20. Chapter 20. Perimeters and Areas of Plane figures NIOS Class 10 Mathematics Textbook Solutions
Chapter 20. Perimeters and Areas of Plane figures NIOS Class 10 Mathematics Textbook Solutions for Exam Preparation
Chapter 20: PERIMETERS AND AREAS OF PLANE FIGURES
CHECK YOUR PROGRESS 20.1
1. Area of a square field is 225 m² . Find the perimeter of the field.
Solution: Let the side of the square be x (in m).
A/Q, x2=225 
⇒x2=152⇒x=15 m 
Therefore, the perimeter of the field =4x=4×15=60 m 
.
2. Find the diagonal of a square whose perimeter is 60 cm.
Solution: Let the side of the square be x m.
A/Q, 4x=60 
⇒x=604⇒x=15 cm 
Therefore, the diagonal of a square =2x²=x2=152 
cm.
3. Length and breadth of a rectangular field are 22.5 m and 12.5 m respectively. Find:
(i) Area of the field
(ii) Length of the barbed wire required to fence the field .
Solution: Here, l=22.5 m and b=12.5 m 
(i) Area of the rectangular field =l×b=22.5×12.5=281.25 m² 
(ii) The length of the barbed wire required to fence the field =2l+b 
=222.5+12.5=2×35=70 m
4. The length and breadth of rectangle are in the ratio 3 : 2. If the area of the rectangle is 726 m² , find its perimeter.
Solution: Let, 3x and 2x 
be the length and breadth of rectangle .
A/Q, 3x×2x=726 
⇒6x2=726 
⇒x2=7266=121
⇒x2=112
⇒x=11 m
Therefore, the perimeter of the rectangle =23x+2x 
=2×5x=10×11=110 m 
5. Find the area of a parallelogram whose base and corresponding altitude are respectively 20 cm and 12 cm.
Solution: Here , b=20 cm and h=12 cm 
We know that, Area of the parallelogram = base × corresponding altitude
Therefore, the area of a parallelogram =b×h=20×12=240 cm² 
6. Area of a triangle is 280 cm² . If base of the triangle is 70 cm, find its corresponding altitude.
Solution: Here, b=70 cm 
and Area of a triangle = 280 cm²
Let the corresponding altitude be h (in cm).
A/Q,
280=12×b×h
⇒12×70×h=280 
⇒35h=280 
⇒h=28035=8 
⇒h=8 cm 
Therefore, the corresponding altitude of the triangle is 8 cm .
7. Find the area of a trapezium, the distance between whose parallel sides of lengths 26 cm and 12 cm is 10 cm.
Solution: We know that,
Area of a trapezium =12sum of the two parallel sides×distance between them 
=12×26+12×10 
=12×38×10=19×10=190 cm² 
8. Perimeter of a rhombus is 146 cm and the length of one of its diagonals is 48 cm. Find the length of its other diagonal.
Solution: Given, the perimeter of rhombus = 146 cm.
Side length = 1464=36.5 cm.
Let the diagonals be d1 and d2 
. Here d1=48 cm
Using Pythagoras theorem,
d122+d222=36.52 
⇒4822+d222=36.52 
⇒242+d222=36.52 
⇒d222=36.52-242=1332.25-576=756.25 
⇒d22=756.25=27.5 
⇒d2=2×27.5 
⇒d2=55 cm
CHECK YOUR PROGRESS 20.2
1. Find the area of a triangle of sides 15 cm, 16 cm and 17 cm.
Solution: Here, a=15 cm , b=16 cm and c=17 cm 
s=a+b+c2=15+16+172=482=24 cm 
s-a=24-15=9 cm 
, s-b=24-16=8 cm 
, s-c=24-17=7 cm
The area of a triangle =ss-as-bs-c 
=24×9×8×7 
=4×6×32×4×2×7 
=42×32×3×2×2×7 
=4×3×23×7=2421 cm² 
2. Using Heron’s formula, find the area of an equilateral triangle whose side is 12 cm. Hence, find the altitude of the triangle.
Solution: Here, a=12 cm , b=12 cm and c=12 cm 
s=a+b+c2=12+12+122=362=18 cm 
s-a=18-12=6 cm 
, s-b=18-12=6 cm 
, s-c=18-12=6 cm
The area of a triangle =ss-as-bs-c
=18×6×6×6 
=3×6×6×6×6 
=3×62×6² 
=6×6×3 =363 cm² 
The altitude of the triangle =32×side=32×12=3×6=63 cm 
CHECK YOUR PROGRESS 20.3
1. There is a 3 m wide path on the inside running around a rectangular park of length 48 m and width 36 m. Find the area of the path.
Solution: Given, Length = 48 m , Width = 36 m and Path width = 3 m (on the inside)
Area of park PQRS =48×36=1728 m² 
Inner length=48-23=48-6=42 m
Inner width=36-23=36-6=30m
Area of the inner rectangle ABCD = 42×30=1260 m² 
Area of the path= Area of PQRS- Area of inner ABCD =1728-1260=468 m² 
Therefore, the area of the path is 468 m².
