Exercise 2.4 - 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 $2.4$
Miscellaneous Practice Problems
Question $1 .$
A wheel of a car covers a distance of $3520 \mathrm{~cm}$ in 20 rotations. Find the radius of the wheel?
Solutions:
Distance covered by circular wheel in 20 rotation $=3520 \mathrm{~cm}$
$\therefore$ Distance covered ini rotation $=\frac{3520}{20} \mathrm{~cm}=176 \mathrm{~cm}$
$\therefore$ Circumference of the wheel $=176 \mathrm{~cm}$
$\therefore 2 \pi \mathrm{r}=176$
$2 \times \frac{-2}{6} \times \mathrm{r}=176$
$\mathrm{r}=\frac{176 \times 7}{2 \times 22}$
$\mathrm{r}=28 \mathrm{~cm}$
Radius of the wheel $=28 \mathrm{~cm}$

Question $2 .$
The cost of fencing a circular race course at the rate of $₹ 8$ per metre is $₹ 2112$. Find the diameter of the race course.
Solution:
Cost of fencing the circumference $=₹ 2112$
Cost of fencing one meter $=₹ 8$
$\therefore$ Circumference of the circle $=\frac{2112}{8}=264 \mathrm{~m}$
$\pi \mathrm{d}=264 \mathrm{~m}$
$\frac{22}{7} \times \mathrm{d}=264$
$\mathrm{d}=\frac{264 \times 7}{22}=12 \times 7 \mathrm{~m}=84 \mathrm{~m}$
$\therefore$ Diameter of the race cource $=84 \mathrm{~m}$

Question $3 .$
A path $2 \mathrm{~m}$ long and $1 \mathrm{~m}$ broad is constructed around a rectangular ground of dimensions $120 \mathrm{~m}$ and $90 \mathrm{~m}$ respectively. Find the area of the path.
Solution:
Length of the rectangular ground $\mathrm{l}=120 \mathrm{~m}$
Breadth $\mathrm{b}=90 \mathrm{~m}$
Length of the path $\mathrm{W}_{1}=2 \mathrm{~m}$
Length of the path $\mathrm{W}_{2}=1 \mathrm{~m}$
Length of the ground with path $\mathrm{L}=1+2\left(\mathrm{~W}_{2}\right)=120+2(1) \mathrm{m}$
$=120+2=122 \mathrm{~m}$
Breadth of the ground with path $\mathrm{B}=1+2\left(\mathrm{~W}_{1}\right)$ units
$=90+2(2) \mathrm{m}=90+4 \mathrm{~m}=94 \mathrm{~m}$
$\therefore$ Area of the path $=(\mathrm{L} \times \mathrm{B})-(1 \times \mathrm{b})$ sq. units
$=(122 \times 94)-(122 \times 94) \mathrm{m}^{2}=668 \mathrm{~m}^{2}$
$\therefore$ Area of the path $=668 \mathrm{~m}^{2}$

Question $4 .$
The cost of decorating the circumference of a circular lawn of a house at the rate of $₹ 55$ per metre is ₹ 16940 . What is the radius of the lawn?
Solution:
Cost of decorating the circumference $=₹ 16,940$
Cost of decorating per meter $=₹ 55$
$\therefore$ Length of the circumference $=\frac{16940}{55} \mathrm{~m}=308 \mathrm{~m}$
Circumference of the circular lawn $=308 \mathrm{~m}$
$2 \times \pi \mathrm{r}=308 \mathrm{~m}$
$2 \times \frac{22}{7} \times \mathrm{r}=308 \mathrm{~m}$
$\mathrm{r}=\frac{308 \times 7}{2 \times 22}$
$\mathrm{r}=49 \mathrm{~m}$
Radius of the lawn $=49 \mathrm{~m}$
 

Question $5 .$
Four circles are drawn side by side in a line and enclosed by a rectangle as shown below.
If the radius of each of the circles is $3 \mathrm{~cm}$, then calculate:
(i) The area of the rectangle.
(ii) The area of each circle.
(iii) The shaded area inside the rectangle.

