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=b×h
Area of the triangle =

1

2

×b×h’
Since the area of the parallelogramis double the area of triangle
∴b×h=2[

1

2

×b×h’]
⇒h=h’
Hence, both altitudes are equal.