Since each leg of the large trapezoid is bisected, then HJ is the average of the lengths of KL and NM.
HJ = (KL + NM)/2
2(HJ) = KL + nm
2(5x + 2) = 4x + 1 + 27
10x + 4 = 4x + 28
6x = 24
x = 4
KL = 4x + 1 = 4(4) + 1 = 16 + 1 = 17
Answer: KL = 17
There are 10 portions to the circle, and 7 of the 10 are shaded. 7/10 is 70%
A
100+89.50×36
=3,322
b
3,322−2,900
=422
c
2,900÷89.50
=33