Solution:
Given radius of a circle $\mathrm{r}=3 \mathrm{~cm}$
Diameter of the circle $=2 \mathrm{r}=2 \times 3=6 \mathrm{~cm}$
Breadth of the rectangle = Diamter of the circle
$B=6 \mathrm{~cm}$
Length of the rectangle $L=4 \times$ diameter of a circle
$\mathrm{L}=4 \times 6$
$\mathrm{L}=24 \mathrm{~cm}$
(i) Area of the rectangle $=\mathrm{L} \times \mathrm{B}$ sq. units
$=24 \times 6 \mathrm{~cm}^{2}$
Area of the rectangle $=144 \mathrm{~cm}^{2}$
(ii) Area of the circle $=\pi \mathrm{r}^{2}$ sq. units
$=\frac{22}{7} \times 3 \times 3 \mathrm{~cm}^{2}$
$=\frac{198}{7} \mathrm{~cm}^{2}$
$=28.28 \mathrm{~cm}^{2}$
(iii) Area of the shaded area = Area of the rectangle - Area of the 4 circles $=144-\left(4 \times \frac{198}{7}\right) \mathrm{cm}^{2}=144-\frac{792}{7} \mathrm{~cm}^{2}$
$=144-113.14 \mathrm{~cm}^{2}=30.85 \mathrm{~cm}^{2}$

Challenge Problems

Question $6 .$
A circular path has to be constructed around a circular lawn. If the outer and inner circumferences of the path are $88 \mathrm{~cm}$ and $44 \mathrm{~cm}$ respectively, find the width and area of the path.
Solution:
Outer circumference of the circular lawn $=88 \mathrm{~cm}$
$2 \pi \mathrm{R}=88 \mathrm{~cm}$
Inner circumference of the lawn $2 \pi \mathrm{r}=44 \mathrm{~cm}$
$2 \pi R-2 \pi r=88-44$ $2 \times \frac{22}{7}(R-r)=44$
$(\mathrm{R}-\mathrm{r})=\frac{44 \times 7}{2 \times 22}$
Outer radius $-$ Inner radius $=7 \mathrm{~cm}$
$\therefore$ Width of the lawn $=7 \mathrm{~cm}$
Also $2 \pi \mathrm{R}+2 \pi \mathrm{r}=88+44$
$2 \pi(\mathrm{R}+\mathrm{r})=132$
$\pi(\mathrm{R}+\mathrm{r})=\frac{132}{2}=66 \mathrm{~cm}$
Area of the path $=\pi R^{2}-\pi r^{2}$ sq. units
$=\pi(\mathrm{R}+\mathrm{r})(\mathrm{R}-\mathrm{r})=66 \times 7$
Area of the path $=462 \mathrm{~cm}^{2}$

Question $7 .$
A cow is tethered with a rope of length $35 \mathrm{~m}$ at the centre of the rectangular field of length $76 \mathrm{~m}$ and breadth $60 \mathrm{~m}$. Find the area of the land that the cow cannot graze?
Solution:
Length of the field $\mathrm{l}=76 \mathrm{~m}$
Breadth of the field $b=60 \mathrm{~m}$
Area of the field $\mathrm{A}=1 \times \mathrm{b}$ sq. units $=76 \times 60 \mathrm{~m}^{2}$
Area of the field $\mathrm{A}=4560 \mathrm{~m}^{2}$
Length of the rope $=35 \mathrm{~m}$
Radius of the land that the cow can graze $=35 \mathrm{~m}$
Area of the land tha the cow can graze $=$ circle of radius $35 \mathrm{~m}=\pi \mathrm{r}^{2}$ sq.units
$\pi \times 35 \times 35 \mathrm{~m}^{2}=\frac{22}{7} \times 35 \times 35 \mathrm{~m}^{2}$
$=3850 \mathrm{~m}^{2}$
Area of the land the cow cannot graze $=$ Area of the field - Area that the cow can graze $=4560-3860 \mathrm{~m}^{2}=710 \mathrm{~m}^{2}$
Area of the land that the cow cannot graze $=710 \mathrm{~m}^{2}$

