[5 marks sum]

Question 15 Marks

The area between the circumferences of two concentric circles is $2464 cm^2$. If the inner circle has circumference of $132 cm$, calculate the radius of outer circle.

Answer

For the inner circle:
Circumference $=2 \pi r=132$
$2 \pi r=132$
$2 \times \frac{22}{7} \times r =132$
$r =\frac{132 \times 7}{2 \times 22}$
$r =21 cm$
Therefore, radius of inner circle $=21 cm$
Area of inner cirel$e=\pi r^2$
$=\frac{22}{7} \times 21 \times 21$
$=1386 cm ^2$
Area of outer circle $=$ area of inner circle + area of concentric circles
$=(1386+2464) cm ^2=3850 cm ^2$
$\Rightarrow \pi r ^2=3850$
$\Rightarrow R^2=3850 \times \frac{7}{22}$
$\Rightarrow R^2=1225$
$\Rightarrow R =35 cm$
Hence, radius of outer circle $=35 cm$

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Question 25 Marks

Find the area of the biggest circle that can be cut from a rectangular piece of 44cm by 28 cm. Also, find the area of the paper left after cutting out the circle.

Answer

Area of piece of paper $=44 \times 28=1232 cm ^2$
The biggest circle that can be rut from rectangular piece of paper is of diameter $28 cm$ (ie. Radius $14 cm$ ).
Area of circle $=\pi r^2$
$=\frac{22}{7} \times 14 \times 14$
$=616 cm ^2$
Area of the cirde $=616 cm ^2$
Area of paper left $=$ area of paper - area of circle
$=(1232-616) cm ^2$
$=616 cm ^2$
Therefore, area of paper left $=616 cm ^2$

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Question 35 Marks

Find the area enclosed between two concentric cirdes of radii 6.3 cm and 8.4 cm. A third concentric circle is drawn outside the 8.4 cm circle, so that the area enclosed between it and 8.4 cm cirde is the same as that between two inner circles. Find the radii of the third circle correct to two decimal places .

Answer

Radius of innermost circle $= r 1=6.3 cm$
Radius of central circle $= r 2=8.4 cm$
Area between two inner circles $=\pi r 2^2-\pi r 1^2$
$=\pi(8.4)^2-\pi(6.3)^2$
$=70.56 \pi-39.69 \pi . .\ldots(i)$
$=221.76-124.74$
$=97.02 cm ^2$
Area between two inner circles $=97.02 cm ^2$
Let radius of third circle be $r$
Area between next two circles $=\pi r ^2-\pi r 2^2$
$=\pi r ^2-\pi(8.4)^2$
$=\pi r ^2-70.56 \pi\ldots(ii)$
Given that (i) and (ii) are equal
$\Rightarrow \pi r^2-70.56 \pi=70.56 \pi-39.69 \pi$
$\Rightarrow r^2-70.56=70.56-39.69$
$\Rightarrow r^2=70.56-39.69+70.56$
$\Rightarrow r^2=101.43$
$\Rightarrow r=10.07 cm$
Therefore, radius of third circle is $10.07 cm$

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Question 45 Marks

A lawn is in the shape of a semi-circle of diameter $42 m$. The lawn is surrounded by a flower bed of width $7 m$ all around. Find the area of the flower bed in $m^2$ .

Answer

Diameter of the inner semi-drde $=42 m$
Hence, radius of the inner semi-circle $=r=21 m$
Width of the flower bed $=7 m$
Diameter of the outer semi-circle $=(42+2 \times 7) m =56 m$
Hence, radius of the outer semi-circle $= R =28 m$
Area of the semicircular flower bed =Area of outer semi-cirde - area of inner semi-drde
$=\frac{1}{2} \pi R ^2-\frac{1}{2} \pi r ^2$
$=\frac{1}{2} \pi\left( R ^2- r ^2\right)$
$=\frac{1}{2} \times \frac{22}{7} \times\left(28^2-21^2\right)$
$=\frac{1}{2} \times \frac{22}{7} \times(784-441)$
$=\frac{1}{2} \times \frac{22}{7} \times 343$
$=539 m ^2$

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Question 55 Marks

The sum of diameter of two circles is $112 cm$ and the sum of their areas is $5236 cm^2$ Find the radii of the two circles.

Answer

Sum of the diameters $=112 cm$
Therefore, sum of the radii $=\frac{112}{2} cm =56 cm$
If $r$ is the radius of one circle, radius of other circle is $(56-r)$
Sum of the areas $=\pi r^2+\pi(56-r)^2$
$\Rightarrow \pi r^2+\pi(56-r)^2=5236$
$\Rightarrow \pi r^2+\pi\left(3136-112 r+r^2\right)=5236$
$\Rightarrow\left(r^2+3136-112 r+r^2\right) \times \frac{22}{7}=5236$
$\Rightarrow 2 r^2-112 r+3136=1666$
$\Rightarrow 2 r^2-112 r+3136-1666=0$
$\Rightarrow 2 r^2-112 r+1470=0$
$\Rightarrow r^2-56 r+735=0$
$\Rightarrow( r -35)( r -21)=0$
$\Rightarrow r=35, r=21$
Therefore, radii of the two circles is $35 cm$ and $21 cm$

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Question 65 Marks

Two circles touch each other externally. The sum of their areas is $5811 cm^2$ and the distance between their centres is $10 \ cm$. Find the radii of the two circles.

