Exercise 16.3
Question 1
(i) r = 7cm
Circumference of Circle = $2 \pi r$
$=2 \times \frac{22}{7} \times 7$
Circumference of circle = 44cm
(ii) r= 21cm
Circumference of circle = $2 \pi r$
$=2 \times \frac{22}{7} \times 21$
=$2 \times 22 \times 3$
Circumference of circle = 132cm
(iii)r= 28m
Circumference of circle = $2 \pi r$
$=2 \times \frac{22}{1} \times 28$
$=2 \times 22 \times 4$
Circumference of circle = 174m
(iv) r = 3.5cm
Circumference of circle = $2 \pi r$
$=2 \times \frac{22}{7} \times 3.5$
Circumference of circle $=22 \mathrm{cm}$
Question 2
(i) $r=14 \mathrm{~mm}$
Area of circle = $\pi r^{2}$
$=\frac{22}{7} \times 14 \times 14$
Area of circle = 616$m^{2}$
(ii) $d=49 m . \Rightarrow r=\frac{d}{2}$
$r= \frac{49}{2}=24.5 \mathrm{~m}$
Area of Circle $=\pi r^{2}$
$=\frac{22}{7} \times 24.5 \times 24.5$
Area of circle = 1886.5$m^{2}$
(iii) Dimension (d)= 9.8m
$r=\frac{d}{2}$
$r=\frac{9.8}{2}$
$r=4.9 \mathrm{~m}$
Area of circle = $\pi r^{2}$
$=\frac{22}{7} \times 4.9 \times 4.9$
Area of circle= 75.46 $m^{2}$
(iv) r=5cm
Area of circle = $\pi r^{2}$
$=\frac{22}{7} \times 5 \times 5$
Area of circle $=78.57 \mathrm{~cm}^{2}$
Question 3
Given
Radius of circle (r) =20cm
Circumference of circle = $2 \pi r$
$=2 \times 3.14 \times 20$
Circumference of circle = $125.6 \mathrm{cm}$
Area of circle = $ \pi r^{2}$
=$3.14 \times 20 \times 20$
Area of circle $=1256 \mathrm{~cm}^{2}$
Question 4
Radius of minute hand = 1.4m
distance covered by minute hand tip in 1 hour= Circumference of circle of radius 1.4m
= $2 \pi r$
=$2 \times \frac{22}{7} \times 1.4$
=$8.8 \mathrm{~m}$
Question 5
Diameter of garden = 21m
Radius (r)= $\frac{21}{2}=10.5 m$
Circumference of garden = $2 \pi r$
$=2 \times 22 \times 10.5$
$=66 \mathrm{~m}$
Length of rope =
$66 \times2=132 m(\because$ For 2 rounds $)$
Rate of cost of rope = Rs 4\m
Total cost of rope = $4 \times 132$
$=₹ 528$
∴Cost of rope = Rs 528
Question 6
Given
Circumference of circle exceeds diameter by 30cm
$2 \pi r=d+30$
$2 \pi r=2 r+30$
$2 \pi r-2 r=30$
$2 r(11-1)=30$
$2 r(n-1)=30$
$2 r\left(\frac{23}{7}-1\right)=30$
$2 \gamma\left(\frac{15}{7}\right)=30$
$r=\frac{30 \times 7}{2 \times 15}$
$r=7 \mathrm{cim}$
∴ Radius of circle = 7cm
Question 7
Given Circumference of circle = 44cm
$\begin{aligned} 2 \pi r &=44 \\ \pi r=& \frac{44}{2} \\ \pi r &=22 \\ r &=\frac{22}{\pi} \\ r &=\frac{22}{\frac{22}{7}} \\ r &=7 \mathrm{~cm} \end{aligned}$
Diameter $=2 r=2 \times 7=14 \mathrm{~cm}$
Question 8
Given Circumference of Circle= 31.4cm
$\begin{aligned} 2 \pi r &=31.4 \\ r &=\frac{31.4}{2 \times \pi} \\ &=\frac{31.4}{2 \times 314} \\ r &=5 \mathrm{~cm} \end{aligned}$
Radius =5cm
$\begin{aligned} \text { Area } &=\pi r^{2} \\ &=3.14 \times 5^{2} \\ \text { Area } &=78.5 \mathrm{~cm}^{2} \end{aligned}$
Question 9
Given
Area of Circle = $144 \pi \mathrm{Cm}^{2}$
$\begin{aligned} r^{2} &=144 \\ r &=\sqrt{144} \\ r &=12 \mathrm{~cm} \end{aligned}$
radius = 12cm
Circumference of circle =$2 \pi r$