Question $8 .$
A path $5 \mathrm{~m}$ wide runs along the inside of the rectangular field. The length of the rectangular field is three times the breadth.of the field. If the area of the path is $500 \mathrm{~m}^{2}$ then find the length and breadth of the field.
Solution:
Let the length of the rectangular field $=$ ' $\mathrm{L}$ ' $\mathrm{m}$
Breadth of the rectangular field $==$ 'B' $\mathrm{m}$
Area of the rectangular field $=(\mathrm{L} \times \mathrm{B}) \mathrm{m}^{2}$
Also given length $=3 \times$ Breadth
$\mathrm{L}=3 \mathrm{~B}$
Width of the path $(\mathrm{W})=5 \mathrm{~m}$
Lenth of the inner rectangle $=\mathrm{L}-2 \mathrm{~W}=1-2(5)$
$=3 B-10 m$
Breadth of the inner rectangle $=\mathrm{B}-2 \mathrm{~W}$
$=\mathrm{B}-2(5)$
$=\mathrm{B}-10 \mathrm{~m}$
Area of the inner rectangle $=(3 \mathrm{~B}-10)(\mathrm{B}-10)$
$=3 \mathrm{~B}^{2}-10 \mathrm{~B}-30 \mathrm{~B}+100$

Area of the inner rectangle $=(3 \mathrm{~B}-10)(\mathrm{B}-10)$
$=3 \mathrm{~B}^{2}-10 \mathrm{~B}-30 \mathrm{~B}+100$
Area of the path = Area of outer rectangle
- Area of inner rectangle
$=3 \mathrm{~B}^{2}-3 \mathrm{~B}^{2}+40 \mathrm{~B}-100$ Area of the path $=40 \mathrm{~B}-100$
$\begin{aligned}
&=(\mathrm{L} \times \mathrm{B})-\left(3 \mathrm{~B}^{2}-10 \mathrm{~B}-30 \mathrm{~B}+100\right) \\
&3 \mathrm{~B} \times \mathrm{B}-\left(3 \mathrm{~B}^{2}-40 \mathrm{~B}+100\right) \\
&=3 \mathrm{~B}^{2}-3 \mathrm{~B}^{2}+40 \mathrm{~B}-100
\end{aligned}$
Given area of the path $=500 \mathrm{~m}^{2}$
$\begin{aligned}
&40 \mathrm{~B}-100=500 \\
&40 \mathrm{~B}=500+100=600 \\
&\mathrm{~B}=\frac{600}{40} \\
&\mathrm{~B}=15 \mathrm{~m}
\end{aligned}$
Length of the field $=45 \mathrm{~m}$; Breadth of the field $=15 \mathrm{~m}$
 

Question $9 .$
A circular path has to be constructed around a circular ground. If the areas of the outer and inner circles are $1386 \mathrm{~m} 2$ and $616 \mathrm{~m} 2$ respectively, find the width and area of the path.
Solution:
Area of the outer circle $=1386 \mathrm{~m}^{2}$
$\pi \mathrm{R}^{2}=1386 \mathrm{~m}^{2}$
Area of the inner circle $=616 \mathrm{~m}^{2}$
$\pi r^{2}=616 \mathrm{~m}^{2}$
Area of the path = Area of outer circle $-$ Area of the inner circle
$1386 \mathrm{~m}^{2}-616 \mathrm{~m}^{2}$

$1386 \mathrm{~m}^{2}-616 \mathrm{~m}^{2}$
Area of the path $=770 \mathrm{~m}^{2}$
Also $\pi R^{2}=1386$
$\begin{aligned}
&\mathrm{R}^{2}=\frac{1386 \times 7}{22} \\
&\mathrm{R}^{2}=63 \times 7 \\
&\mathrm{R}^{2}=9 \times 7 \times 7 \\
&\mathrm{R}^{2}=32 \times 72 \\
&\mathrm{R}=3 \times 7
\end{aligned}$
Outer Radius $\mathrm{R}=21 \mathrm{~m}$
Again $\pi \mathrm{r}^{2}=616$
$\begin{aligned}
&\frac{22}{7} \times r^{2}=616 \\
&r^{2}=28 \times 7 \\
&r^{2}=4 \times 7 \times 7 \\
&r^{2}=22 \times 72 \\
&r=2 \times 7
\end{aligned}$
Inner radius $\mathrm{r}=14 \mathrm{~m}$
Width of the path = Outer radius $-$ Inner radius $=21-14$
Width of the path $=7 \mathrm{~m}$

