Chords

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ISEE Upper Level Quantitative Reasoning › Chords

Questions 1 - 10

Chords

Figure NOT drawn to scale

In the above diagram, evaluate .

$t = 4 \sqrt{6 }$

$t = 6 \sqrt{2}$

Explanation

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:

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

Chords

Figure NOT drawn to scale

In the above diagram, evaluate .

$t = 4 \sqrt{6 }$

$t = 6 \sqrt{2}$

Explanation

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:

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

Chords

Figure NOT drawn to scale

In the figure above, evaluate .

Explanation

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:

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

Chords

Figure NOT drawn to scale

In the figure above, evaluate .

Explanation

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:

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

Chords

Figure NOT drawn to scale

In the above diagram, evaluate .

$t = 5 \sqrt{2}$

$t = 6 \frac{2}{3 }$

$t = 10 \sqrt{2}$

Explanation

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:

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

Chords

Figure NOT drawn to scale

In the above diagram, evaluate .

$t = 5 \sqrt{2}$

$t = 6 \frac{2}{3 }$

$t = 10 \sqrt{2}$

Explanation

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:

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

Secant

Figure NOT drawn to scale

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

Evaluate .

Explanation

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 :

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

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

Secant

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

Evaluate .

$6 \sqrt{10}$

$6\sqrt{6}$

Explanation

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:

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

Secant

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

Evaluate .

Explanation

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

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