# SEGMENT LENGTHS IN CIRCLES WORKSHEET

## About "Segment Lengths in Circles Worksheet"

Segment Lengths in Circles Worksheet :

Worksheet given in this section is much useful to the students who would like to practice problems on segment lengths in circles.

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## Segment Lengths in Circles Worksheet - Problems

Problem 1 :

In the diagram shown below, prove the following.

EA ⋅ EB  =  EC  ED

Problem 2 :

Chords ST and PQ intersect inside the circle.  Find the value of x.

Problem 3 :

Find the value of x in the diagram shown below.

Problem 4 :

Find the value of x in the diagram shown below.

Problem 5 :

You are standing at point C,  about 8 feet from a circular aquarium tank.  The distance from you to a point of tangency on the tank is about 20 feet.  Estimate the radius of the tank.

## Segment Lengths in Circles Worksheet - Solutions

Problem 1 :

In the diagram shown below, prove the following.

EA ⋅ EB  =  EC  ED

Solution :

We can use similar triangles to prove the Theorem.

Given : AB, CD are chords that intersect at E.

To Prove : EA · EB =  EC · ED

Draw DB and AC in the above diagram.

Because mC and mB intercept the same are C    B.  Likewise  A    D.

By the AA Similarity Postulate.  ∆AEC  ∆DEB.

So, the lengths of corresponding sides are proportional.

EA/ED  =  EC/EB

EA ⋅ EB  =  EC  ED

Problem 2 :

Chords ST and PQ intersect inside the circle.  Find the value of x.

Solution :

Using Theorem, we have

RQ · RP  =  RS · RT

Substitute.

9 · x  =  3 · 6

9x  =  18

Divide each side by 9.

9x/9  =  18/9

x  =  2

Problem 3 :

Find the value of x in the diagram shown below.

Solution :

Using Theorem, we have

RP · RQ  =  RS · RT

Substitute.

9 · (11 + 9)  =  10 · (x + 10)

Simplify.

180  =  10x + 100

Subtract 100 from each side.

80  =  10x

Divide each side by 10.

80/10  =  10x/10

8  =  x

Problem 4 :

Find the value of x in the diagram shown below.

Solution :

Using Theorem, we have

(BA)2  =  BC · BD

Substitute.

62  =  x · (x + 5)

Simplify.

36  =  x2 + 5x

Subtract 36 from each side.

0  =  x2 + 5x - 36

or

x2 + 5x - 36  =  0

Factor.

(x + 9)(x - 4)  =  0

x + 9  =  0        or        x - 4  =  0

x  =  - 9        or        x  =  4

We can use only positive value for x. because lengths cannot be negative.

So, we have

x  =  4

Problem 5 :

You are standing at point C,  about 8 feet from a circular aquarium tank.  The distance from you to a point of tangency on the tank is about 20 feet.  Estimate the radius of the tank.

Solution :

Using Theorem, we have

(CB)2  =  CE · CD

Substitute.

202    8 · (2r + 8)

Simplify.

400    16r + 64

Subtract 64 from each side.

336    16r

Divide each side by 16.

336/16    16r/16

21    r

Hence, the radius of the tank is about 21 feet.

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