Exercise 2.3 - Chapter 2 Measurements Term 2 7th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $2.3$
Question $1 .$
Find the area of a circular pathway whose outer radius is $32 \mathrm{~cm}$ and inner radius is $18 \mathrm{~cm}$.
Solution:
Radius of the outer circle $R=32 \mathrm{~cm}$
Radius of the inner circle $\mathrm{r}=18 \mathrm{~cm}$
Area of the circular pathway $=\pi\left(\mathrm{R}^{2}-\mathrm{r}^{2}\right)$ sq. units $=\frac{22}{7}\left(32^{2}-18^{2}\right) \mathrm{cm}^{2}$
$=\frac{22}{7} \times(32+18) \times(32-18) \mathrm{cm}^{2}$
$=\frac{22}{7} \times 50 \times 14 \mathrm{~cm}^{2}=2,200 \mathrm{~cm}^{2}$
Area of the circular pathway $=2,200 \mathrm{~cm}^{2}$

Question $2 .$
There is a circular lawn of radius $28 \mathrm{~m}$. A path of $7 \mathrm{~m}$ width is laid around the lawn. What will be the area of the path?
Solution:
Radius of the circular lawn $\mathrm{r}=28 \mathrm{~m}$
Radius of the lawn with path $=28+7 \mathrm{~m}=35 \mathrm{~m}$
Area of the circular path $=\pi\left(\mathrm{R}^{2}-\mathrm{r}^{2}\right)$ sq. units
Area of the path $=\frac{22}{7}\left(35^{2}-28^{2}\right) \mathrm{m}^{2}=\frac{22}{7} \times(35+28)(35-28) \mathrm{m}^{2}$
$=\frac{22}{7} \times 63 \times 7 \mathrm{~m}^{2}=1386 \mathrm{~m}^{2}$
Area of the path $=1386 \mathrm{~m}^{2}$

Question 3 .
A circular carpet whose radius is $106 \mathrm{~cm}$ is laid on a circular hall of radius $120 \mathrm{~cm}$. Find the area of the hall uncovered by the carpet.

Solution:
Radius of the circular hall $\mathrm{R}=120 \mathrm{~cm}$
Radius of the circular carpet $\mathrm{r}=106 \mathrm{~cm}$
Area of the hall uncovered = Area of the hall - Area of the carpet
$=\pi\left(\mathrm{R}^{2}-\mathrm{r}^{2}\right) \mathrm{cm}^{2}$
$=\frac{22}{7} \times\left(120^{2}-106^{2}\right) \mathrm{cm}^{2}$
$=\frac{22}{7} \times(120+106) \times(120-106) \mathrm{cm}^{2}$
$=\frac{22}{7} \times 226 \times 14 \mathrm{~cm}^{2}=9,944 \mathrm{~cm}^{2}$
Area of the hall uncovered $=9,944 \mathrm{~cm}^{2}$
 

Question $4 .$
A school ground is in the shape of a circle with radius $103 \mathrm{~m}$. Four tracks each of $3 \mathrm{~m}$ wide has to be constructed inside the ground for the purpose of track events. Find the cost of constructing the track at the rate of ₹ 50 per sq.m.

Solution:
Radius of the ground $\mathrm{R}=103 \mathrm{~m}$
Width of a track W $=3 \mathrm{~m}$
Width of 4 tracks $=4 \times 3=12 \mathrm{~m}$
Radius of the ground without track
$\mathrm{r}=(103-12) \mathrm{m}$
$\mathrm{r}=91 \mathrm{~m}$
Area of 4 tracks $=$ Area of the ground
- Area of the ground without crack
$=\pi \mathrm{R}^{2}-\pi \mathrm{r}^{2}$ sq.units
$=\pi\left(\mathrm{R}^{2}-\mathrm{r}^{2}\right)$ sq.units
$=\frac{22}{7}\left[103^{2}-91^{2}\right]$
$=\frac{22}{7}[103+91][103-91] \mathrm{m}^{2}$
$=\frac{22}{7} \times 194 \times 12=\frac{51216}{7}=7316.57 \mathrm{~m}^{2}$
$\therefore$ Area of 4 tracks $=7316.57 \mathrm{~m}^{2}$
Cost of constructing $7316.57 \mathrm{~m}^{2}=₹ 50$
$\therefore$ Cost of constructing $7316.57 \mathrm{~m}^{2}=₹ 50 \times 7316.57=₹ 3,65,828,57$
Cost of constructing the track ₹ $3,65,828,57$

Question $5 .$
The figure shown is the aerial view of the pathway. Find the area of the pathway.

Solution:
Area of the rectangle $=($ Lenght $\times$ Breadth $)$ sq. units
Area of the outer rectangle $=(\mathrm{L} \times \mathrm{B})$ sq. units
Length of the outer rectangle $\mathrm{L}=80 \mathrm{~m}$
Breadth of the outer rectangle $B=50 \mathrm{~m}$
Length of the inner rectangle $\mathrm{l}=70 \mathrm{~m}$
Breadth of the inner rectangle $b=40 \mathrm{~m}$
Area of the outer rectangle $=80 \times 50 \mathrm{~m}^{2}=4000 \mathrm{~m}^{2}$
Area of the inner rectangle $=1 \times$ b sq. unit $=70 \times 40 \mathrm{~m}^{2}=2800 \mathrm{~m}^{2}$
Area of the pathway $=$ Area of the outer rectangle
- Area of the inner rectangle
$=4000-2800 \mathrm{~m}^{2}=1200 \mathrm{~m}^{2}$
Area of the pathway $=1200 \mathrm{~m}^{2}$

