) must be $45^{\circ}$ each.

Use the 45/45/90 triangle ratios to find the final side. Additionally, you could use Pythagorean Theorem to find the missing side.

45/45/90 side length ratios:

$x/x/\sqrt{2}x$

Segment $BX=15\sqrt{2}:m$

Or, using the Pythagorean Theorem, by rearranging it and solving for , the hypotenuse, which in this case is segment :

$BX=\sqrt{15^2+15^2}=\sqrt{450}=\sqrt{2252}=\sqrt{225}\sqrt{2}=15*\sqrt{2}=15\sqrt2:m$

Compare your answer with the correct one above

Question

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 .

Answer

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:

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

Compare your answer with the correct one above

Question

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

Answer

The figure referenced is below.

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

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

Compare your answer with the correct one above

Question

The chord of a $60^{\circ }$ central angle of a circle with circumference $90 \pi$ has what length?

Answer

A circle with circumference $90 \pi$ has as its radius

$90\pi \div 2\pi = 45$ .

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

By way of the Isosceles Triangle Theorem, $\Delta AOB$ can be proved equilateral, so , the correct response.

Compare your answer with the correct one above

Question

The chord of a $90^{\circ }$ central angle of a circle with circumference $60 \pi$ has what length?

Answer

A circle with circumference $60 \pi$ has as its radius

$60\pi \div 2\pi = 30$ .

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

By way of the Isosceles Triangle Theorem, $\Delta AOB$ can be proved a 45-45-90 triangle with legs of length 30. By the 45-45-90 Theorem, its hypotenuse - the chord of the central angle - has length $\sqrt{ 2}$ times this, or $30 \sqrt{2}$ . This is the correct response.

Compare your answer with the correct one above

Question

The chord of a $90^{\circ }$ central angle of a circle with area $50 \pi$ has what length?

Answer

The radius of a circle with area $50 \pi$ can be found as follows:

$\pi r^{2} = A$

$\pi r^{2} = 50 \pi$

$r^{2} = 50$

$r = \sqrt{50 } = \sqrt{25} \cdot \sqrt{2} = 5 \sqrt{2}$

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

By way of the Isosceles Triangle Theorem, $\Delta AOB$ can be proved a 45-45-90 triangle with legs of length $5 \sqrt{2}$ . Its hypotenuse has length $\sqrt{ 2}$ times this, or

$5 \sqrt{2} \times \sqrt{2}= 5 \times 2 = 10$

This is the correct response.

Compare your answer with the correct one above

Question

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

Answer

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:

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

Compare your answer with the correct one above

Question

The chord of a $120^{\circ }$ central angle of a circle with area $32 \pi$ has what length?

Answer

The radius of a circle with area $32 \pi$ can be found as follows:

$\pi r^{2} = A$

$\pi r^{2} = 32 \pi$

$r^{2} = 32$

$r = \sqrt{32} = \sqrt{16} \cdot \sqrt{2} = 4 \sqrt{2}$

The circle, the central angle, and the chord are shown below, along with $\overline{AM}$ , which bisects isosceles $\Delta AOB$

We concentrate on $\Delta AOM$ , a 30-60-90 triangle. By the 30-60-90 Theorem,

$OM = \frac{1}{2}AO = \frac{1}{2} \cdot 4 \sqrt{2} = 2\sqrt{2}$

and

$AM = OM \sqrt{3} = 2 \sqrt{2} \cdot \sqrt{3} = 2 \sqrt{6}$

The chord $\overline{AB}$ has length twice this, or

$2 \cdot 2 \sqrt{6} = 4 \sqrt{6}$

Compare your answer with the correct one above

Question

The chord of a $120^{\circ }$ central angle of a circle with circumference $40 \pi$ has what length?

Answer

A circle with circumference $40 \pi$ has as its radius

$40 \div 2\pi = 20$ .

The circle, the central angle, and the chord are shown below, along with $\overline{AM}$ , which bisects isosceles $\Delta AOB$ :

We concentrate on $\Delta AOM$ , a 30-60-90 triangle. By the 30-60-90 Theorem,

$\overline{OM}$ has half the length of $\overline{AO}$ , so

and

$AM = OM \sqrt{3} =10 \sqrt{3}$

The chord $\overline{AB}$ has length twice this, or

$2 \cdot 10 \sqrt{3}= 20 \sqrt{3}$

Compare your answer with the correct one above

Question

Consider the Circle :

(Figure not drawn to scale.)

If $\angle BOX$ is a $90^{\circ}$ angle, what is the measure of segment ?

Answer

This is a triangle question in disguise. We have a ninety-degree triangle with two sides made up of the radii of the circle. This means the other two angles ( $\angle OXB$ and $\angle OBX$ ) must be $45^{\circ}$ each.

Use the 45/45/90 triangle ratios to find the final side. Additionally, you could use Pythagorean Theorem to find the missing side.

45/45/90 side length ratios:

$x/x/\sqrt{2}x$

Segment $BX=15\sqrt{2}:m$

Or, using the Pythagorean Theorem, by rearranging it and solving for , the hypotenuse, which in this case is segment :

$BX=\sqrt{15^2+15^2}=\sqrt{450}=\sqrt{2252}=\sqrt{225}\sqrt{2}=15*\sqrt{2}=15\sqrt2:m$

Compare your answer with the correct one above

Question

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 .

Answer

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:

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

Compare your answer with the correct one above

Question

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

Answer

The figure referenced is below.

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

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

Compare your answer with the correct one above

Question

The chord of a $60^{\circ }$ central angle of a circle with circumference $90 \pi$ has what length?

Answer

A circle with circumference $90 \pi$ has as its radius

$90\pi \div 2\pi = 45$ .

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

By way of the Isosceles Triangle Theorem, $\Delta AOB$ can be proved equilateral, so , the correct response.

Compare your answer with the correct one above

Question

The chord of a $90^{\circ }$ central angle of a circle with circumference $60 \pi$ has what length?

Answer

A circle with circumference $60 \pi$ has as its radius

$60\pi \div 2\pi = 30$ .

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

By way of the Isosceles Triangle Theorem, $\Delta AOB$ can be proved a 45-45-90 triangle with legs of length 30. By the 45-45-90 Theorem, its hypotenuse - the chord of the central angle - has length $\sqrt{ 2}$ times this, or $30 \sqrt{2}$ . This is the correct response.

Compare your answer with the correct one above

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