Question 14 Marks
A wire bent in the form of an equilateral triangle has an area of $121 \sqrt{3} cm ^2$. If the same wire is bent into the form of a circle, find the area enclosed by the wire.
AnswerArea of equi $I$ ater al triangle $=121 \sqrt{3} cm ^2$
$\Rightarrow \frac{ s ^2 \sqrt{3}}{4}=121 \sqrt{3}$
$\Rightarrow s ^2-484$
$\Rightarrow s =22 cm$
Side of the triangle $=22 cm$
Perimeter of the triangle $=3 \times 22=66 cm$
Perimeter of the circle $=$ perimeter of the triangle
$\Rightarrow 2 \pi r =66$
$\Rightarrow r =66 \times \frac{7}{22} \times \frac{1}{2}$
$\Rightarrow r =10.5 cm$
Radius of circle $=10.5 cm$
Area of cirde $=\pi r^2$
$=\frac{22}{7} \times 10.5 \times 10.5$
$=346.5 cm ^2$
Hence, area of the circle $=346.5 cm ^2$
View full question & answer→Question 24 Marks
A wire when bent in the form of a square encloses an area of $484 cm^2$. If the same wire is bent into the form of a circle, find the area of the circle.
AnswerArea of the square $=484 cm ^2$
Side of the square $=\sqrt{484}=22 cm$
Perimeter of the square $=4 \times 22=88 cm$
Perimeter of square $=$ perimeter of circle
$21 \pi r=88$
$r=88 \times \frac{7}{22} \times \frac{1}{2}$
$r =14 cm$
Radius of circle $=14 cm$
Area of cirde $=11 \pi r^2$
$=\frac{22}{7} \times 14 \times 14$
$=616 cm ^2$
Hence, area of the circle $=616 cm ^2$
View full question & answer→Question 34 Marks
Find the area enclosed between two concentric circles, if their radii are $6$ cm and $13$ cm respectively.
AnswerArea between two concentric circles $=$ Area of larger cirde - area of smaller circle
Area of the drde $=\pi r^2$
Radius of bigger drde $=r 1=13 cm$
Radius of smaller drde $=r 2=6 cm$
$\text { Required area }=\pi r 1^2-\pi r 2^2 $
$=\left(\frac{22}{7} \times 13 \times 13\right)-\left(\frac{22}{7} \times 6 \times 6\right) $
$=531.1429-113.1429 $
$=418 cm ^2$
Hence, required area $=418 cm ^2$
View full question & answer→Question 44 Marks
The area of a circular ring enclosed between two concentric circles is $88\ cm^2.$ Find the radii of the two circles, if their difference is $1\ cm.$
AnswerLet $r cm$ be the radius of inner circle.
Radius of outer circle $=( r +1) cm$
Area of circular ring $=$ Area of outer cirde - area of inner circle
$\Rightarrow \pi(r+1)^2-\pi r^1=88 $
$\Rightarrow \pi\left(r^1+2 r+1\right)-\pi r^2=88 $
$\Rightarrow \pi r^1+2 \pi r+\pi-\pi r^1=88 $
$\Rightarrow \frac{22}{7}(2 r+1)=88 $
$\Rightarrow 2 r+1=88 \times \frac{7}{22} $
$\Rightarrow 2 r+1=28 $
$\Rightarrow 2 r=27$
$\Rightarrow r =\frac{27}{2}=13.5 cm$
$(r+1)=13.5+1=14.5 cm$
Therefore, radii are $13.5 cm$ and $14.5 cm$.
View full question & answer→Question 54 Marks
Find the area grazed by a horse tied with a $11. 2$ m rope to a corner of a field measuring $25$ m by $15$ m.
Answer
Area grazed by the horse $=\frac{\pi r^2}{4}$
$=\frac{\frac{22}{7} \times 11.2 \times 11.2}{4}$
$=\frac{394.24}{4} $
$=98.56$
Area grazed by the horse $=98.56 m ^2$ View full question & answer→Question 64 Marks
A bucket is raised from a well by means of a rope wound round a wheel of diameter $35$ cm. If the bucket ascends in $2$ minutes with a uniform speed of $1.1$ m per sec, calculate the number of complete revolutions the wheel makes in raising the bucket.
AnswerDiameter of the wheel $=35 cm$
Circumference $=2 \pi r=\pi d$
$=\frac{22}{7} \times 35 $
$=110\ cm$
Circumference $=110\ cm$
Time taken by the bucket $=2\ min .=2 \times 60=120$ seoonds
Speed of the rope $=1.1\ m / s$
Length of the rope $=120 \times 1.1=132\ m$
No.of. revolutions $=\frac{132 \times 100}{110}=120$
No. of revolutions made by rope $=120$
View full question & answer→Question 74 Marks
The sum of the radii of two circles is $10.5 cm$ and the difference of their circumferences is $13. 2 cm.$ Find the radii of the two circles.
