How to find the length of a chord - ISEE Upper Level: Quantitative Reasoning

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Question

In the above figure, $\overline{NT}$ is a tangent to the circle.

Evaluate .

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Answer

If a secant segment and a tangent segment are constructed to a circle from a point outside it, the square of the distance to the circle along the tangent is equal to the product of the distances to the two points on the circle along the secant; in other words,

$NA \cdot NB = (NT)^{2}$ ,

and, substituting,

$t(t+24) = 16 ^{2}$

Distributing and writing in standard quadratic polynomial form,

$t^{2} + 24t=256$

$t^{2} + 24t- 256 =256 - 256$

$t^{2} + 24t- 256 =0$

We can factor the polynomial by looking for two integers with product and sum 24; through some trial and error, we find that these numbers are 32 and , so we can write this as

By the Zero Product Principle,

, in which case - impossible since is a (positive) distance; or,

, in which case - the correct choice.

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Question

Refer to the above figure. Which is the greater quantity?

(a) $\frac{a}{d}$

(b) $\frac{c}{b}$

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Answer

If two chords intersect inside a circle, both chords are cut in a way such that the products of the lengths of the two chords formed in each are the same - in other words,

Divide both sides of this equation by , then cancelling:

$\frac{ab }{bd}= \frac{cd}{bd}$

$\frac{a }{d}= \frac{c}{b}$

The two quantities are equal.

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Question

Refer to the above figure. Which is the greater quantity?

(a)

(b) 3

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Answer

If two chords intersect inside a circle, both chords are cut in a way such that the products of the lengths of the two chords formed in each are the same - in other words,

$a \cdot b = 1.5 \cdot 1.5$

or

Therefore, .

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Question

Figure NOT drawn to scale.

Refer to the above figure. Which is the greater quantity?

(a)

(b) 7

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Answer

If two chords intersect inside a circle, both chords are cut in a way such that the products of the lengths of the two chords formed in each are the same - in other words,

$t \cdot 4t = 8 \cdot 24$

Solving for :

$4t ^{2} = 192$

$4t ^{2} \div 4 = 192 \div 4$

$t ^{2} = 48$

Since $t ^{2} < 49$ , it follows that $t < \sqrt{49 }$ , or .

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Question

In the above figure, $\overline{NT}$ is a tangent to the circle.

Which is the greater quantity?

(a)

(b) 32

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Answer

If a secant segment and a tangent segment are constructed to a circle from a point outside it, the square of the distance to the circle along the tangent is equal to the product of the distances to the two points on the circle along the secant; in other words,

$NA \cdot NB = (NT )^{2}$

$NA \cdot (NA + AB)= (NT )^{2}$

$t \cdot (t+t) = 16 ^{2}$

Simplifying, then solving for :

$t \cdot (2t) = 256$

$2t ^{2} = 256$

$2t ^{2} \div 2 = 256 \div 2$

$t ^{2} = 128$

To compare to 32, it suffices to compare their squares:

, so, applying the Power of a Product Principle, then substituting,

$(NB)^{2} = (2t) ^{2} = 2^{2} \cdot t^{2} = 4 t ^{2} = 4 \cdot 128 = 512$

$32^{2} = 1,024$

, so

$(NB)^{2} < 32 ^{2}$ ;

it follows that

.

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Question

Figure NOT drawn to scale

In the above figure, $\overline{NT}$ is a tangent to the circle.

Which is the greater quantity?

(a)

(b) 8

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Answer

If a secant segment and a tangent segment are constructed to a circle from a point outside it, the square of the distance to the circle along the tangent is equal to the product of the distances to the two points on the circle intersected by the secant; in other words,

$(NT)^{2}= NA \cdot NB$

$(NT)^{2}= NA \cdot (NA+AB)$

$(2t) ^{2} = t (t + 24)$

Simplifying and solving for :

$2^{2} \cdot t^{2} = t \cdot t + t \cdot 24$

$4 t^{2} = t ^{2} + 24t$

$4 t^{2} - t ^{2} - 24t = t ^{2} + 24t - t ^{2} - 24t$

$3 t^{2}- 24t = 0$

Factoring out :

Either - which is impossible, since must be positive, or

, in which case .

