ML Aggarwal Solution Class 9 Chapter 16 Mensuration EXERCISE 16.3

 EXERCISE 16.3

Question 1

Find the length of the diameter of a circle whose circumference is 44 cm.

Sol :
Let radius of the circle =r then circumference =2πr
2πr=44⇒=2×227r=44

r=44×72×22=7 cm

Diameter =2r=2×7=14 cm


Question 2

Find the radius and area of a circle if its circumference is 18π cm.

Sol :
Let r be the radius of the circle  
Circumference =2πr
2πr=18π2r=18r=182=9 cm
Area =πr2=π×9×9=81πcm2

Question 3

Find the perimeter of a semicircular plate of radius 3.85 cm.

Sol :
Radius of semicircular plate =385 cm 
Length of semicircular plate =πr









Perimeter =πr+2r=r(π+2)

=385(227+2)=385×367

=055×36=1980=198 cm


Question 4

Find the radius and circumference of a circle whose area is 144π cm2.

Sol :
Area of the circle =144πcm2
Let radius =r

πr2=144πr2=144

r=144=12 cm

Circumference =2πr=2×12×π

=24πcm


Question 5

A sheet is 11 cm long and 2 cm wide. Circular pieces 0.5 cm in diameter are cut from it to prepare discs. Calculate the number of discs that can be prepared.

Sol :
Length of sheet =11 cm
Width =2 cm
First of all, we have to cut the sheet is squares of side 05 cm.
No. of squares =1105×205
=11×105×2×10522×4=88
No. of discs will be equal to number of squares cut out =88

Question 6

If the area of a semicircular region is 77cm2, find its perimeter.
Sol :
Area of semicircular region =77cm2







Let r be the radius of the region

Then area =12πr2
12πr2=77

12×227(r)2=77
r2=77×2×722=49=(7)2

r=7 cm

Now, perimeter of the region
=πr+2r
=227×7+2×7
=22+14=36 cm

Question 7

(a) In the figure (i) given below, AC and BD are two perpendicular diameters of a circle ABCD. Given that the ara of the shaded portion is 308 cm2, calculate
(i) the length of AC and
(ii) the circumference of the circle.
(b) In the figure (ii) given below, AC and BD are two perpendicular diameters of a circle with centre O. If AC = 16 cm, calculate the area and perimeter of the shaded part. (Take π = 3.14)












Sol :
(a) Area of shaded portion
=Area of semicircle =308 cm2

Let r be the radius of the circle, then
12πr2=30812×227r2=308
r2=308×2×722r2=196=(14)2

(i) Now AC=2r=2×14=28 cm
(ii) Circumference of the circle =2πr
=28×227 cm=4×22=88 cm


(b) Diameters of circle=16 cm
Radius =162=8 cm
Area of shaded part
Area =2× area of one quadrant
=12πr2=12×3.14×8×8=100.48 cm2

Perimeter of shaded part =12 of circumference +4r
=12×2πr+4r=πr+4r=r(π+4)
=8(3.14+4)=8×7.14=57.12 cm

Question 8

A bucket is raised from a well by means of a rope which is wound round a wheel of diameter 77 cm. Given that the bucket ascends in 1 minute 28 seconds with a uniform speed of 1.1 m/sec, calculate the number of complete revolutions the wheel makes in raising the bucket.

Sol :
Diameter of wheel=77 cm
radius =772 cm

Circumference =2πr=2×227×772=242 cm


Question 9

The wheel of a cart is making 5 revolutions per second. If the diameter of the wheel is 84 cm, find its speed in km/hr. Give your answer correct to the nearest km.

Sol :
Diameter of wheel =84 cm
Radius =842=42 cm

Circumference of the wheel
=2πr=2×227×42=264cm

Distance covered in 5 reductions
=264×5=1320 cm.
Time =1 second

speed of the wheel =13201×60×60100×1000 km/hr

=47.52 km/hr=48 km/hr


Question 10

The circumference of a circle is 123.2 cm. Calculate :

(i) the radius of the circle in cm.

