In the verge of coronavirus pandemic, we are providing FREE access to our entire Online Curriculum to ensure Learning Doesn't STOP!

Ex.11.2 Q5 Perimeter and Area - NCERT Maths Class 7

Go back to  'Ex.11.2'


\(PQRS\) is a parallelogram (Fig 11.23). \(QM\) is the height from \(\, Q \, \) to \(SR\) and \(QN\) is the height from\(\, Q \, \) to \(PS\). If \(SR\) \(= 12\,\rm cm\) and \(QM\) \(= 7.6 \,\rm cm\).


(a) the area of the parallelogram \(PQRS\)

(b) \(QN\), if \(PS\) \(= 8 \,\rm cm\)

 Video Solution
Perimeter And Area
Ex 11.2 | Question 5

Text Solution

What is known?

\(PQRS\) is a parallelogram. \(QM\) is the height from \(\, Q \, \) to \(SR\) and \(QN\) is the height from \(\, Q \, \) to \(PS\). Also, \(SR\) \(= 12\,\rm cm\) and \(QM\) \(= 7.6 \,\rm cm\).

What is unknown?

(a) The area of the parallelogram \(PQRS\) and (b) \(QN\), if \(PS\) \(= 8 \,\rm cm\)


First, using the given information (base \(SR\) = 12cm and perpendicular height \(QM\) = 7.6cm), find out the area of parallelogram \(PQRS\). Now, area of \(PQRS\) and base \(PS\) is known. Again, by using the formula of area of parallelogram, find out height \(QN\).


(a) Given, \(SR\) \(= 12\,\rm cm\) and

\(QM\) \(= 7.6 \,\rm cm\).

\(\begin{align} & \text{Area of parallelogram} \; PQRS \\ &= {\rm{Base}} \times {\rm{Height}} \end{align} \)

\(\begin{align}&= SR \times QM \\&= 12 \times 7.6\\&= 91.2 \, \rm cm^2\end{align}\)

(b) Given, Base \(PS\) \(= 8 \,\rm cm\)

Area of the parallelogram \(= 91.2 \rm\, cm^2\) [calculated in part (a)]

\(\begin{align} & \text{Area of parallelogram}\; PQRS \\ &= {\rm{Base}} \times {\rm{Height}} \end{align} \)

\(\begin{align} 91.2 &= 8 \times QN \\{{QN}} &= \frac{91.2}{8}\\{{QN}} &= 11.4\;\rm cm\end{align}\)

Learn from the best math teachers and top your exams

  • Live one on one classroom and doubt clearing
  • Practice worksheets in and after class for conceptual clarity
  • Personalized curriculum to keep up with school