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Question

Figure NOT drawn to scale

In the above figure, is the center of the circle, and $\overline{NT}$ is a tangent to the circle. Also, the circumference of the circle is $14 \pi$ .

Which is the greater quantity?

(a)

(b) 25

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Answer

$\overline{OT}$ is a radius of the circle from the center to the point of tangency of $\overline{NT}$ , so

$\overline{OT} \perp \overline{NT}$ ,

and $\bigtriangleup NTO$ is a right triangle. The length of leg $\overline{NT}$ is known to be 24. The other leg $\overline{OT}$ is a radius radius; we can find its length by dividing the circumference by $2 \pi$ :

$OT = r = \frac{C}{2\pi} = \frac{14 \pi}{2\pi} = 7$

The length hypotenuse, $\overline{NO}$ , can be found by applying the Pythagorean Theorem:

$NO = \sqrt{(NT)^{2} + (OT)^{2}}= \sqrt{24^{2} + 7^{2}}=\sqrt{576 + 49} = \sqrt{625} = 25$ .

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Question

Which equation is the formula for chord length?

Note: is the radius of the circle, and is the angle cut by the chord.

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Answer

The length of a chord of a circle is calculated as follows:

Chord length = $2rsin(\frac{C}{2})$

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Question

The radius of a circle is $5\ cm$ , and the perpendicular distance from a chord to the circle center is $4\ cm$ . Give the chord length.

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Answer

Chord length = $2\sqrt{r^2-d^2}$ , where is the radius of the circle and is the perpendicular distance from the chord to the circle center.

Chord length = $2\sqrt{r^2-d^2}=2\sqrt{5^2-4^2}$

$=2\sqrt{25-16}=2\sqrt{9}=2\times 3$

$\Rightarrow$ Chord length = $6\ cm$

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Question

In the circle below, the radius is $5\ cm$ and the chord length is $8\ cm$ . Give the perpendicular distance from the chord to the circle center (d).

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Answer

Chord length = $2\sqrt{r^2-d^2}$ , where is the radius of the circle and is the perpendicular distance from the chord to the circle center.

Chord length = $2\sqrt{r^2-d^2}\Rightarrow 8=2\sqrt{5^2-d^2}\Rightarrow 4=\sqrt{5^2-d^2}$

$\Rightarrow 16=25-d^2\Rightarrow d^2=25-16=9$

$\Rightarrow d=3\ cm$

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Question

Give the length of the chord of a $60^{\circ }$ central angle of a circle with radius 20.

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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 answer is not among the choices given.

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Question

Give the length of the chord of a $90^{\circ }$ central angle of a circle with radius 18.

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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 legs of length 18, so its hypotenuse - the desired chord length - is $\sqrt{2}$ times this, or $18\sqrt{2}$ .

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Question

A $60^{\circ}$ central angle of a circle intercepts an arc of length $7 \pi$ ; it also has a chord. What is the length of that chord?

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Answer

The arc intercepted by a $60^{\circ}$ central angle is $\frac{60}{360}= \frac{1}{6}$ of the circle, so the circumference of the circle is $7 \pi \cdot 6 = 42\pi$ . The radius is the circumference divided by $2 \pi$ , or $42 \pi \div 2 \pi = 21$ .

The figure below shows a $60^{\circ}$ central angle $\angle AOB$ , along with its chord $\overline{AB}$ :

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

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Question

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

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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. By the 30-60-90 Theorem,

$OM = \frac{1}{2} AO = \frac{1}{2} \cdot 12 = 6$

and

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

is the midpoint of $\overline{AB}$ , so

$AB = 2 \cdot AM = 2 \cdot 6 \sqrt{3}= 12 \sqrt{3}$

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Question

A $120^{\circ }$ central angle of a circle intercepts an arc of length $\frac{4 \pi}{3}$ ; it also has a chord. What is the length of that chord?