$2 \times \pi \times 12$
$24 \pi \mathrm{cm}$
∴ Circumference of Circle = $24 \pi \mathrm{cm}$
Question 10
Diameter of Wheel = 56cm
Radius of Wheel = $\frac{56}{2}=28 \mathrm{~cm}$
Circumference of wheel = $2 \pi r$
$=2 \times \pi \times 28$
$=2 \times \frac{22}{7} \times 28$
Circumference of wheel = $176 \mathrm{~cm}$
No. of rotations = $\frac{Distance covered by car}{Distance for one rotation}$
=$\frac{88 \times 10^{3} \times 10^{2}}{176}$
No. of rotations = $45454.54$ =45455
Question 11
(Diagram to be added)
Given
Square with side = 21cm
Circle with maximum area diameter
d= 21cm
Shaded area = Square area - Circle area
$=(21)^{2}-\pi\left(\frac{21}{2}\right)^{2}$
$=441-\frac{22}{7} \times \frac{441}{4}$
$=441\left(1-\frac{22}{28}\right)$
$=441\left(\frac{6}{28}\right)$
Shaded area = $94.5 \mathrm{~cm}^{2}$
Question 12
Side of equilateral triangle = 4.4cm
Perimeter of triangle = $3 \times 4.4$
=$13.2 \mathrm{~cm}$
∴ Perimeter of circle = Perimeter of equilateral triangle
$\begin{aligned} 2 \pi r &=13.2 \\ r &=\frac{13.2}{2 \times 22} \times 7 \\ r &=2.1 \mathrm{~cm} \end{aligned}$
Radius of circle= 2.1cm
Area of circle = $\pi r^{2}$
$=\frac{22}{7} \times(2.1)^{2}$
Area of circle = 13.86 $\mathrm{~cm}^{2}$
Question 13
Wire is bent in the form of square of side = 27.5cm
Perimeter of square = Length of wire
Length of wire = $4 \times 275$
Length of wire = 110cm
Now same wire bent in the shape of circle
Length of wire = Perimeter of circle
$110=2 \pi r$
$r=\frac{110}{2 \times 22} \times 7$
$r=\frac{35}{2}$
$r=17.5 \mathrm{~cm}$
Radius of circle = 17.5cm
Area of circle = $\pi r^{2}$
$=\frac{22}{7} \times(17.5)^{2}$
Area of circle = $962.5 \mathrm{~cm}^{2}$
Question 14
Wire is initially in the form of rectangle of length , breadth = 187cm, 14.3cm
Length of wire = Perimeter of rectangle
$\begin{aligned} & 2(18.7+14.3) \\=& 2(33) \\=& 66 \mathrm{~cm} \end{aligned}$
Now same wire is bent into circle
Length of wire = Perimeter of circel
$66=2 \pi r$
$r=\frac{66}{2 \times 22} \times 7$
$r=\frac{21}{2}$
$r=10.5 \mathrm{~cm}$
Radius of circle = 10.5cm
Area of circle = $\pi r^{2}$
$=\frac{22}{7} \times(10.5)^{2}$
Area of circle = 346.5$\mathrm{Cm}^{2}$
Question 15
(diagram to be added)
Diameter of circular park = 84m
Radius of circular park = 42m
Radius outer circle = 42+3.5
=45.5m
∴ Area of road = Outer circle area - Inner circle area
$=\pi(45.5)^{2}-\pi(42)^{2}$
Area of road $=962.5 \mathrm{~m}^{2}$
Cost of constructing the road = Rs $240 / m^{2}$
Cost of Constructing the road = $962.5 \times 240$
= Rs 231000
Question 16
(diagram to be added)
Outer circle Circumference = 44m
$2 \pi R=44$
$R=\frac{44}{2 \times 22} \times 7$
$R=7 \mathrm{~m}$
Outer circle radius =7m
Inner circle radius= 7-3=5m
Circumference of inner circle = $2 \pi r=2 \times \frac{22}{7} \times 5=31.42 \mathrm{~m}$
Area of inner circle = $\pi r^{2}=\pi(5)^{2}$
=$78.57 m^{2}$
Question 17
(diagram to be added)
