Exercise 2.2 - Chapter 2 Measurements 8th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Question 1.
Find the perimeter and area of the figures given below. $\left(\pi=\frac{22}{7}\right)$

$\begin{aligned}
&\text { Length of the arc of the semicircle }=\frac{1}{2} \times 2 \pi \mathrm{r} \text { units } \\
&=\frac{22}{7} \times \frac{7}{2} \mathrm{~m}=11 \mathrm{~m} \\
&\therefore \text { Perimeter }=\text { Sum of all lengths of sides that form the closed boundary } \\
&\mathrm{P}=11+10+7+10 \mathrm{~m} \\
&\text { Perimeter }=38 \mathrm{~m} \\
&\text { Areu }-\text { Area of the rectangle }-\text { Area of semicircle } \\
&=(1 \times \mathrm{b})-\frac{1}{2} \pi^{2} \text { squ units }^{2} \\
&=(10 \times 7)-\frac{1}{2} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \\
&=70-\frac{11 \times 7}{2 \times 2}=\frac{280-77}{4}=\frac{203}{4}
\end{aligned}$

$=50.75 \mathrm{~m}^{2} \text { (approx) }$
Area of the figure $=50.75 \mathrm{~m}^{2}$ approx.

Perimeter = sum of outside lengths
Length of the arc of quadrant circle $=\frac{1}{4} \times 2 \pi \mathrm{r}$ units
$=\frac{1}{2} \times \frac{22}{7} \times 3.5 \mathrm{~cm}$
$=11 \times 0.5 \mathrm{~cm}=5.5 \mathrm{~cm}$
$\therefore$ Length of arc of 2 sectors $2 \times 5.5 \mathrm{~cm}$
$=11 \mathrm{~cm}$
$\therefore$ Perimeter $\mathrm{P}=11+6+3.5+6+35 \mathrm{~cm}$
$\mathrm{P}=30 \mathrm{~cm}$
Area Area of 2 quadrant circle + Area of a rectangle.
$=2 \times \frac{1}{4} \pi \mathrm{r}^{2}+\mathrm{lb}$ sq. units
$=\left(\frac{1}{2} \times \frac{22}{7} \times 3.5 \times 3.5\right)+(6 \times 3.5) \mathrm{cm}^{2}$
$=(11 \times 3.5 \times 0.5)+21 \mathrm{~cm}^{2}$
$=(19.25+21) \mathrm{cm}^{2}=40.25 \mathrm{~cm}^{2}$
$\therefore$ Area $=40.25 \mathrm{~cm}^{2}$ approx

Question 2.
Find the area of the shaded part in the following figures. $(\pi=3.14)$

Answer:
Area of the shaded part = Area of 4 quadrant circles of radius $\frac{10}{2} \mathrm{~cm}$ $=4 \times \frac{1}{4} \times \pi \mathrm{r}^{2}=3.14 \times \frac{10}{2} \times \frac{10}{2} \mathrm{~cm}^{2}$ $=\frac{314}{4} \mathrm{~cm}^{2}=78.5 \mathrm{~cm}^{2}$
Area of the shaded part $=78.5 \mathrm{~cm}^{2}$
Area of the unshaded part = Area of the square - Area of shaded part $=\mathrm{a}^{2}-78.5 \mathrm{~cm}^{2}=(10 \times 10)-78.5 \mathrm{~cm}^{2}$
$=100-78.5 \mathrm{~cm}^{2}=21.5 \mathrm{~cm}^{2}$
Area of the unshaded part $=21.5 \mathrm{~cm}^{2}$ (approximately)

Answer:
Area of the shaded part = Area of semicircle $-$ Area of the triangle
$\begin{aligned}
&=\left(\frac{1}{2} \pi r^{2}\right)-\left(\frac{1}{2} b h\right) \mathrm{cm}^{2} \\
&=\frac{1}{2} \times 3.14 \times 7 \times 7-\frac{1}{2} \times 14 \times 7 \mathrm{~cm}^{2} \\
&=\frac{153.86}{2}-49 \mathrm{~cm}^{2}=76.93-49 \mathrm{~cm}^{2} \\
&=27.3 \mathrm{~cm}^{2}
\end{aligned}$
$\therefore$ Area of the shaded part $=27.93 \mathrm{~cm}^{2}$ (approximately)

Question $3 .$
Find the area of the combined figure given which is got by joining of two parallelograms

Answer:
Area of the figure $=$ Area of 2 parallelograms with base $8 \mathrm{~cm}$ and height $3 \mathrm{~cm}$ $=2 \times$ (bh) sq. units
$=2 \times 8 \times 3 \mathrm{~cm}^{2}=48 \mathrm{~cm}^{2}$
$\therefore$ Area of the given figure $=48 \mathrm{~cm}^{2}$
 

Question $4 .$
Find the area of the combined figure given, formed by joining a semicircle of diameter $6 \mathrm{~cm}$ with a triangle of base $6 \mathrm{~cm}$ and height $9 \mathrm{~cm} .(\pi=3.14)$