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Answer

The arc intercepted by a $120^{\circ}$ central angle is $\frac{120}{360}= \frac{1}{3}$ of the circle, so the circumference of the circle is $\frac{4 \pi}{3}\cdot 3= 4\pi$ . The radius is the circumference divided by $2 \pi$ , or $4\pi \div 2 \pi = 2$ .

The figure below shows a $120^{\circ}$ central angle $\angle AOB$ , along with its chord $\overline{AB}$ and triangle bisector $\overline{OM}$ .

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

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

and

$AM = OM \cdot \sqrt{3}= 1\sqrt{3}= \sqrt{3}$

is the midpoint of $\overline{AB}$ , so

$AB = 2 \cdot AM = 2 \sqrt{3}$

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Question

Figure NOT drawn to scale

In the above diagram, evaluate .

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Answer

If two chords of a circle intersect inside the circle, the product of the lengths of the parts of each chord is the same. In other words,

$2 t \cdot t = 10 \cdot 10$

$2 t ^{2}= 10 0$

Solving for :

$2 t ^{2} \div 2 = 10 0 \div 2$

$t ^{2} = 50$

$t = \sqrt{50 }$

Simplifying the radical using the Product of Radicals Principle, and noting that 25 is the greatest perfect square factor of 50:

$t = \sqrt{25 } \cdot \sqrt{2} = 5 \sqrt{2}$

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Question

Figure NOT drawn to scale

In the above diagram, evaluate .

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Answer

If two chords of a circle intersect inside the circle, the product of the lengths of the parts of each chord is the same. In other words,

$t \cdot t = 8 \cdot 12$

$t ^{2} = 96$

Solving for :

$t = \sqrt{96 }$

Simplifying the radical using the Product of Radicals Principle, and noting that the greatest perfect square factor of 96 is 16:

$t = \sqrt{16 } \cdot \sqrt{6 } = 4 \sqrt{6 }$

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Question

Figure NOT drawn to scale

In the figure above, evaluate .

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Answer

If two chords of a circle intersect inside the circle, the product of the lengths of the parts of each chord is the same. In other words,

Solving for - distribute:

$40 t = 20 t+ 20 \cdot 12$

Subtract from both sides:

Divide both sides by 20:

$20 t \div 20 = 240 \div 20$

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Question

In the above figure, $\overline{NT}$ is a tangent to the circle.

Evaluate .

Tap to reveal answer

Answer

If a secant segment and a tangent segment are constructed to a circle from a point outside it, the square of the distance to the circle along the tangent is equal to the product of the distances to the two points on the circle along the secant; in other words,

$t^{2} = 12 \cdot (12+ 18) = 12 \cdot 30 = 360$

Solving for :

$t^{2} = 360$

$t = \sqrt{360}$

Simplifying the radical using the Product of Radicals Principle, and noting that 36 is the greatest perfect square factor of 360:

$t = \sqrt{36} \cdot \sqrt{10} = 6 \sqrt{10}$

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Question

Figure NOT drawn to scale

In the above figure, $\overline{NT}$ is a tangent to the circle.

Evaluate .

Tap to reveal answer

Answer

If a secant segment line and a tangent segment are constructed to a circle from a point outside it, the square of the length of the tangent is equal to the product of the distances to the two points on the circle intersected by the secant; in other words,

$NA \cdot NB = (NT)^{2}$

$NA \cdot (NA+AB )= (NT)^{2}$

Substituting:

$20 \cdot (20+t) = 25 ^{2}$

Distributing, then solving for :

$20 \cdot 20+ 20 \cdot t= 6 25$

$20t \div 20 = 225 \div 20$

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