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**Joy of Mathematics Class 8 Solutions Chapter 20 Perimeter and Area of Plane Figures**

Welcome to NCTB Solution. Here with this post we are going to help 8th class students for the Solutions of Joy of Mathematics Class 8 Book, Chapter 20 Perimeter and Area of Plane Figures. Here students can easily find step by step solutions of all the problems for Perimeter and Area of Plane Figures. Exercise wise proper solutions for every problems. All the problem are solved with easily understandable methods so that all the students can understand easily. Here students will find solutions for Exercise 20.1, 20.2 and 20.3

**Perimeter and Area of Plane Figures Exercise 20.1 Solution**

**Question no – (1)**

**Solution : **

The ratio of the sides = 1 : 3 : 4

Let, the sides be x, 3x and 4x

**∴** Perimeter of the triangle of △PQR = (x + 3x + 4x) cm

According to the problem,

x + 3x + 4x = 40

=> 8x = 40

=> x = 40/8

= 5

Therefore the sides of the triangles are 5 cm 15 cm and 20 cm.

**Question no – (2)**

**Solution :**

Here, the length of the rectangular = 100 m

Breadth of the rectangular = 55 m

**∴** Area of the rectangular

= 100 × 55 m^{2}

= 5500 m^{2}

Now, the side of the square = 156 m

**∴** Area of the square

= (156)^{2}

= 24336 m^{2}

**∴** Area is left

= (24336 – 5500) m^{2}

= 18836 m^{2}

**∴** Area in hectares

= 18836/10000

= 1.8836 hectares.

**Question no – (3)**

**Solution :**

**(a)** Let, length of the each side of the square is = x cm

**∴** Diagonal of the square = length of the side × √2

According to the question,

x × √2 = 4√2

=> x = 4√2/√2

=> x = 4

**∴** Length of the each side = 4 cm

**(b)** Here, length of the each side = 4 cm

**∴** Perimeter of the square = 4 × x cm

= 4 × 4 cm

= 16 cm

**Question no – (4)**

**Solution :**

Here, uniform path = 2 cm

Now,

New length,

= 50 + 2 + 2

= 54 m

New breadth,

= 36 + 2 + 2

= 40 m

**∴ **Area of the track

= {(54 × 40) – (50 × 36)} m

= {2160 – 1800} m

= 360 m

**∴** Thus the area of the track is 360 m.

**Question no – (5)**

**Solution :**

Here in Inward rectangular, EFGH

l_{1} = 27 – (6.25 + 6.25) m

= (27 – 12.5) m

= 14.5 m

b_{1} = 17 – (4 + 4) m

= (17 – 78) m

= 9 m

And, the out word rectangular ABCD,

l_{2} = 27m

b_{2} = 17m

**∴** Area of the shaded portion

= (27 × 17) – (14.5 × 9)

= (459 – 130.5) m^{2}

= 328.5 m^{2}

Therefore the area of the shaded portion 328.5m^{2}

**Question no – (6)**

**Solution :**

Here, length of the carpet = 4m

Breadth of the carpet = 2.5m

**∴** Area of the carpet

= (4 × 2.5) m^{2}

= 10 m^{2}

**∴** Area of the rectangular room = 16m^{2}

**∴** Now, Area of the floor

= (16 – 10) m^{2}

= 6 m^{2}

**Question no – (7)**

**Solution :**

Here, the length of the field = 70 m

Breadth of the field = 40 m

Let, wide of path = x m

Area of path = 1000m^{2}

Now, according to the question,

40 × 70 – {(40 – x – x) × (70 – x – x)} = 1000

=> 2800 – {(40 – 2x) (70 – 2x)} = 1000

=> 2800 – {2800 – 80x – 140x + 4x^{2}} = 1000

=> 2800 – 2800 + 80x + 140x – 4x^{2} = 1000

=> – 4x^{2} + 220x – 1000 = 0

=> x^{2} – 55x + 250 = 0

=> x^{2} – (50 + 5)x + 250 = 0

=> x^{2} – 50x – 5x (x – 50) = 0

=> (x – 50) (x – 5) = 0

**∴** x – 50 = 0

=> x = 50

x – 5 = 0

=> x = 5

Hence, width of the path will be 5 m.

