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

## Question

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

Find:

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

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

## 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 \(PS\) \(= 8 \,\rm cm\)

**Reasoning:**

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 \(QN\).

**Steps:**

(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)]

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

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