Question

16. Find the perimeter of AJKL.

J
8x - 35
N
5x - 8
77-9
K
Р
M M
2y + 11
o
32
L

Answer 1

Explanation:

g

Answer 2

Parameter OF Δ JKL = 138 unit

Explanation:

Given:

JN = 8x - 35

NK = 7y - 9

OK = 2y + 11

OL = 32 - [2y + 11]

JM = 5x - 8

Find:

Parameter OF Δ JKL

Computation:

We know that tangent from same points are always equal

So,

ML = OL

JM = JN

5x - 8 = 8x - 35

8x - 5x = 35 - 8

3x = 27

x = 9

So,

JN = 8x - 35

JN = 8(9) - 35

JN = 37 unit

JM = 5x - 8

JM = 5(9) - 8

JM = 37 unit

NK = OK

7y - 9 = 2y + 11

5y = 20

y = 4

NK = 7y - 9

NK = 7(4) - 9

NK = 19 unit

OK = 2y + 11

OK = 2(4) + 11

Ok = 19 unit

OL = 32 - [2(y) + 11]

OL = 32 - [2(4) + 11]

OL = 13 unit

So,

LM = OL = 13 unit

Parameter OF Δ JKL = JN + NK + OK + OL + LM + JM

Parameter OF Δ JKL = 37 + 19 + 19 + 13 + 13 + 37

Parameter OF Δ JKL = 138 unit