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Ex.11.2 Q3 Perimeter and Area - NCERT Maths Class 7

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Question

Find the missing values:

S. No. Base Height Area of Parallelogram
a. \(20 \,\rm cm\)   \(246 \,\rm cm^2\)
b.   \(15 \,\rm cm\) \(154.5 ^\,\rm cm^2\)
c.   \(8.4 \,\rm cm\) \(48.72 \,\rm cm^2\)
d. \(15.6 \,\rm cm\)   \(16.38\,\rm cm^2\)

 Video Solution
Perimeter And Area
Ex 11.2 | Question 3

Text Solution

What is known?

Area of the parallelogram.

What is unknown?

Base and height of the parallelogram.

Reasoning:

Area of all the parallelograms is given but in some parts of this question base and in some other parts height of the parallelogram is given. Put the values of the given part directly into the equation of area of parallelogram and find the unknown value.

Steps:

(a) Given,

Base of parallelogram \(= 20 \,\rm cm\)

Area of parallelogram \(= 246 \,\rm cm^2\)

Height of parallelogram \(= \,?\)

We know that,

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

\(\begin{align}246 &= 20 \times {\rm{Height}}\\{\rm{Height}} &= \frac{{246}}{{20}}\\{\rm{Height}} &= 12.3\;\rm cm\end{align}\)

(b) Given

Height of parallelogram \(= 15\,\rm cm\)

Area of parallelogram \(= 154.5 \,\rm cm^2\)

Base of parallelogram \(= \,?\)

We know that,

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

\(\begin{align}154.5 &= {\rm{Base}} \times {\rm{15}}\\{\rm{Base}} &= \frac{{154.5}}{{15}}\\{\rm{Base}} &= 10.3\;\rm cm\end{align}\)

(c) Given

Height of parallelogram \(= 8.4\,\rm cm\)

Area of parallelogram \(= 48.72 \,\rm cm^2\)

Base of parallelogram \(= \,?\)

We know that,

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

\(\begin{align}48.72 &= {\rm{Base}} \times {\rm{8}}{\rm{.4}}\\{\rm{Base}} &= \frac{{48.72}}{{8.4}}\\{\rm{Base}} &= 5.8\;\rm cm\end{align}\)

(d) Given

Base of parallelogram \(= 15.6 \,\rm cm\)

Height of parallelogram \(= \,?\)

Area of parallelogram \(= 16.38 \,\rm cm^2\)

We know that,

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

\(\begin{align} 16.38 &= {\rm{Base}} \times {\rm{15}}{\rm{.6}}\\{\rm{Base}} &= \frac{{16.38}}{{15.6}}\\{\rm{Base}} &= 1.05\;\rm cm\end{align}\)

  
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