SSC CPO SI and ASI Model Paper 7

(1) Given that the base of theparallelogram and the triangle arethe same, so let the base be 'b'.
Altitude of the parallelogram = h
Altitude of the triangle = h'
Area of parallelogram$=\mathit{b}\times \mathit{h}$
Area of the triangle $=\frac{1}{2}\times \mathit{b}\times \mathit{h}’$
Since the area of the parallelogramis double the area of triangle
$\therefore \mathit{b}\times \mathit{h}=2[\frac{1}{2}\times \mathit{b}\times \mathit{h}’]$
⇒h=h’
Hence, both altitudes are equal.