2. There are two paths of width 2 m each in the middle of a rectangular garden of length 80 m and breadth 60 m such that one path is parallel to the length and the other is parallel to the breadth. Find the area of the paths.
Solution: Given, Length = 80 m Breadth = 60 m
Two paths, each 2 m wide, one parallel to the length and one parallel to the breadth.
Area of the path parallel to the length: Length = 80 m, Width = 2 m
Area of the first path =80×2=160 m² 
Area of the path parallel to the breadth: Length = 60 m, Width = 2 m
Area of the second path =60×2=120 m² 
Overlap area of the path = 2 × 2 = 4 m²
Total area of the paths=160 + 120 – 4 = 276 m²
The total area of the paths is 276 m².
3. Find the area of the rectangular figure ABCDE given in Fig. 20.6, where EF, BG and DH are perpendiculars to AC, AF = 40 m, AG = 50 m, GH = 40 m and CH = 50 m.
Solution:
4. Find the area of the figure ABCDEFG in Fig. 20.7, where ABEG is a trapezium, BCDE is a rectangle, and distance between AG and BE is 2 cm.
Solution:
CHECK YOUR PROGRESS 20.4
1. The radii of two circles are 9 cm and 12 cm respectively. Find the radius of the circle whose area is equal to the sum of the areas of these two circles.
Solution: Let r
be the radius of new circle .
Here, r1=9 
cm and r2=12 
cm .
A/Q, πr2=πr12+πr22 
⇒r2=r12+r22
⇒r=92+122
⇒r=81+144
⇒r=225=15²=15 
cm
Therefore, the radius of the new circle is 15 cm .
2. The wheels of a car are of radius 40 cm each. If the car is travelling at a speed of 66 km per hour, find the number of revolutions made by each wheel in 20 minutes.
Solution: Here , r=40 cm
The perimeter of the wheels of a car =2πr
=2×227×40 cm
=44×407cm=17607cm
The speed of the car
=66kmhr =66×1000×10060cmmin.
=110000cmmin.
The distances travel by the car =
Speed of the car × time [ Time =20
min.]
=110000×20cmmin.×min.
=2200000 cm
The number of complete revolution of the wheels of the car
=220000017607
=154000001760=8750 
3. Around a circular park of radius 21 m, there is circular road of uniform width 7 m outside it. Find the area of the road.
Solution: Here, r=21 m and R=21+7=28 m 
The area of the road =πR2-πr2=πR2-r2 
=227 282-212 
=227 28+2128-21 
=227×49×7=22×49=1078 m² 
CHECK YOUR PROGRESS 20.5
1. Find the perimeter and area of the sector of a circle of radius 14 cm and central angle 30° .
Solution: Here, r=14 cm and θ=30° 
The perimeter of the sector =2r+θ180°×πr 
=2×14+30°180°×227×14 
=28+16×22×2=28+223=84+223=1063 cm 
Area of the sector =θ360°×πr2
=30°360°×227×14×14 
=112×22×2×14=11×143=1543 cm² 
2. Find the perimeter and area of the sector of a circle of radius 6 cm and length of the arc as 11 cm.
Solution: Here, r=6 cm and l=11 cm 
The perimeter = l + 2 × r 
=11+2×6=11+12=23 cm 
We have,
l=θ180°×πr 
⇒11=θ180°×227×6 
⇒θ=11×180°×722×6=180°×72×6=180°×712
⇒θ=15°×7=105° 
⇒θ=105° 
So, area of the sector
=θ360°×πr2
=105°360°×227×6×6 
=105°360°×227×36 
=105°10°×227=51×117=3×11=33 cm² 
CHECK YOUR PROGRESS 20.6
1. A square ABCD of side 6 cm has been inscribed in a quadrant of a circle of radius 14 cm (See Fig. 20.14). Find the area of the shaded region in the figure.
Solution: Given, ABCD is a square where AB = BC = CD = AD = 6 cm .
Area of the square ABCD = AB × BC = 6 × 6 = 36 cm²
For quadratic : Here, r=14 cm 
Area of the quadrant of a circle =14πr²=14×227×14×14=11×14=154 cm² 
Therefore, the area of the shaded region = Area of the quadrant of a circle – Area of the square.
=154-36=118 cm² 
2. A shaded design has been formed by drawing semicircles on the sides of a square of side
length 10 cm each as shown in Fig. 20.15. Find the area of the shaded region in the design.
Solution: Let us mark the four unshaded regions as I, II, III and IV (see Fig. 12.18).
Area of I + Area of III = Area of ABCD – Areas of two semicircles of each of radius 5 cm
=10×10-2×12πr2
=100-πr²=100-227×5²=100-3.14×25 
= (100-78.5) cm² = 21.5 cm²
Also, Area of II + Area of IV = 21.5 cm²
So, area of the shaded design = Area of ABCD – Area of (I + II + III + IV)
= (100 – 2 × 21.5) cm² = (100 – 43) cm² = 57 cm²
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