(ii) the area of the circle in cm2, correct to the nearest cm2.

(iii) the effect on the area of the circle if the radius is doubled.

Sol :
Circumference of a circle =123.2 cm

Let radius = r

2πr=12322×227r=123210

(i) Radius =156 cm

(ii) Area of the circle =πr2

=227×196×196 cm2

=120736 cm2=1207 cm2

(iii) If radius is doubled i.e.=196×2

=39.2 cm

Then area of the circle =πr2

=227×392×392 cm3=482944 cm2

Effect on area =4829.441207=4 times


Question 11

(a) In the figure (i) given below, the area enclosed between the concentric circles is 770 cm2. Given that the radius of the outer circle is 21 cm, calculate the radius of the inner circle.

(b) In the figure (ii) given below, the area enclosed between the circumferences of two concentric circles is 346.5 cm2. The circumference of the inner circle is 88 cm. Calculate the radius of the outer circle.










Sol :
(i) Radius of the outer circle (R)=21 cm 
Let radius of inner circle =r cm 
Area of the ring =π(R2r2)

=227(212r2)=227(441r2)

But area of the ring =770 cm2

227(441r2)=770

441r2=770×722=245

r2=441245=196r=196=14

radius of inner circle =14 cm


(ii) Area of the ring =3465 cm2

Circumference of inner circle =88 cm

radius =88×72×22=14 cm

Let radius of outer circle =R 

Area of ring =π(R2r2)

=227(R2142)cm2=227(R2196)cm2

227(R2196)=3465

R2196=3465×722=11025

R2=11025+196=30625

R=30625=175

Radius of outer circle =175 cm


Question 12

A road 3.5 m wide surrounds a circular plot whose circumference is 44 m. Find the cost of paving the road at ₹50 per m2.

Sol :

Area of the Road =π(R2r2)

=227(105272)m2

=227(105+7)(1057)m2

=227×175×35=1925 m2

Rate of paving the road =50 per m2.

Total cost =1925×50

=9625


Question 13

The sum of diameters of two circles is 14 cm and the difference of their circumferences is 8 cm. Find the circumference of the two circles.

Sol :
Sum of the diameters of two circles =14 cm 
Let R and r be the radii of two circles
2R+2r=14
R+r=7...(i)
(Dividing by 2)
Difference of their circumference =8 cm
2πR2πr=8
2π(Rr)=82×227(Rr)=8
Rr=8×72×22=1411...(ii)

Adding (i) and (ii)
2R=7+1411=77+1411=9111
R=9111×2=9122

From (i)
R+r=7
9122+r=7r=79122

r=1549122=6322

Now , circumference of first circle

=2πR=2×227×9122 cm

=26 cm

and the circumference of second circle

=2πR=2×227×6322=18 cm


Question 14

Find the circumference of the circle whose area is equal to the sum of the areas of three circles with radius 2 cm, 3 cm and 6 cm.

Sol :
Radius of first circle =2 cm Area =πr2=π(2)2=4πcm2
Radius of second circle =3 cm Area =πr2=π(3)2=9πcm2
Radius of third circle =6 cm
Area =πr2=π(6)2=36πcm2
Total area of the three circles =4π+9π+36π=49πcm2
or Area of the given circle =49πcm2
radius =49ππ=49=7 cm
and circumference =2πr=2×227×7
=44 cm

Question 15

A copper wire when bent in the form of a square encloses an area of 121 cm2. If the same wire is bent into the form of a circle, find the area of the circle.

Sol :
Area of the square =121 cm2 
side =121=11 cm
Perimeter =4a=4×11=44 cm

Now, circumference of the circle =44 cm
radius =44×72×22=7 cm
and area of the circle =πr2=227(7)2
=227×7×7
=154 cm2


Question 16

A copper wire when bent into an equilateral triangle has area 121√3 cm2. If the same wire is bent into the form of a circle, find the area enclosed by the wire.

