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5) Find the perimeter of JKL. j 5x-8 8r-35 I ㅏ N 7y-9 2y + 11 -32- K​

5) Find the perimeter of JKL. j 5x-8 8r-35 I ㅏ N 7y-9 2y + 11 -32- K​-example-1
User Anudeep
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1 Answer

7 votes

Answer: 128 units

=======================================================

Step-by-step explanation:

Segments JK, KL, and JL are tangent to the circle.

We can break those segments into smaller pieces and notice we have these 3 pairs of congruent tangents:

  • MJ = NJ
  • KN = KO
  • LM = LO

Let's use the first equation to solve for x.

MJ = NJ

5x-8 = 8x-35

5x-8x = -35+8

-3x = -27

x = -27/(-3)

x = 9

Use the same idea to find y.

KN = KO

7y-9 = 2y+11

7y-2y = 11+9

5y = 20

y = 20/5

y = 4

Now compute the length of segment KO.

KO = 2y+11 = 2*4+11 = 19 units

Subtract this from LK to find LO

LO = LK - KO

LO = 32 - 19

LO = 13

This makes segment LM to be also 13 units long.

Now use the value of x to find sides MJ and NJ

MJ = 5x-8 = 5*9-13 = 32

NJ = 32 as well

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We have these pieces:

  • MJ = NJ = 32
  • KN = KO = 19
  • LM = LO = 13

which give us

  • JK = NJ+KN = 32+19 = 51
  • KL = KO+LO = 19+13 = 32
  • JL = MJ+LM = 32+13 = 45

Therefore, the perimeter of triangle JLK is:

JK+KL+JL = 51+32+45 = 128 units

User HackToHell
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5.7k points