**Question no – (8)**

**Solution :**

In 60 per m rate = 120600

In 1 per m rate = 120600/60

= 2010

Let, the breadth of the rectangular = x m

**∴** Perimeter of the rectangular

= 2(l + x) m

= 2(150 + x) m

Now, according to question,

=> 2 (150 + m) = 2010

=> 300 + 2x = 2010

=> 2x = 2010 – 300

=> 2x = 1710

=> x = 1710/2

= 855

Now, the area of the rectangular

= (150 × 855) m^{2}

= 128,250m^{2}

**Perimeter and Area of Plane Figures exercise 20.2 Solution : **

**Question no – (1)**

**Solution :**

Here, the base of length = 25 cm

Altitude of length = 8.4 cm

**∴** Thus the area of a parallelogram is

= (base × altitude) cm²

= (25 × 8 . 4) cm²

= (25 × 84/10) cm²

= 210 cm²

**∴** The area of the parallelogram is 210 cm²

**Question no – (2)**

**Solution :**

**Required Figure,**

Here , ABCD is a rhombus

AB = 13 CM

BC = 10 CM

Now , Area of the rhombus = 2 × area of △ABC

= 2 × 1/2 × AB × BC

= 2 × 1/2 × 13 × 10

= 130 cm²

So, the area of the rhombus 130 cm²

**Question no – (3)**

**Solution :**

Required Figure,

Here, area of parallelogram is same

AF × BC = AE × CD

Now , AE = 6.4 CM ,

AB = 18 cm ,

BC = 12 cm

AF × BC = AE × CD

= AE × 12 = 6.4 × 18

= AF = 9.6 cm

**Question no – (4)**

**Solution :**

Here, the perimeter of the parallogram = 60 cm

Let, the length of the side be 3x and 2x cm

Now, according to the question ,

2 (3x × 2x ) = 60

= 2 × 5x = 60

= 10x = 60

= x = 60/10

= x = 6

**∴** The side of the parallelogram are 18 and 12 cm

And, altitude corresponding to the larger side = 5 cm

**∴** Area of the parallelogram,

= base × altidude

= 18 × 5

= 90 cm²

Let, y be the altitude corresponding to the smaller side .

Now , Area of the parallelogram

= base × altitude

= 90 = 12 × y

= y = 90/12

= 7.5

**∴** The altitude corresponding to the smaller side is 7.5 cm .

**Question no – (5)**

**Solution** :

Here the side of triangle are 12 cm,

35 cm and 37 cm .

**∴** Now , a = 12 cm

B = 35 cm

C = 37 cm

And the area of triangle

= √s (s – a) (s – b) (s – c)

S = a + b + c/2 (From Her or′ s formula)

= 12 + 35 + 37/2

= 84/2

**∴** s = 42

Thus area of triangle,

= √42 (42 – 12) (42 – 35) (42 – 37) cm²

= √42 × 30 × 7 × 5 cm²

= √6 × 7 × 5 × 6 × 7 × 5 cm²

= 6 × 7 × 5 cm²

= 30 × 7 cm²

= 210 cm²

**∴** Therefore the area of triangle 210 cm²

**Question no – (6)**

**Solution :**

Here, side of the equilateral triangle is (x) = 12 cm

**∴** Thus the area of the equilateral triangle

= √3/4 x² cm²

= √3/4 × (12)² cm²

= √3/4 × 12 × 12 cm²

= 36√3 cm²

**∴** Area of the equilateral triangle is 36√3 cm²

**Question no – (7)**

**Solution :**

Let, the length of each side of the triangle is x cm .

**∴** Area of an equilateral triangle is

= √3/4 × x² cm²

According to the question ,

√3/4 × x² = 16 √3

= x² = 16 × 4 √3

= x² =16 × 4

= x = √4 × 4 × 2 × 2

= x = 4 × 2

= x = 8

Therefore, the length of each side of an equilateral triangle is 8 cm .

