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ISEE Upper Level Math : Plane Geometry

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6 Diagnostic Tests 244 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #2 : How To Find The Length Of A Chord

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.

Possible Answers:

$6\ cm$

$7\ cm$

$7.5\ cm$

$6.5\ cm$

$8\ cm$

Correct answer:

$6\ cm$

Explanation:

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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Example Question #1 : How To Find The Length Of A Chord

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).

Chord

Possible Answers:

$3.5\ cm$

$4.5\ cm$

$5\ cm$

$4.7\ cm$

$3\ cm$

Correct answer:

$3\ cm$

Explanation:

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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Example Question #1 : Chords

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

Possible Answers:

$9\sqrt{2}$

$18\sqrt{3}$

$18\sqrt{2}$

$9\sqrt{3}$

Correct answer:

$18\sqrt{2}$

Explanation:

The figure below shows $\angle AOB$ , which matches this description, along with its chord $\overline{AB}$ :

Chord

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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Example Question #1 : How To Find The Length Of A Chord

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?

Possible Answers:

$14\sqrt{3}$

$\frac{21\sqrt{2}}{2}$

$14\sqrt{2}$

$7\sqrt{3}$

Correct answer:

Explanation:

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 $\overline{AB}$ :

Chord

By way of the Isoscelese Triangle Theorem, $\Delta AOB$ can be proved equilateral, so AB = OA = 21 .

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Example Question #1 : Chords

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?

Possible Answers:

$\sqrt{3}$

$2 \sqrt{3}$

$\sqrt{2}$

$2 \sqrt{2}$

Correct answer:

$2 \sqrt{3}$

Explanation:

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 $\overline{OM}$ .

Chord

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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Example Question #3 : How To Find The Length Of A Chord

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

Possible Answers:

$10\sqrt{2}$

$10\sqrt{3}$

$20\sqrt{3}$

$20\sqrt{2}$

The correct answer is not among the other choices.

Correct answer:

The correct answer is not among the other choices.

Explanation:

The figure below shows $\angle AOB$ , which matches this description, along with its chord $\overline{AB}$ :

Chord

By way of the Isosceles Triangle Theorem, $\Delta AOB$ can be proved equilateral, so AB = OA = 20 .

This answer is not among the choices given.

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Example Question #2 : Chords

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

Possible Answers:

$12 \sqrt{3}$

$18 \sqrt{2}$

$12 \sqrt{2}$

Correct answer:

$12 \sqrt{3}$

Explanation:

The figure below shows $\angle AOB$ , which matches this description, along with its chord $\overline{AB}$ and triangle bisector $\overline{OM}$ .

Chord

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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Example Question #5 : Chords

Chords

Figure NOT drawn to scale

In the figure above, evaluate .

Possible Answers:

t = 18

t= 12

t= 20

t = 16

Correct answer:

t= 12

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,

40 t = 20 (t+12)

Solving for - distribute:

40 t = 20 (t+12)

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

40 t = 20 t+ 240

Subtract 20t from both sides:

40 t- 20 t = 20 t+ 240 - 20 t

20 t = 240

Divide both sides by 20:

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

t = 12

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Example Question #1 : Chords

Secant

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

Evaluate .

Possible Answers:

$6 \sqrt{10}$

$6\sqrt{6}$

Correct answer:

$6 \sqrt{10}$

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:

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

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Example Question #153 : Plane Geometry

Chords

Figure NOT drawn to scale

In the above diagram, evaluate .

Possible Answers:

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

$t = 5 \sqrt{2}$

$t = 10 \sqrt{2}$

t = 5

Correct answer:

$t = 5 \sqrt{2}$

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:

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

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All ISEE Upper Level Math Resources

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ISEE Upper Level Math : Plane Geometry

Study concepts, example questions & explanations for ISEE Upper Level Math

All ISEE Upper Level Math Resources

Example Questions

Example Question #2 : How To Find The Length Of A Chord

Example Question #1 : How To Find The Length Of A Chord

Example Question #1 : Chords

Example Question #1 : How To Find The Length Of A Chord

Example Question #1 : Chords

Example Question #3 : How To Find The Length Of A Chord

Example Question #2 : Chords

Example Question #5 : Chords

Example Question #1 : Chords

Example Question #153 : Plane Geometry

All ISEE Upper Level Math Resources

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