Area between the circles = $770 \mathrm{~cm}^{2}$
Radius of outer circle = 21cm
Area between the circles= Outer circle area - Inner circle area
770= $\pi (21)^{2}$ - \pi r^{2}$
$770=\frac{22}{7}\left(441-r^{2}\right)$
$\frac{770 \times 7}{22}=441-r^{2}$
$245=441-r^{2}$
$r^{2}=441-245$
$r^{2}=196$
$r=\sqrt{196}$
$r=14 cm$
Inner circle radius = 14cm
Question 18
(diagram to be added)
Radius of big circle = 14cm
Shaded region area = $\pi r^{2}-$[Rectangle area + $2 \times$ Circle area]
=$\pi(14)^{2}-\left[3 \times 1+\pi(3.5)^{2}\right]$
=$616-[3+38.5]$
=$616-41.5$
∴ Shaded region area =$574.5 \mathrm{~cm}^{2}$ ∴ Shaded region area
Question 19
(diagram to be added)
(i) Length of boundary = Semi- circle length +10+7+10
$=\frac{2 \pi r}{2}+10+7+10$
$=\pi r+10+7+10$
$=\frac{22}{7} \times \frac{7}{2}+10+7+10$
$=11+10+7+10$
$=38 \mathrm{~cm}$
∴ Length of boundary = 38cm
Area of shaded Region = Rectangle Area - Semi circle Area
$=10 \times 7-\frac{\pi r^{2}}{2}$
=$10 \times 7-\frac{22}{7} \times \frac{7^{2} }{4 \times 2}$
$=70-19.25$
Area of Shaded Region = $50.75 \mathrm{~cm}^{2}$
(ii) (diagram to be added)
From the Figure
Side of square = $2 \times$ radius of Circle
$\begin{aligned} r+2 r+r &=28 \\ 4 r &=28 \\ r &=\frac{28}{4} \\ r &=7 \mathrm{cm} \end{aligned}$
Radius of semi- circle = 7cm
Side of square = $2 \times 7=14 \mathrm{cm}$
Length of boundary = $4 \times$ perimeter of semi circle
$\begin{aligned} & 4 \times \frac{2 \pi r}{2} \\=& 4 \times \frac{2 \times 22 \times 7}{1 \times 2} \end{aligned}$
Length of boundary= 88cm
Area of shaded region = $4 \times$ semi circle area + Square Area
$=4 \times \frac{\pi \times 7^{2}}{2}+14^{2}$
$=\frac{4 \times 22 \times 7^{2}}{7 \times{2}}+14^{2}$
$=308+196$
Area of shaded region = $504 \mathrm{~cm}^{2}$
Question 20
(diameter to be added)
From figure
Diameter of semicircle = 10.5-3.5
Diameter of semicircle = 7cm
Radius of semi- circle = 3.5cm
Length of boundary = $4+35+$ semicircle perimeter +$4+10.5$
$=7.5+\frac{2 \pi \times 3.5}{2}+14.5$
$=7.5-1 \frac{22}{7} \times 3.5+14.5$
$=7.5+11+14.5$
Length of boundary $=33 \mathrm{~cm}$
Area of shaded region = Rectangle Area+ Semi circle
$=4 \times 10.5+\frac{\pi(3.5)^{2}}{2}$
$=42+\frac{22}{7} \times \frac{(3.5)^{2}}{2}$
$=42+19.25$
Area of shaded region = $61.25 \mathrm{Cm}^{2}$
(ii) (diameter to be added)
Consider $\triangle O A B$
$O A^{r}=O B^{2}+A B^{2}$
$10^{r}=r^{2}+8^{2}$
$r^{2}=100-64$
$r^{r}=36$
$r= \sqrt{36}$
$r=6 \mathrm{~cm} .$
Length of boundary = 10+ Semicircle length +10
=$10+\frac{2 \pi \times r}{2} +10$
=$10+2 \times \frac{22}{7} \times \frac{6}{2}+10$
=$10+18.857+10$
Length of boundary = 38.857cm
Area of shaded region = Area of $\triangle A O C+$ Area of semi - circle
$=\frac{1}{2} \times 2 \times r \times 8+\frac{\pi r^{2}}{2}$
$=\frac{1}{2} \times 2 \times 6 \times 8+ \frac{22}{7} \times \frac{6^{2}}{2}$
$=48+56.57$
Area of shaded Region = 104.57$\mathrm{cm}^{2}$
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