**Question no – (8)**

**Solution** :

Let, the one of the parallel side is x other sides = 10 cm .

Here, height = 11 cm

**∴** Area of the trapezium = 1/2 × (sum of the parallel side) × height

= 1/2 × (x + 10) × 11

According to the question ,

= 1/2 (10 + x) × 11 = 121

= 1/2 (10 + x) = 121/11

= 1/2 (10 + x) = 11

= 10 + x/2 = 11

= 10 + x = 22

= x = 22 – 10

= x = 12

Therefore, the length of the parallel side is 12 cm .

**Question no – (9)**

**Solution :**

Here, the parallel sides of the trapezium is ,

16 cm and 20 cm .

**∴** Height = 12 cm

**∴ **Area of the trapezium

= 1/2 × (16 + 20) × 12

= 1/2 × 36 × 12

= 216 cm²

**∴** Thus the area of the trapezium is 216 cm²

**Question no – (10)**

**Solution** :

Here , the parallel sides of the trapezium is = 12 m and 18 m

**∴** Distance between them = 6.4 m

**∴** Thus the area of the trapezium is = 1/2 × (sum of parallel sides) × height

= 1/2 × (18+12) × 6.4 cm²

= 1/2 × 30 × 6.4/10

= 92 cm²

Therefore, the area of the trapezium is 96 cm²

**Question no – (11)**

**Solution :**

Let , the length of the parallel side is = x cm

**∴** Other side of the trapezium is = 64 cm

**∴** Distance between two parallel side = 17 cm

**∴** Area of the trapezium = 1/2 × (64 + x) × 17 cm² according to the question ,

= 1/2 (64+x) × 17 = 850

= 64 + x/2 = 850/17

= 64 + x/2 = 50

= 64 + x = 100

= x = 100 – 64

= 36

Thus, length of the parallel side is 36 cm.