Answer:
Area of the figure $=$ Area of the semicircle of radius $3 \mathrm{~cm}+2$ (Area of triangle with $\mathrm{b}=9 \mathrm{~cm}$ and $\mathrm{h}=3 \mathrm{~cm})$
$=\left(\frac{1}{2} \pi r^{2}\right)+\left(2 \times \frac{1}{2} b h\right)$ sq. units
$\begin{aligned}
&=\frac{1}{2} \times 3.14 \times 3 \times 3+\left(2 \times \frac{1}{2} \times 9 \times 3\right) \mathrm{cm}^{2} \\
&=\frac{28.26}{2}+27 \mathrm{~cm}^{2}=14.13+27 \mathrm{~cm}^{2}=41.13 \mathrm{~cm}^{2}
\end{aligned}$
$\therefore$ Area of the figure $=41.13 \mathrm{~cm}^{2}$ (approximately)
 

Question $5 .$
The door mat which is in a hexagonal shape has the following measures as given in the figure. Find its area.

Answer:
Area of the doormat $=$ Area of 2 trapezium
Height of the trapezium $\mathrm{h}=\frac{70}{2} \mathrm{~cm}: \mathrm{a}=90 \mathrm{~cm} ; \mathrm{b}=70 \mathrm{~cm}$
$\therefore$ Area of the trapezium $=\frac{1}{2} \mathrm{~h}(\mathrm{a}+\mathrm{b})$ sq. units
Area of the door mat $=2 \times \frac{1}{2} \times 35(90+70) \mathrm{cm}^{2}$
$=35 \times 160 \mathrm{~cm}^{2}=5600 \mathrm{~cm}^{2}$
$\therefore$ Area of the door mat $-5600 \mathrm{~cm}^{2}$
 

Question 6.
A rocket drawing has the measures as given in the figure. Find its area.

Answer:
Area - Area of a rectangle $+$ Area of a triangle $+$ Area of a trapezium
For rectangle length $1=120-20-20 \mathrm{~cm}=80 \mathrm{~cm}$
Breadth $\mathrm{b}=30 \mathrm{~cm}$
For the triangle base $=30 \mathrm{~cm}$
Height $=20 \mathrm{~cm}$
For the trapezium height $h=20 \mathrm{~cm}$
Parallel sided $\mathrm{a}=50 \mathrm{~cm}$
$\mathrm{b}=30 \mathrm{~cm}$
$\therefore$ Area of the figure $(1 \times \mathrm{b})+\left(\frac{1}{2} \times\right.$ base $\times$ height $)+\frac{1}{2} \times \mathrm{h} \times(\mathrm{a}+\mathrm{b})$ sq. units $=(80 \times 30)+\left(\frac{1}{2} \times 30 \times 20\right)+\frac{1}{2} \times 20 \times(50+30) \mathrm{cm}^{2}$ $=2400+300+800 \mathrm{~cm}^{2}=3500 \mathrm{~cm}^{2}$
Area of the figure $=3500 \mathrm{~cm}^{2}$

Question $7 .$
Find the area of the irregular polygon shaped fields given below.

Answer:
Area of the field $=$ Area of trapezium $\mathrm{FBCH}+$ Area of $\triangle \mathrm{DHC}+$ Area of $\triangle \mathrm{EGD}+$
Area of $\triangle \mathrm{EGA}+$ Area of $\triangle \mathrm{BFA}$
Area of the triangle $=\frac{1}{2}$ bhsq.units
Area of the trapezium $=\frac{1}{2} \times \mathrm{h} \times(\mathrm{a}+\mathrm{b})$ sq.units
Area of the trapezium $\mathrm{FBCH}=\frac{1}{2} \times(10+8) \times(8+3) \mathrm{m}^{2}=9 \times 11=99 \mathrm{~m}^{2} \ldots . .$ (1)
Area of the $\Delta \mathrm{DHC}=\frac{1}{2} \times 8 \times 5 \mathrm{~m}^{2}=20 \mathrm{~m}^{2} \ldots . .(2)$
Area of $\Delta \mathrm{EGD}=\frac{1}{2} \times 8 \times 15 \mathrm{~m}^{2}=60 \mathrm{~m}^{2} \ldots \ldots . .(3)$
Area of $\triangle \mathrm{EGA}=\frac{1}{2} \times 8 \times(8+6) \mathrm{m}^{2}=4 \times 14 \mathrm{~m}^{2}$
$=56 \mathrm{~m}^{2}$
Area of $\triangle B F A=\frac{1}{2} \times 3 \times 6 \mathrm{~m}^{2}=9 \mathrm{~m}^{2}$
$\therefore$ Area of the field $=99+20+60+56+9 \mathrm{~m}^{2}$
$=244 \mathrm{~m}^{2}$
Area of the field $=244 \mathrm{~m}^{2}$