Exercise 9.1 (Revised) - Chapter 11 - Perimeter & Area - Ncert Solutions class 7 - Maths

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Chapter 9 - Perimeter & Area | NCERT Solutions for Class 7 Maths

Ex 9.1 Question 1.

Find the area of each of the following parallelograms:

Answer:

We know that the area of parallelogram = base $x$ height
(a) Here base $=7 \mathrm{~cm}$ and height $=4 \mathrm{~cm}$
$\therefore$ Area of parallelogram $=7 \times 4=28 \mathrm{~cm}^2$
(b) Here base $=5 \mathrm{~cm}$ and height $=3 \mathrm{~cm}$
$\therefore$ Area of parallelogram $=5 \times 3=15 \mathrm{~cm}^2$
(c) Here base $=2.5 \mathrm{~cm}$ and height $=3.5 \mathrm{~cm}$
$\therefore$ Area of parallelogram $=2.5 \times 3.5=8.75 \mathrm{~cm}^2$
(d) Here base $=5 \mathrm{~cm}$ and height $=4.8 \mathrm{~cm}$

$\therefore$ Area of parallelogram $=5 \times 4.8=24 \mathrm{~cm}^2$
(e) Here base $=2 \mathrm{~cm}$ and height $=4.4 \mathrm{~cm}$
$\therefore$ Area of parallelogram $=2 \times 4.4=8.8 \mathrm{~cm}^2$

Ex 9.1 Question 2.

Find the area of each of the following triangles:

Answer:

We know that the area of triangle $=\frac{1}{2} \mathrm{x}$ base $\mathrm{x}$ height
(a) Here, base $=4 \mathrm{~cm}$ and height $=3 \mathrm{~cm}$
$\therefore$ Area of triangle $=\frac{1}{2} \times 4 \times 3=6 \mathrm{~cm}^2$
(b) Here, base $=5 \mathrm{~cm}$ and height $=3.2 \mathrm{~cm}$
$\therefore$ Area of triangle $=\frac{1}{2} \times 5 \times 3.2=8 \mathrm{~cm}^2$
(c) Here, base $=3 \mathrm{~cm}$ and height $=4 \mathrm{~cm}$
$\therefore$ Area of triangle $=\frac{1}{2} \times 3 \times 4=6 \mathrm{~cm}^2$

(d) Here, base $=3 \mathrm{~cm}$ and height $=2 \mathrm{~cm}$
$\therefore$ Area of triangle $=\frac{1}{2} \times 3 \times 2=3 \mathrm{~cm}^2$

Ex 9.1 Question 3.

Find the missing values:

Answer:

We know that the area of parallelogram = base $x$ height
(a) Here, base $=20 \mathrm{~cm}$ and area $=246 \mathrm{~cm}^2$
$\therefore$ Area of parallelogram $=$ base $\mathrm{x}$ height
$\Rightarrow 246=20 \mathrm{x}$ height $\Rightarrow$ height $=\frac{246}{20}=12.3 \mathrm{~cm}$
(b) Here, height $=15 \mathrm{~cm}$ and area $=154.5 \mathrm{~cm}^2$
$\therefore$ Area of parallelogram $=$ base $\mathrm{x}$ height
$\Rightarrow 154.5=$ base $\mathrm{x} 15 \Rightarrow$ base $=\frac{154.5}{15}=10.3 \mathrm{~cm}$
(c) Here, height $=8.4 \mathrm{~cm}$ and area $=48.72 \mathrm{~cm}^2$
$\therefore$ Area of parallelogram $=$ base $\mathrm{x}$ height
$\Rightarrow 48.72=$ base $x 8.4 \Rightarrow$ base $=\frac{48.72}{8.4}=5.8 \mathrm{~cm}$
(d) Here, base $=15.6 \mathrm{~cm}$ and area $=16.38 \mathrm{~cm}^2$
$\therefore$ Area of parallelogram $=$ base $\mathrm{x}$ height

$
\Rightarrow 16.38=15.6 \times \text { height } \Rightarrow \text { height }=\frac{16.38}{15.6}=1.05 \mathrm{~cm}
$

Thus, the missing values are:

Ex 9.1 Question 4.

Find the missing values:

Answer

We know that the area of triangle $=\frac{1}{2} \mathrm{x}$ base $\mathrm{x}$ height
In first row, base $=15 \mathrm{~cm}$ and area $=87 \mathrm{~cm}^2$
$
\therefore 87=\frac{1}{2} \times 15 \times \text { height } \Rightarrow \text { height }=\frac{87 \times 2}{15}=11.6 \mathrm{~cm}
$

In second row, height $=31.4 \mathrm{~mm}$ and area $=1256 \mathrm{~mm}^2$
$
\therefore 1256=\frac{1}{2} \mathrm{x} \text { base } \mathrm{x} 31.4 \Rightarrow \text { base }=\frac{1256 \times 2}{31.4}=80 \mathrm{~mm}
$

In third row, base $=22 \mathrm{~cm}$ and area $=170.5 \mathrm{~cm}^2$
$
\therefore 170.5=\frac{1}{2} \times 22 \times \text { height } \Rightarrow \text { height }=\frac{170.5 \times 2}{22}=15.5 \mathrm{~cm}
$

Thus, the missing values are:

Ex 9.1 Question 5.