Answer

Let r be radius of one drde, so the radius of other drde is
$(10-r)$
Sum of the areas $= \pi r^2 + \pi (10 - r)^2$
$\Rightarrow \pi r^2+ \pi (10 - r)^2 = 58 \pi$
$\Rightarrow \pi r^2 + \pi (100 - 20 r + r^2) = 58 \pi$
$\Rightarrow r^2 + 100 - 20 r + r^2 = 58$
$\Rightarrow 2r^2 + 100 - 20 r - 58 = 0$
$\Rightarrow 2r^2 - 20 r + 42 = 0$
$\Rightarrow r^2 - 10 r + 21 = 0$
$\Rightarrow (r - 7)(r - 3) = 0$
$\Rightarrow r = 7 , r = 3$
Therefore, radii of the two circles is ?cm and $3 \ cm$

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Question 75 Marks

A $4.2 m$ wide road surrounds a circular plot whose circumference is $176 m$. Find the cost of paving the road at Rs $75 m ^2$.

Answer

Circumference of the plot $=176 m$
Radius of the circularplot =
$2 \pi r=176$
$r=176 \times \frac{7}{22} \times \frac{1}{2}$
$r=28 m$
Radius of circular plot $=28 m$
Area of circular plot $=\pi r^2$
$=\frac{22}{7} \times 28 \times 28$
$=2464 m ^2$
Area of circular plot $=2464 m ^2$
Radius of circular plot with surrounding road $=R=(28+4.2) m =32.2 m$
Area of circular plot with surrounding road $=\pi R^2$
$=\frac{22}{7} \times 32.2 \times 32.2$
$=3258.64 m ^2$
Area of circular plot with surrounding road $=3258.64 m ^2$
Area of road =Area of circular plot with surrounding road- Area of circular plot
$=(3258.64-2464) m ^2$
$=794.64 m ^2$
Area of road $=794.64 m ^2$
Cost of paving one sq $m = Rs .75$
Cost of paving the road $=$ Rs $(75 \times 794.64)=$ Rs. 59,598
Cost of paving the road $=$ Rs. 59,598

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Question 85 Marks

A $7 m$ road surrounds a circular garden whose area is $5544 m^2$. Find the area of the road and the cost of tarring it at the rate of Rs.$150$ per sq m.

Answer

Area of garden $=5544 m ^2$
Radius of cirOJlar garden=
$\pi r^2=5544$
$r^2=5544 \times \frac{7}{22}$
$r^2=1764$
$r=42 m$
Radius of cirOJlar garden $=42 m$
Radius of garden with surrounding road $=R=42+7 m =49 m$
Area of garden with surrounding road $=\pi R^2$
$=\frac{22}{7} \times 49 \times 49$
$=7546 m ^2$
Area of garden with surrounding road $=7546 m ^2$
Area of road =Area of garden with surrounding road - area of garden
$=7546-5544 m ^2$
$=2002 m ^2$
Area of road $=2002 m ^2$
Cost of tarring one sq $m=$ Rs. 150
Cost of tarring area of road $=$ Rs. $(150 \times 2002)=$ Rs 3,00,300
Cost of tarring the road $=$ Rs. $3,00,300$

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Question 95 Marks

A wire is in the form of a circle of radius 42 cm. It is bent into a square.
Determine the side of the square and compare the area of the regions enclosed in two cases.

Answer

Radius of the circle $=42 cm$
Area of the cirde $=\pi r^2$
$=\frac{22}{7} \times 42 \times 42$
$=5544 cm ^2$
Area of the cirde $=5544 cm ^2$
Circumference of the circle $=2 \pi r$
$=2 \times \frac{22}{7} \times 42$
$=264 cm$
Perimeter of the square $=$ Circumference of the circle $=264 cm$
$4 \times$ side $=264$
side $=\frac{264}{4}=66 cm$
Side of square $=66 cm$
Area of square $=$ side $^2=66 \times 66=4356 cm ^2$
Area of a circle : Area of square $=\frac{5544}{4356}=\frac{14}{11}$
Side of square $=66 cm$
Area of the cirde: Area of square $=14: 11$

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Question 105 Marks

Find the circumference of the circle whose area is 25 times the area of the circle with radius 7 cm.

Answer

Area of the circle with radius $7 cm =\pi r^2$
$=\frac{22}{7} \times 7 \times 7$
$=154 cm ^2$
Area of new circle $=25 x$ area of the drde with radius $7 cm$
$=25 \times 154 cm ^2$
$=3850 cm ^2$
Therefore ,
$\pi R^2=3850$
$R^2=3850 \times \frac{7}{22}$
$R^2=1225$
$R=35 cm$
Radius of new circle $=35 cm$
Circumference of new circle $=2 \pi R$
$=2 \times \frac{22}{7} \times 35$
$=220 cm$
Hence, Circumference of new circle $=220 cm$

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Mathematics [5 marks sum]

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