Sol :
Area of the equilateral triangle =1213 cm2
Let side of the triangle =a
area =34a2

34a2=1213

a2=121×3×43

a2=484

a=484=22 cm

Length of wire=66 cm

radius of the circle =662π=66×72×22=212 cm

Hence area of the circle =πr2

=227×(212)2

=227×212×212 cm2

=6932=3465 cm2


Question 17

(a) Find the circumference of the circle whose area is 16 times the area of the circle with diameter 7 cm.

(b) In the given figure, find the area of the unshaded portion within the rectangle. (Take π = 3.14)








Sol :

(a) Diameter of the circle =7cm

Radius =72 cm

and area =πr2

=227×72×72=772 cm2

Now, area of the bigger circle

=772×16=616 cm2

Let radius=r

πr2=616

227r2=616

r2=616×722

r2=196 cm2

r=196=14 cm.

∴Circumference 

=2πr=2×227×14

=88 cm


(b) In the figure radius of each circle=3 cm

∴Diameter=2×3 cm=6 cm

Length of rectangle (l)=6+6+3=15 cm and breadth (b)=6 cm

Area of rectangle = length × breadth

=15×6=90 cm2

and area of 212 circles =52πr2

=52×3.14×3×3 cm2

=5×1.57×9 cm2

=70.65 cm2

Area of unshaded portion

=90 cm270.65 cm2

=19.35 cm2


Question 18

In the adjoining figure, ABCD is a square of side 21 cm. AC and BD are two diagonals of the square. Two semicircle are drawn with AD and BC as diameters. Find the area of the shaded region. Take π = 227

Sol :








We have , side=21 cm

Area of square =Side2=212=441 cm2

We know,

AOD+COD+AOB+BOC=441 cm2

x+x+x+x=441 cm2

4x=441 cm2

x=4414=110.25 cm2

In this question, we have to find the area of shaded portion in square 

ABCD which is AOD and BOC

AOD+BOC=110.25+110.25 cm2

=220.5 cm2

Now,

Area of two semcircle =πr2

=227×10.5×10.5=346.50 cm2

Area of shaded portion =220.5+346.5 cm2

=567 cm2


Question 19

(a) In the figure (i) given below, ABCD is a square of side 14 cm and APD and BPC are semicircles. Find the area and the perimeter of the shaded region.

(b) In the figure (ii) given below, ABCD is a square of side 14 cm. Find the area of the shaded region.

(c) In the figure (iii) given below, the diameter of the semicircle is equal to 14 cm. Calculate the area of the shaded region. Take π = 227




















Sol :

(a) ABCD is a square whose each side (a) =14 cm

APD and BPC are semi-circle with diameter 14 cm each

Radius of each semi circle (a)=142=7 cm


(i) Area of square =a2=(14)2=196 cm2

and area of two semicircles

=2×12πr2=πr2

=227×7×7=154 cm2


(ii) Length of arcs of two semicircles =2πr

=2×227×7=44 cm

Perimeter of shaded portion =44+14+14 cm=72 cm


(b) ABCD is a square whose each side (a)

=14 cm

Four circles are drawn which touch each other and also touch the sides of the square as shown

Radius of each circle (r)=72=3.5 cm

(i) Area of square ABCD=a2=(14)2 cm2

=196 cm2

and area of 4 circles =4×πr2

=4×227×72×72 cm2=154 cm2

(ii) Perimeter of 4 circles =2×2πr

=4×2×227×72=88 cm

Perimeter of shaded portion

=Perimeter of 4 circles + Perimeter of square 

=88+4×14=88+56=144 cm


(c) Area of a rectangle ACDE=ED×AE =14×7=98 cm2

Area of semicircle DEF=πr22

=22×7×77×2=77 cm2

Area of shaded region =77+(982×14
×227×7×7)cm2
=77+21=98 cm2

Question 20

(a) Find the area and the perimeter of the shaded region in figure (i) given below. The dimensions are in centimetres.
(b) In the figure (ii) given below, area of ∆ABC = 35 cm2. Find the area of the shaded region.