AnswerThe sum of the radii $=10.5 cm$
$r_1+r_2=10.5 cm\ldots(i)$
The difference of circumferences $=13.2 cm$
$2 \pi\left(r_1-r_1\right)=13.2$
$\left(r_1-r_1\right)=\frac{13.2}{2 \pi}$
$\left(r_1-r_1\right)=\frac{13.2 \times 7}{2 \times 22}$
$r_1-r_1=2.1\ldots(ii)$
Adding (i) and (ii)
$2 r_1=12.6$
$r_1=6.3 cm$
Therefore, $r_2=(10.5-6.3) cm =4.2 cm$
Hence, the two radii are $6.3 cm$ and $4.2 cm$
View full question & answer→Question 84 Marks
Find the area and perimeter of the circle with the following:
Diameter $= 35\ cm$
AnswerDiameter $=2 r$
$\Rightarrow r =\frac{\text { diameter }}{2}=\frac{35}{2}=17.5 cm$
Area of the drde $=\pi r^2$
$=\frac{22}{7} \times 17.5 \times 17.5$
$=962.5 cm ^2$
Perimeter of the circle $=2 \pi r$
$=2 \times \frac{22}{7} \times 17.5$
$=110 cm$
Therefore, Area $=962.5 cm ^2$ and Perimeter $=110 cm$
View full question & answer→Question 94 Marks
Find the area and perimeter of the circle with the following :
Diameter $= 77 cm$
AnswerDi am eter $=2 r$
$\Rightarrow r =\frac{\text { diameter }}{2}=\frac{77}{2}=38.5 cm$
Area of the circle $=\pi r^2$
$=\frac{22}{7} \times 38.5 \times 38.5 $
$=4658.5 cm ^2$
Perimeter of the circle $=2 \pi r$
$=2 \times \frac{22}{7} \times 38.5 $
$=242 cm$
Therefore, Area $=4658.5 cm ^2$ and Perimeter $=242 cm$
View full question & answer→Question 104 Marks
The circumference of a circle is equal to the perimeter of a square. The area of the square is 484 sq. m. Find the area of the circle.
AnswerArea of the square $=484 m ^2$
Side of the square $=\sqrt{484}=22 m$
Perimeter of the square $=4 \times 22=88 m$
Perimeter of square $=$ circumference of circle
$2 \pi r=88$
$r =88 \times \frac{7}{22} \times \frac{1}{2}$
$r =14 m$
Radius of circle $=14 m$
Area of the drde $=\pi r^2$
$=\frac{22}{7} \times 14 \times 14$
$=616 m ^2$
Hence, area of the circle $=616 m ^2$
View full question & answer→Question 114 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.
AnswerFor 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 circle $=\pi r^2$
$=\frac{22}{7} \times 21 \times 21$
$=1386 \ cm ^2$
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 $=35 \ cm$
View full question & answer→Question 124 Marks
Find the area of the biggest circle that can be cut from a rectangular piece of $44\ cm$ 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 $=616 \ cm ^2$ View full question & answer→Question 134 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 .
AnswerRadius 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 $10.07 \ cm$
View full question & answer→Question 144 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
$=\frac{1}{2} \pi R ^2-\frac{1}{2} \pi r ^2$
$=\frac{1}{2} \pi\left( R ^2- r ^2\right)$
$=\frac{1}{2} \times \frac{22}{7} \times\left(28^2-21^2\right)$
$=\frac{1}{2} \times \frac{22}{7} \times(784-441)$
$=\frac{1}{2} \times \frac{22}{7} \times 343$
$=539 \ m ^2$ View full question & answer→Question 154 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.
AnswerSum 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 $21 \ cm$
View full question & answer→Question 164 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 $3 \ cm$ View full question & answer→Question 174 Marks
A $4.2\ m$ wide road surrounds a circular plot whose circumference is $176\ m.$ Find the cost of paving the road at $Rs. 75\ m ^2.$
AnswerCircumference 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 $= Rs. 59,598$
View full question & answer→Question 184 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 $sq\ m.$
AnswerArea 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 $= Rs. 3,00,300$
View full question & answer→Question 194 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.
AnswerRadius 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 $=14: 11$
View full question & answer→Question 204 Marks
Find the circumference of the circle whose area is $25$ times the area of the circle with radius $7 \ cm.$
AnswerArea 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 $=220 \ cm$
View full question & answer→