Question $10 .$
A goat is tethered with a rope of length $45 \mathrm{~m}$ at the centre of the circular grass land whose radius is $52 \mathrm{~m}$. Find the area of the grass land that the goat cannot graze.
Length of the rope $=45 \mathrm{~m}=$ Radius of the inner circle
$\therefore$ Area of the circular area that the goat graze $=\pi r^{2}$ sq. units
$=\frac{22}{7} \times 45 \times 45 \mathrm{~m}^{2}=6364.28 \mathrm{~m}^{2}$
Radius of the gross land $=52 \mathrm{~m}$
Area of the grass land $=\frac{22}{7} \times 52 \times 52=8,498.28 \mathrm{~m}^{2}$
Area that the goat cannot graze
$=$ Area of the outer circle - Area of the inner circle
$=8498.28-6364.28=2134 \mathrm{~m}^{2}$
Area of the goat cannot grass $=2134 \mathrm{~m}^{2}$
 

Question $11 .$
A strip of $4 \mathrm{~cm}$ wide is cut and removed from all the sides of the rectangular cardboard with dimensions $30 \mathrm{~cm} \times 20 \mathrm{~cm}$. Find the area of the removed portion and area of the remaining cardboard.
Solution:
Area of the outer rectangular cardboard
$=\mathrm{L} \times \mathrm{B}$ sq.units $=30 \times 20 \mathrm{~cm}^{2}=600 \mathrm{~cm}^{2}$
Width of the stip $=4 \mathrm{~cm}$
Length of the inner rectangle $=\mathrm{L}-2 \mathrm{~W}$
$l=30-2(4)=30-8$
$\mathrm{l}=22 \mathrm{~cm}$
Breadth of the inner rectangle $\mathrm{B}=2 \mathrm{~W}=20-2(4)=20-8$
$\mathrm{b}=12 \mathrm{~cm}$

$\mathrm{b}=12 \mathrm{~cm}$
Area of the inner rectangle $=1 \times$ b sq.units $=22 \times 12 \mathrm{~cm}^{2}=264 \mathrm{~cm}^{2}$
Area of the remaining cardboard $=264 \mathrm{~cm}^{2}$
Area of the removed portion $=$ Area of outer rectangle
- Area of the inner rectangle
$=600-264 \mathrm{~cm}^{2}$
Area of the removed portion $=336 \mathrm{~cm}^{2}$
 

Question $12 .$
A rectangular field is of dimension $20 \mathrm{~m} \times 15 \mathrm{~m}$. Two paths run parallel to the sides of the rectangle through the centre of the field. The width of the longer path is $2 \mathrm{~m}$ and that of the shorter path is $1 \mathrm{~m}$. Find (i) the area of the paths (ii) the area of the remaining portion of the field (iii) the cost of constructing the roads at the rate of $₹ 10$ per sq.m.
Solution:
Length of the rectangular field $\mathrm{L}=20 \mathrm{~m}$
Breadth $\mathrm{B}=15 \mathrm{~m}$
Area $=\mathrm{L} \times \mathrm{B}$
$20 \times 15 \mathrm{~m}^{2}$
Area of outer rectangle $=300 \mathrm{~m}^{2}$

Area of inner small rectangle $=\frac{19}{2} \times \frac{13}{2}=61.75 \mathrm{~cm}^{2}$
(i) Area of the path = Area of the outer rectangle
- Area of 4 inner small rectangles
$=300-4(61.75)=300-247=53 \mathrm{~m}^{2}$
Area of the paths $=53 \mathrm{~m}^{2}$
(ii) Area of the remaining portion of the field
$=$ Area of the outer rectangle - Area of the paths $=300-53 \mathrm{~m}^{2}=247 \mathrm{~m}^{2}$
Area of the remaining portion $=247 \mathrm{~m}^{2}$
(iii) Cost of constructing $1 \mathrm{~m}^{2}$ road $=₹ 10$
$\therefore$ Cost of constructing $53 \mathrm{~m}^{2}$ road $=₹ 10 \times 53=₹ 530$
$\therefore$ Cost of constructing road $=₹ 530$