Sol :

(a) There are two semicircle, smaller is inside the larger radius of larger semicircles
(R) 14 cm and radius of smaller circle
(r)=142=7 cm

(i) Area of shaded portion
=Area of larger semicircle-Area of smaller circle

=12πR212πr2
=12π(R2r2)=12×227[14272]cm2
=117[14+7][147]cm2
=117×21×7 cm2=231 cm2


(ii) Perimeter of shaded portion = Circumference of larger semicircle+ circumference of smaller semicircle + Radius of larger semicircle =πR+πr+R
=227×14+227×7+14 cm
=44+22+14=80 cm

(b) Area of ΔABC which is formed in a semicircle =3.5 cm
Altitude CD=5 cm

Base AB= Area ×2 Altitude =35×25 cm=14 cm

Diameter of semicircle =14 cm

and then radius (R)=142=7 cm

Area of semicircle =12πR2=12×227×7×7 cm2=77 cm2

Area of shaded portion = Area of semicircle Area of triangle =77 cm235 cm2=42 cm2


Question 21

(a) In the figure (i) given below, AOBC is a quadrant of a circle of radius 10 m. Calculate the area of the shaded portion. Take π = 3.14 and give your answer correct to two significant figures.

(b) In the figure, (ii) given below, OAB is a quadrant of a circle. The radius OA = 3.5 cm and OD = 2 cm. Calculate the area of the shaded portion.












Sol :

(a) In the figure, shaded portion = quadrant ΔAOB

Radius of the quadrant =10 m

Now Area of quadrant

=14πr2=14×314×10×10

=314×1004=3144=785 m2

and area of ΔAOB=12 AO×OB

=12×10×10=50 m2

Area of shaded portion =78550=285 m2


(b) In the figure (ii) radius of quadrant =35 cm

(i) Area of quadrant =14×πr2

=14×227×3.5×3.5=9.625 cm2

(ii) Area of ΔAOD=12×AO×OD

=12×3.5×2=3.5 cm2

Area of shaded portion = Area of quadrant Area of ΔAOD

=9.6253.6 cm2=6.125 cm2


Question 22

A student takes a rectangular piece of paper 30 cm long and 21 cm wide. Find the area of the biggest circle that can be cut out from the paper. Also find the area of the paper left after cutting out the circle. (Take π = 227)

Sol :

Length of rectangle =30 cm. and width =21 cm. Area of rectangle =l×b

=30×21=630 cm2

Radius of the biggest circle =212 cm










(i) Area of the circle =πr2

=227×212×212=6932 cm2=3465 cm2

Area of remaining part =6303465

=2835 cm2


Question 23

A rectangle with one side 4 cm is inscribed in a circle of radius 2.5 cm. Find the area of the rectangle.

Sol :
In rectagle ABCD,AB=4 cm
AC= diameter of circle =25 cm×2=50 cm










BC=AC2AB2

=(5)2(4)2=2516=9=3 cm

Area of rectangle =AB×BC

=4×3=12 cm2


Question 24

(a) In the figure (i) given below, calculate the area of the shaded region correct to two decimal places. (Take π = 3. 142).

(b) In the figure (ii) given below, ABC is an isosceles right angled triangle with ∠ABC = 90°. A semicircle is drawn with AC as diameter. If AB = BC = 7 cm, find the area of the shaded region. Take π = 227 







Sol :

(a) In the figure, ABCD is a rectangle inscribed in a circle whose length =12 cm and width =5 cm.

AC=AB2+BC2=(12)2+(5)2

=144+25=169=13 cm

Diameter of the circle =AC=13 cm

radius =132 cm=65 cm

Area of the circle =πr2 =3142×(65)2 cm2=3142×4225

=13275 cm2

Area of rectangle =l×b =12×5=60 cm2

Area of the shaded portion =132756000=7275 cm2


(b) Area of ΔABC=12×AB×BC

=12×7×7 cm2=492 cm2

=AC2=AB2+BC2=49+49

AC=72

So, radius of the semi-circle =722 cm

Area of the semi-circle =π2×(722)2 cm2

So, area of the shaded region = Area of the semi-circle Area of ABC

=(772492)cm2=282=14 cm2


Question 25

A circular field has perimeter 660 m. A plot in the shape of a square having its vertices on the circumference is marked in the field. Calculate the area of the square field.