Question $6 .$
A rectangular garden has dimensions $11 \mathrm{~m} \times 8 \mathrm{~m}$. A path of $2 \mathrm{~m}$ wide has to be constructed along its sides. Find the area of the path.
Solution:

Area of the rectangular garden $\mathrm{L} \times \mathrm{B}=11 \mathrm{~m} \times 8 \mathrm{~m}=88 \mathrm{~m}^{2}$
Length of the inner rectangle $L=L-2 W=11-2(2)=11-4=7 \mathrm{~m}$ Breadth of the inner rectangle $b=B-2 W=8-2(2)=8-4=4 \mathrm{~m}$
Area of the inner rectangle $=1 \times$ b sq. units $=7 \times 4 \mathrm{~m}^{2}=28 \mathrm{~m}^{2}$
Area of the path = Area of the outer rectangular garden
- Area of the inner rectangle
$=88 \mathrm{~m}^{2}-28 \mathrm{~m}^{2}=60 \mathrm{~m}^{2}$
Area $\Omega f$ the path $=60 \mathrm{~m}^{2}$

Question $7 .$
A picture is painted on a ceiling of a marriage hall whose length and breadth are $18 \mathrm{~m}$ and $7 \mathrm{~m}$ respectively. There is a border of $10 \mathrm{~cm}$ along each of its sides. Find the area of the border.
Solution:
Length of the ceiling $\mathrm{L}=18 \mathrm{~m}$
Breadth of the ceiling $\mathrm{B}=7 \mathrm{~m}$
Area of the ceiling $=\mathrm{L} \times \mathrm{B}$ sq. units $=18 \times 7 \mathrm{~m}^{2}=126 \mathrm{~m}^{2}$
Width of the boarder $W=10 \mathrm{~cm}=\frac{10}{100} \mathrm{~m}=0.1 \mathrm{~m}$
Length of the ceiling without border $=\mathrm{L}-2 \mathrm{~W}=18-2(0.1) \mathrm{m}$
$=18-0.2 \mathrm{~m}=17.8 \mathrm{~m}$
Breadth of the ceiling without border $=\mathrm{B}-2 \mathrm{~W}=7-2(0.1) \mathrm{m}$
$=7-0.2 \mathrm{~m}=6.8 \mathrm{~m}$
Area of the ceiling without border $=1 \times b$ sq.units
$=17.8 \times 6.8 \mathrm{~m}^{2}=121.04 \mathrm{~m}^{2}$
$\therefore$ Area of the border $=$ Area of the ceiling
- Area of the ceiling without border
$=126-121.04 \mathrm{~m}^{2}=4.96 \mathrm{~m}^{2}$
Area of the border $=4.96 \mathrm{~m}^{2}$

Question $8 .$
A canal of width $1 \mathrm{~m}$ is constructed all along inside the field which is $24 \mathrm{~m}$ long and $15 \mathrm{~m}$ wide. Find (i) the area of the canal (ii) the cost of constructing the canal at the rate of $₹ 12$ per sq.m.
Solution:

Length of the field $\mathrm{L}=24 \mathrm{~m}$
Width (Breadth) of the field $\mathrm{B}=15 \mathrm{~m}$
(i) Area of the field $=\mathrm{L} \times \mathrm{B}$ sq. units $=24 \times 15 \mathrm{~m}^{2}=360 \mathrm{~m}^{2}$
(ii) Width of the canal $(\mathrm{W})=1 \mathrm{~m}$
Length of the field without canal $(1)=L-2(W)=24-2(1) \mathrm{m}$ $=24-2 \mathrm{~m}=22 \mathrm{~m}$
Width of the field without canal $(b)=B-2 W=15-2(1) \mathrm{m}$ $=15-2 \mathrm{~m}=13 \mathrm{~m}$
Area of the field without canal $=1 \times b$ sq. units $=22 \times 13 \mathrm{~m}^{2}=286 \mathrm{~m}^{2}$
Area of the canal $=360-286=74 \mathrm{~m}^{2}$
Cost of constructing $1 \mathrm{~m}^{2}$ canal $=₹ 12$
Cost of the constructing $74 \mathrm{~m}^{2}$ canal $=₹ 12 \times 74=₹ 888$

Objective Type Question
Question $9 .$
The formula to find the area of the circular path is
(i) $\pi\left(\mathrm{R}^{2}-\mathrm{r}^{2}\right)$ sq. units
(ii) $\pi \mathrm{r}^{2}$ sq. units
(iii) $2 \pi \mathrm{r}^{2}$ sq. units
(iv) $\pi \mathrm{r}^{2}+2 \mathrm{r}$ sq. units
Answer:
(i) $\pi\left(\mathrm{R}^{2}-\mathrm{r} 2\right)$ sq. units
 

Question $10 .$
The formula used to find the area of the rectangular path is
(i) $\mathrm{p}\left(\mathrm{R}^{2}-\mathrm{r}^{2}\right)$ sq. units
(ii) $(\mathrm{L} \times \mathrm{B})-(\mathrm{l} \times \mathrm{b})$ sq. units
(iii) LB sq. units
(iv) lb sq. units
Answer:
(ii) $(\mathrm{L} \times \mathrm{B})-(\mathrm{l} \times \mathrm{b})$ sq. units

Question $11 .$
The formula to find the width of the circular path is
(i) $(\mathrm{L}-\mathrm{l})$ units
(ii) $(B-b)$ units
(iii) $(\mathrm{R}-\mathrm{r})$ units
(iv) $(\mathrm{r}-\mathrm{R})$ units
Answer:
(iii) $(R-r)$ units