**Perimeter and Area of Plane Figures Exercise 20.3 Solution**

**Question no – (1)**

**Solution :**

**(a)** Here , the radius of the circle (r) = 14 dm

= 4 × 14 dm

= 5.6 dm

**∴** Area of the circle = πr² dm²

= π × (14)² dm²

= 22/7 × 14 × 14 dm²

= 616 dm²

**∴** circumference and area of the circle is 56 dm and respectively

**(b)** Here, radius (r) = 2.1 m

**∴** Circumference of the circle = 4 × r m

= 4 × 2.1 m

= 8.4 m

**∴** Area of the circle = πr²m²

= 22/7 × (2.1)² m²

= 22/7 × 2.1 × 2.1 m²

= 13.86 m²

**(c)** Here, radius (r) = 4.9 cm

**∴** Circumference of the circle = 4r cm

= 4 × 4.9 cm

= 19 .6 cm

**∴** Area of the circle = πr²

= π × 4.9 × 4.9 cm²

= 22/7 × 4.9 × 4.9 cm²

= 75.46 cm²

**Question no – (2)**

**Solution :**

Let, the radius of the circle is = x cm

**∴** Circumference of the circle is = 4x cm

According to the question,

4x = 154

=> x =154/4

= 38.5

**∴** Thus the radius of the circle is 38.5 cm

**Question no – (3)**

**Solution :**

Let, the radius of the circle = x cm

Now, circumference of the circle = 4x cm

According to the question,

4x = 44

=> x = 44/4 = 11

**∴** Thus the diameter of the circle

= (2 × 11) cm

= 22 cm

**Question no – (4)**

**Solution :**

Here, diameter of two circle is 3.5 cm and 4.2 cm

**∴** Radius of two circle is = 3.5/2 cm and 4.2/2 cm

**∴** Ratio of the area of the smaller circle to that of the bigger circle

= π (3.5/2)² : π (4.2/2)²

= 12.25/4 : 17.64/4

= 12.25 : 17.64

=1225 : 1764

= 175 : 252

= 25 : 36

**Question no – (5)**

**Solution** :

Let, the length of the square encloses is x

**∴** Area of the square enclose = x² , m²

Now, x² = 121

=> x = √121

= >X = 11

**∴** same wire is repent to from a circle and area of the circle

= π × (11)² cm²

= π × 11 × 11 cm²

=22/7 × 11 × 11 cm²

= 380.28 cm²

**∴** Thus the area of the circle is 380.28 cm²

**Question no – (6)**

**Solution** :

20,000 revolution going over a distance = 15.7 km

1 revolution going over a distance

= 15.7 × 10000/20000 × 10 cm

= 157/2 cm

Let, the radius of the wheel = r cm

**∴** Circumference of the wheel =2πr cm

Now, 2πr = 157 / 2

=> r = 157/2 × 2 × 3.14

= 12.5

So, the radius of the wheel 12.5 cm

**Question no – (7)**

**Solution** :

The radius of circle 12.5 cm

**∴** perimeter of the circle

= 2 × π × 56 cm

And, Let the side of the square = x cm

Now, 2 × π × 56 cm = 4 × x

=> x = 2 × π × 56 / 4

=> x = 2 × 22/7 × 14 [∵ π = 22/7]

=> x = 88

**∴** Thus the side of the square = 88 cm

**Question no – (8)**

**Solution** :

Let, the radius of the circle = r m

**∴** Circumference of the circle = 2πr m

According to the question,

2πr = 35.2

=> r = 35. 2 / 2π

=> 35.2 × 7 /10 × 2 × 22 [**∵** π = 22/7]

=> 56 / 10

=> r = 5.6

**∴** Thus the diameter of the circle = 2r

= 2 × 5.6

= 11.2 m

**∴** Area of the circle = πr²m²

= π.(5.6)²m²

= 22/7 × 5.6 × 5.6/10 × 10 m²

= 176 × 56/100 m²

= 9856/100 m²

= 98.56 m²

Therefore, area of the circle 98. 56 m²

**Question no – (9)**

**Solution** :

Let, the radius of the circle = r cm

**∴** Area of the circle = πr²cm^{2}

Now, according to the question,

=> r² = 2464 / π

=> r^{2} = 2464/5

=> r² = 2464 × 7 / 22 [∵ π = 22/7]

=> r = √112 × 7

=> √2 × 7 × 2 × 2 × 2 × 7

=> r = 2 × 2 × 7

=> r = 28

**∴** Thus the radius of the circle = 28 cm

**∴** Circumference of the circle

= 2 × π × 28 cm

= 2 × 22/7 × 28 cm

= 8 × 22 cm

= 176 cm

**Question no – (10)**

**Solution : **** **

Here, Diameter of the wheel (D) = 77 cm

**∴** Circumference of the wheel = πD cm

= π × 77

= 22/7 × 77

= 242 cm

**∴** Length of the journey = 121 × 1000 × 100 cm

**∴** Number of revolution will it make to travel

= 121 × 1000 × 100 / 242

= 50000

Thus, the required revolution 50000

**Question no – (12)**

**Solution** **:**

Let, the radius of one circle is r and the radius of another circle is R

∴ Area of another circle = πR²

∴ Area of one circle = = πr²

Now, according to the question,

πR² = 100 × πr²

=> R² / r² = 100/1

= R / r = √100/1

=> R/r = 10/1

=> R : r = 10:1

**∴** Circumference of the circle =2πR : 2πr

= 10 : 1

**Question no – (13)**

**Solution : **** **

Total cost of paving = 4435.20 Rs

Rate of paving = 20 Rs

**∴** Area of the room

= 4435.20/20

= 221.76

Let, r be the radius of the circle room

**∴** πr² = 221.76

=> r² = 221.76/π

=> r² = 221.76/22 × 7

=> r² = 70.56

=> r = 8.4

**∴** Now circumference of the room = 2πr

= 2 × π × 8.4

= 2 × 22/7 × 8.4

= 52.8 cm

**∴** Cost of fencing

= (52.8 × 5) Rs

= 264 Rs

**Previous Chapter Solution : **