PQRS is a parallelogram. QM is the height from Q to SR and QN is the height from $Q$ to $P S$. If $S R=12 \mathrm{~cm}$ and $\mathrm{QM}=7.6 \mathrm{~cm}$. Find:
(a) the area of the parallelogram PQRS
(b) $\mathrm{QN}$, if $\mathrm{PS}=8 \mathrm{~cm}$

Answer:

Given: $\mathrm{SR}=12 \mathrm{~cm}, \mathrm{QM}=7.6 \mathrm{~cm}, \mathrm{PS}=8 \mathrm{~cm}$.
(a) Area of parallelogram $=$ base $x$ height $=12 \times 7.6=91.2 \mathrm{~cm}^2$
(b) Area of parallelogram $=$ base $x$ height
$
\Rightarrow 91.2=8 \times \mathrm{QN} \Rightarrow \mathrm{QN}=\frac{91.2}{8}=11.4 \mathrm{~cm}
$

Ex 9.1 Question 6.

DL and BM are the heights on sides $\mathrm{AB}$ and $\mathrm{AD}$ respectively of parallelogram $\mathrm{ABCD}$. If the area of the parallelogram is $1470 \mathrm{~cm}^2, \mathrm{AB}=35 \mathrm{~cm}$ and $\mathrm{AD}=49 \mathrm{~cm}$, find the length of BM and DL.

Answer:

Given: Area of parallelogram $=1470 \mathrm{~cm}^2$
Base $(\mathrm{AB})=35 \mathrm{~cm}$ and base $(\mathrm{AD})=49 \mathrm{~cm}$
Since Area of parallelogram = base $\mathrm{x}$ height

$
\begin{aligned}
& \Rightarrow 1470=35 \times \mathrm{DL} \Rightarrow \mathrm{DL}=\frac{1470}{35}=42 \mathrm{~cm} \\
& \text { Again, Area of parallelogram }=\text { base } \times \text { height } \\
& \Rightarrow 1470=49 \times \mathrm{BM} \Rightarrow \mathrm{BM}=\frac{1470}{49}=30 \mathrm{~cm}
\end{aligned}
$

Thus, the lengths of DL and BM are $42 \mathrm{~cm}$ and $30 \mathrm{~cm}$ respectively.

Ex 9.1 Question 7.

$\Delta \mathrm{ABC}$ is right angled at $\mathrm{A}$. $\mathrm{AD}$ is perpendicular to $\mathrm{BC}$. If $\mathrm{AB}=5 \mathrm{~cm}, \mathrm{BC}=13 \mathrm{~cm}$ and $\mathrm{AC}=12 \mathrm{~cm}$, find the area of $\Delta \mathrm{ABC}$. Also, find the length of $\mathrm{AD}$.

Answer:

In right angled triangle $\mathrm{BAC}, \mathrm{AB}=5 \mathrm{~cm}$ and $\mathrm{AC}=12 \mathrm{~cm}$
Area of triangle $=\frac{1}{2} \times$ base $\times$ height $=\frac{1}{2} \times$ AB $\times$ AC $=\frac{1}{2} \times 5 \times 12=30 \mathrm{~cm}^2$
Now, in $\triangle \mathrm{ABC}$,
Area of triangle $\mathrm{ABC}=\frac{1}{2} \times \mathrm{BC} \times \mathrm{AD}$
$
\Rightarrow 30=\frac{1}{2} \times 13 \times \mathrm{AD} \Rightarrow \mathrm{AD}=\frac{30 \times 2}{13}=\frac{60}{13} \mathrm{~cm}
$

Ex 9.1 Question 8.

$\Delta \mathrm{ABC}$ is isosceles with $\mathrm{AB}=\mathrm{AC}=7.5 \mathrm{~cm}$ and $\mathrm{BC}=9 \mathrm{~cm}$. The height $\mathrm{AD}$ from $\mathrm{A}$ to $\mathrm{BC}$, is $6 \mathrm{~cm}$. Find the area of $\triangle \mathrm{ABC}$. What will be the height from $\mathrm{C}$ to $\mathrm{AB}$ i.e., $\mathrm{CE}$ ?

Answer:

In $\Delta \mathrm{ABC}, \mathrm{AD}=6 \mathrm{~cm}$ and $\mathrm{BC}=9 \mathrm{~cm}$
Area of triangle $=\frac{1}{2} \times$ base $\times$ height $=\frac{1}{2} \times B C \times A D=\frac{1}{2} \times 9 \times 6=27 \mathrm{~cm}^2$
Again, Area of triangle $=\frac{1}{2} \times$ base $\times$ height $=\frac{1}{2} \times$ AB x CE
$
\Rightarrow 27=\frac{1}{2} \times 7.5 \times \mathrm{CE} \Rightarrow \mathrm{CE}=\frac{27 \times 2}{7.5}=7.2 \mathrm{~cm}
$

Thus, height from $\mathrm{C}$ to $\mathrm{AB}$ i.e., $\mathrm{CE}$ is $7.2 \mathrm{~cm}$.