Sol :
Perimeter of circular field=660 m










Radius of the field =6602π

=660×72×22=105 m

ABCD is a square which is inscribed in the circle whose diagonal is AC, which is the diameter of the circular field.

Let a be the side of the square

AC=2aa=AC2=105×22

a=105×2×22×2

=105×2×22=1052 m

Area of the square =a2=(1052)2

=1052×1052 m2

=22050 m2


Question 26

In the adjoining figure, ABCD is a square. Find the ratio between









(i) the circumferences

(ii) the areas of the incircle and the circumcircle of the square.

Sol :
Let the side of the square =2a 
Area =(2a)2=4a2
and diagonal of AC=2AB

(i) The radius of the circumcircle =12AC

=12(2×AB)

=22×2a=2a

Circumference =2πr=2×π×2a=22πa

The radius of the incircle =AB=12×2a=a

Circumference =2πr=2πa

Ratio between the circumference incircle and circuin circle

=2πa:22πa=1:2


(ii) Area of incircle =πr2=πa2

Area of circumcircle =πR2=π(2a)2

=π2a2=2πa2

Ratio =πa2:2πa2=1:2


Question 27

(a) The figure (i) given below shows a running track surrounding a grassed enclosure PQRSTU. The enclosure consists of a rectangle PQST with a semicircular region at each end.

PQ = 200 m ; PT = 70 m.













(i) Calculate the area of the grassed enclosure in m2.

(ii) Given that the track is of constant width 7 m, calculate the outer perimeter ABCDEF of the track.

(b) In the figure (ii) given below, the inside perimeter of a practice running track with semi-circular ends and straight parallel sides is 312 m. The length of the straight portion of the track is 90 m. If the track has a uniform width of 2 m throughout, find its area.

Sol :

(a) Length of PQ=200 m and width PT=70 m

(i) Area of rectangle PQST=l×b

=200×70=14000 m2

Radius of each semi-circular part on either side of the rectangle

=702=35 m

Area of both semi-circular parts

=2×12πr2=227×35×35 m2

=3850 m2

Total area of grassed enclosure

=14000+3850=17850 m2


(ii) Width of track around the enclosure =7 m Outer length =200 m

and  width =70+7×2

=70+14=84 m

outer radius =842=42 m

Circumference of both semi-circular part

=2×πr=2×227×42=264 m

Outer perimeter =264+200×2 m

=264+400=664 m


(b) Inside perimeter =312 m 

Total length of the parallel sides

=90+90=180 m

Circumference of two semi-circles

=312180=132 m

Radius of each semi-circle

=1322π=66314 m=2102 m

Diameter =66π×2=132π=132314 m

=132×100314=4204 m

Width of track=2 m

∴Outer diameter=42.04+4

=46.04 m

Radius=46042=2302 m

Now area of two semi-circles

=2×12×πR2

=πR2=314×(2302)2 m2

=314×2302×2302 m2

=166395 m2

and area of rectangle=90×46.04

=41436 m2

Total area =166395+414360=580755 m2

and area of two inner circles =2×12πr2

=314×2102×2102 m2

=138738 m2

and area of inner rectangle

=90×4204 m2

=37836 m2

Total inner area=3783.60+1387.38

=517098 m2

Area of path =580755517098

=63657 m2


Question 28

(a) In the figure (i) given below, two circles with centres A and B touch each other at the point C. If AC = 8 cm and AB = 3 cm, find the area of the shaded region.

(b) The quadrants shown in the figure (ii) given below are each of radius 7 cm. Calculate the area of the shaded portion.













