Parallelogram A:
Let the diagonal be a1 and a2
a1 = a2+20
Paralelogram B:
Let the diagonal be b1 and b2
b1 = b2+8
Given:
a1+a2 = b1+b2
a2+a2+20=b2+b2+8
2a2+20=2b2+8
a2+10=b2+8
Squaring both sides \((a2+10)^2\) = \((b2+8)^2\)
\(a2^2+20a2+100\) = \(b2^2+16b2+16\)
\(a2^2+20a2-b2^2+16b2 = 84\) - Eq 1
Need:
Area of Parallelogram A - Area of Parallelogram B
Area of A: \(\frac{(a1*a2)}{2}\)
Substituting A1 value in the above equation
\(\frac{a2*(a2+20)}{2}\) = \(\frac{((a2)^2+20a2)}{2} \)
Area of B: \(\frac{(b1*b2)}{2}\)
Substituting B1 value in the above equation
= \(\frac{((b2)^2+8b2)}{2} \)
Area B-A = \(\frac{((a2)^2+20a2)}{2}\) - \(\frac{(b2)^2+8b2}{2} \) - Eq 2
Substituting Eq1 in E2 we get ans as 42
Ans:
D
_________________