How to find the length of a chord - ISEE Upper Level (grades 9-12) Quantitative Reasoning

Card 0 of 6

Question

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

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

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

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.

Compare your answer with the correct one above

Question

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

(a)

(b) 3

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, .

Compare your answer with the correct one above

Question

Figure NOT drawn to scale.

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

(a)

(b) 7

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 .

Compare your answer with the correct one above

Question

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

Which is the greater quantity?

(a)

(b) 32

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

.

Compare your answer with the correct one above

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

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 .

Compare your answer with the correct one above

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

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:

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

Compare your answer with the correct one above

Tap the card to reveal the answer