Sol :

(a) AC=8 cm

BC=AC-AB=8-3=5 cm

Area of big circle of radius AC=πR2

=227×8×8 cm2=64×227 cm2

and area of smaller circle

=πr2=227×5×5=25×227 cm2

∴Area of shaded portion

=64×22725×227

=227(6425)cm2=227×39 cm2

=12257 cm2


(b) Radius of each quadrant=7 cm







Area of shaded region

=Area of square-4 area of the quadrant

=( side )24×14πr2

=(14)2227×7×7 cm2

=196154=42 cm2


Question 29

(a) In the figure (i) given below, two circular flower beds have been shown on the two sides of a square lawn ABCD of side 56 m. If the centre of each circular flower bed is the point of intersection O of the diagonals of the square lawn, find the sum of the areas of the lawn and the flower beds.
















(b) In the figure (ii) given below, a square OABC is inscribed in a quadrant OPBQ of a circle. If OA = 20 cm, find the area of the shaded region. (Use π = 3.14)
Sol :
(a) Side of square lawn ABCD(a)=56 cm
Area =a2=(56)2
=3136 cm2

Length of the diagonal of the square =2a
=2×56 cm















Radius of each quadrant =2×562

=282 cm

Area of each segment

=14πr2areaΔOBC

=14×227×282×28212× 282×282

=282×282(14×22712)

=784×2(111412)

=784×2×414=448 cm2

Area of two segments =448×2=896 cm2

Total area of the lawn and beds

=3136+896=4032 cm2


(b) In the figure OPBQ is a quadrant and OABC is a square which is inscribed in it side of square =20 cm

OB is joined

OB=2a=2×20 cm

Radius of quadrant =OB=202 cm

Now, area of quadrant =14πr2

=14×3.14×(202)2

=14×3.14×800 cm

=314×2=628 cm2

Area of square =a2=(20)2

=400 cm2

Area of shaded portion =628400 =228 cm2


Question 30

(a) In the figure (i) given below, ABCD is a rectangle, AB = 14 cm and BC = 7 cm. Taking DC, BC and AD as diameters, three semicircles are drawn as shown in the figure. Find the area of the shaded portion.











(b) In the figure (ii) given below, O is the centre of a circle with AC = 24 cm, AB = 7 cm and ∠BOD = 90°. Find the area of the shaded region. (Use π = 3.14).












Sol :
(a) In the figure ABCD is a rectangle. Three semicircles are drawn as shown in the figure AB=14 cm, BC=7 cm






Area of rectangle ABCD

=l×b=14×7=98 cm2

The radius of each outer semicircles =72 cm

Area =2×12πr2

=227×72×72=772 cm2

=38.5 cm2

Area of semicircle drawn on CD as diameter

=12πR2=12×227×(7)2

=117×7×7=77 cm2

Area of shaded region

=(98+38.577)cm2=59.5 cm2


(b) In the given figure, AC=24 cm,AB=7 cm

BOD=90

In ΔABC,

BC2=AC2+AB2  (Pythagoras Theorem)
=242+72=576+49
=625=25 cm

Radius of the circle =252 cm

Now area of ΔABC=12AB×AC
=12×7×24
=84 cm2

and area of quadrant COD
=14πr2=14×3.14×252×252 cm2

=1962.516=122.66 cm2

Area of circle =πr2=3.14×252×252 cm2

=1962.54=490.63 cm2

Area of shaded portion = Area of circle

(Area of ΔABC+ area of quad. COD) =490.63(84+122.66)cm2

=490.63206.66 cm2

=283.97 cm2


Question 31

(a) In the figure given below ABCD is a square of side 14 cm. A, B, C and D are centres of the equal circle which touch externally in pairs. Find the area of the shaded region.













(b) In the figure (ii) given below, the boundary of the shaded region in the given diagram consists of three semi circular arcs, the smaller being equal. If the diameter of the larger one is 10 cm, calculate.

(i) the length of the boundary.

