SSC CPO SI and ASI Model Paper 1

(1) According to the given condition,
The base of the isosceles triangle =Base of the parallelogram = 34 cmAnd the isosceles triangle isconstructed between the parallelsides of the parallelogram.
Therefore, the vertical distancebetween the parallel sides is thealtitude of that triangle.
Hence, the area of the triangle$=\frac{1}{2}\times$area of parallelogram
Also, area of triangle
$=\frac{1}{2}\times$ base × altitude$\Rightarrow \frac{1}{2}\times 34\times$altitude =544
⇒altitude $=\frac{544}{17}$altitude 32 cm