Answer:
Parameter OF Δ JKL = 138 unit
Step-by-step 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