(ii) the area of the shaded region. (Take π to be 3.14)








Sol :

(a) Side of square ABCD=14 cm Radius of each circle drawn from A,B,C and D and touching externally in pairs

=142=7 cm

Now area of square =a2=14×14 =196 cm2

Area of 4 sectors of 90 each =4×π×r2

=4×227×7×7×14

=154 cm

Area of each sector of 270 angle =34πr2

=34×227×7×7 cm2

=2312=115.5 cm2

Area of 4 sectors =115.5×4=462 cm2

Area of shaded portion = Area of square +area of 4 bigger sector - area of 4 smaller sector =196+462154

=658154=504 cm2


(b) Radius of big semi-circle =102=5 cm

and radius of each smaller circle =52 cm

(i) Length of the boundary=Circumference of bigger semi-circle+2 circumference of smaller semi-circles

=πR+πr+πr

=314(R+2r)=314(5+2×52)

=314×10=314 cm

(ii) Area of shaded region=Area of bigger semi circle+Area of one smaller semi circle - area of other smaller semi circle 

=area of bigger semi-circle =12πR2

=3142×5×5=157×25

=39.25 cm


Question 32

(a) In the figure (i) given below, the points A, B and C are centres of arcs of circles of radii 5 cm, 3 cni and 2 cm respectively. Find the perimeter and the area of the shaded region. (Take π = 3.14).

(b) In the figure (ii) given below, ABCD is a square of side 4 cm. At each corner of the square a quarter circle of radius 1 cm, and at the centre a circle of diameter 2 cm are drawn. Find the perimeter and the area of the shaded region. Take π = 3.14.








Sol :

(a) Radius of bigger circle =5 cm. Radius of small circle (r1)=3 cm and radius of smaller circle (r2)=2 cm

(i) Perimeter of the shaded region=Circumference of bigger semi-circle+ circumference of small semi-circle+ circumference of smaller semi-circle

=πR+πr1+πr2=π(R+r1+r2)

=π(5+3+2)=3.14×10=31.4 cm2


(ii) Area of the shaded region= Area of bigger semi-circle + Area of smaller semi-circle - Area of small semicircle

=12πR2+12πr2212πr21

=12π(R2+r22r21)

=12π(52+2232)

=12π(25+49)=12π×20 cm2

=10×3.14=31.4 cm2


(b) Side of square ABCD=4 cm Radius of each quadrant circle =1 cm. and radius of circle in the square

=22=1 cm


(i) Perimeter of shaded region = Circumference of four quadrants + Circumference of circle+4×12 side of square.

=4×14(2πr)+(2πr)+4×2 cm

=2πr+2πr+8 cm

=4πr+8=4×314×1 cm+8 cm

=1256 cm+8 cm=2056 cm


(ii) Area of shaded regions=Area of square - area of 4 quadrants - area of circle

=( side )24×14πr2πr2

=(4)2πr2πr2=162πr2 cm2

=162×314×(1)2

=16628 cm2=972 cm2


Question 33

(a) In the figure given below, ABCD is a rectangle. AB = 14 cm, BC = 7 cm. From the rectangle, a quarter circle BFEC and a semicircle DGE are removed. Calculate the area of the remaining piece of the rectangle. (Take π = 22/7)

(b) The figure (ii) given below shows a kite, in which BCD is in the shape of a quadrant of circle of radius 42 cm. ABCD is a square and ∆ CEF is an isosceles right angled triangle whose equal sides are 6 cm long. Find the area of the shaded region.























Sol :

(a) Area of remaining piece

Area of rectangle ABCD-Area of semi circle DGE-Area of quarter BFEC

=14×712×π(72)214π×72

=14×712×227×72×7214×227×7×7

=987741544

=9819.2538.5

=9857.75=40.25 cm2


(b) In the figure, ABCD is a square whose side=radius of the quadrant=42 cm

ΔCEF is an isosceles right triangle whose, each equal side=6 cm

Now, the area of the shaded portion = Area of the quadrant + area of isosceles right triangle

=14πr2+12EC×FC

=14×227×42×42+12×6×6

=1386+18

=1404 cm2


Question 34

(a) In the figure (i) given below, the boundary of the shaded region in the given diagram consists of four semi circular arcs, the smallest two being equal. If the diameter of the largest is 14 cm and of the smallest is 3.5 cm, calculate

(i) the length of the boundary.

(ii) the area of the shaded region.

(b) In the figure (ii) given below, a piece of cardboard, in the shape of a trapezium ABCD, and AB || DC and ∠BCD = 90°, quarter circle BFEC is removed. Given AB = BC = 3.5 cm and DE = 2 cm. Calculate the area of the remaining piece of the cardboard.











Sol :

(a) (i) Length of boundary = Circumference of bigger semi-circle+ Circumference of small semi-circle+2× circumference of the smaller semi-circles

=πR+πr1+2×πr2=π(R+r1)+2πr2

=227(7+35)+2×227×352

=227×105+11

=33+11=44 cm


(ii) Area of shaded region=Area of bigger semi circle+Area of small semicircle -2×Area of smaller semicircle

=12π(7)2+12π(35)22×12π(175)2

=12×227×7×7+12×227×35×35 227(175)×(175)

=770+19259625=86625 cm2


(b) ABCD is a trapezium in which

ABDC and C=90

AB=BC=35 cm,DE=2 cm

Radius of quadrant =35 cm.

Area of trapezium =12(AB+DC)×BC

=12(35+35+2)×35 cm2

=12(9×35)=45×35=1575 cm2

Area of quadrant =14πr2

=14×227×35×35 cm2

=9625 cm2

Area of shaded portions =15759625=6125 cm2


Question 35

(a) In the figure (i) given below, ABC is a right angled triangle, ∠B = 90°, AB = 28 cm and BC = 21 cm. With AC as diameter a semi-circle is drawn and with BC as radius a quarter circle is drawn. Find the area of the shaded region correct to two decimal places.

(b) In the figure (ii) given below, ABC is an equilateral triangle of side 8 cm. A, B and C are the centres of circular arcs of equal radius. Find the area of the shaded region correct upto 2 decimal places.    (Take π = 3.142 and √3 = 1.732).












Sol :
(a) In right ΔABC,B=90
AC2=AB2+BC2
=(28)2+(21)2
=784+441=1225
AC=1225=35 cm
Radius of semi-circle (R)=352
and radius of quadrant (r)=21 cm 

Area of shaded region
=Area of ΔABC+ area of semi-circle- area of quadrant

=12×28×21+12πR214r2 cm2
=294 cm+12×227×352×352 14×227×21×21
=294+192546932
=294+481253465 cm2
=775.25346.50=42875 cm2


(b) ΔABC is an equilateral triangle of side 8 cm. At A,B and C as centre three circular arcs of equal radius.

Radius =84=4 cm
Now area of ΔABC
=34a2=34(8×8)cm2
=34×64=163 cm2
=16(1732)=27712 cm2

Area of 3 equal sectors of 60 whose radius 
=4 cm

=3×πr2×60360

=3×3.142×4×4×16 cm2

=3142×8=25136 cm2

Area of shaded region =2771225136

=2576 cm2=258 cm2


Question 36

A circle is inscribed in a regular hexagon of side 2√3 cm. Find

(i) the circumference of the inscribed circle

(ii) the area of the inscribed circle

Sol :

ABCDEF is a regular hexagon of side 2√3 cm. and a circle is inscribed in it with centre O.












Radius of inscribed circle
=32× side of regular hexagon
=32×23=3 cm

(i) Circumference of the circle =2πr

=2π×3=6×227 cm=1327 cm

(ii) Area of the circle =πr2=π×3×3

=9×227=1987 cm2


Question 37

In the figure (i) given below, a chord AB of a circle of radius 10 cm subtends a right angle at the centre O. Find the area of the sector OACB and of the major segment. Take π = 3.14.










Sol :

Radius of circle=10 cm

Angle at the centre subtended by a chord AB=90

=314×10×10×90360

=314×14=785 cm2

Area of ΔOAB=12×10×10=50 cm2

Area of minor segment 

= Area of sector ΔACB Area of ΔOAB =78550=285 cm2


Area of circle =πr2

=314×10×10=314 cm2

So Area of Major segment

= Area of circle Area of minor segment 

=314285=2855 cm2

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