Selina Solutions Concise Maths Class 7 Chapter 20 Mensuration (Perimeter and Area of Plane Figures) has important formulas and shortcut methods explained in simple language. The motto of providing chapter wise solutions is to boost the exam preparation of students. For a better academic score, Selina Solutions Concise Maths Class 7 Chapter 20 Mensuration (Perimeter and Area of Plane Figures), PDF links are provided below with a free download option.

Chapter 20 deals with the method of finding the perimeter and area of figures like rectangle, square and triangle. The important formulas and the method of solving problems are given in a stepwise manner, for a better conceptual knowledge among students.

## Selina Solutions Concise Maths Class 7 Chapter 20: Mensuration (Perimeter and Area of Plane) Download PDF

### Exercises of Selina Solutions Concise Maths Class 7 Chapter 20 – Mensuration (Perimeter and Area of Plane)

## Access Selina Solutions Concise Maths Class 7 Chapter 20: Mensuration (Perimeter and Area of Plane)

#### Exercise 20A page: 220

**1. The length and the breadth of a rectangular plot are 135 m and 65 m. Find, its perimeter and the cost of fencing it at the rate of ₹ 60 per m.**

**Solution:**

**It is given that**

**Length of a rectangular plot = 135 m**

**Breadth of a rectangular plot = 65 m**

**We know that**

**Perimeter of a rectangular plot = 2 (length + breadth)**

**Substituting the values**

**= 2 (135 + 65)**

**= 2 (200)**

**= 400 m**

**Here the cost of fencing = ₹ 60 per m**

**So the cost of fencing 400 m = 60 × 400 = ₹ 24,000**

**2. The length and breadth of a rectangular field are in the ratio 7 : 4. If its perimeter is 440 m, find its length and breadth. **

**Also, find the cost of fencing it @ ₹150 per m.**

**Solution:**

**It is given that**

**Perimeter of rectangular field = 440 m**

**Consider 7x as the length and 4x as the breadth of rectangular field**

**So we get**

**2 (l + b) = Perimeter**

**Substituting the values**

**2 (7x + 4x) = 440**

**By further calculation**

**2 (11x) = 440**

**22x = 440**

**So we get**

**x = 440/22 = 11 m**

**Here**

**Length = 7x = 7 × 11 = 77m**

**Breadth = 4x = 4 × 11 = 44m**

**We know that**

**Cost of fencing = ₹150 per m**

**So the cost of fencing 440 m = 150 × 440 = ₹ 66,000**

**3. The length of a rectangular field is 30 m and its diagonal is 34 m. Find the breadth of the field and its perimeter.**

**Solution:**

**It is given that**

**Length of a rectangular field = 30 m**

**Diagonal of a rectangular field = 34 m**

**Consider the breadth of rectangular field = b m**

**Using the Pythagoras theorem**

**AC ^{2} = AB^{2} + BC^{2}**

**Substituting the values**

**34 ^{2} = 30^{2} + b^{2}**

**By further calculation**

**1156 = 900 + b ^{2}**

**So we get**

**b ^{2} = 1156 – 900 = 256**

**b = **√256 = 16 m

We know that

Perimeter = 2 (l + b)

Substituting the values

= 2 (30 + 16)

= 2 × 46

= 92 m

**4. The diagonal of a square is 12 **√**2 cm. Find its perimeter.**

**Solution:**

**It is given that**

**Diagonal of a square = 12 **√**2 cm**

**We know that diagonal = side **× √**2**

**Here the side = 12 cm**

**So the perimeter = 4 **× 12 = 48 cm

**5. Find the perimeter of a rectangle whose length = 22.5 m and breadth = 16 dm.**

**Solution:**

**It is given that**

**Length = 22.5 m**

**Breadth = 16 dm = 1.6 m**

**We know that**

**Perimeter of a rectangle = 2 (l + b)**

**Substituting the values**

**= 2 (22.5 + 1.6)**

**So we get**

**= 2 (24.1)**

**= 48.2 m**

**6. Find the perimeter of a rectangle with length = 24 cm and diagonal = 25 cm.**

**Solution:**

**It is given that**

**Length = 24 cm**

**Diagonal = 25 cm**

**Consider the breadth of a rectangle = b m**

**Using Pythagoras theorem in triangle ABC**

**AC ^{2} = AB^{2} + BC^{2}**

**Substituting the values**

**25 ^{2} = 24^{2} + b^{2}**

**625 = 576 + b ^{2}**

**By further calculation**

**b ^{2} = 625 – 576 = 49**

**b = **√49 = 7 cm

**Here the perimeter of rectangle = 2 (l + b)**

**Substituting the values**

**= 2 (24 + 7)**

**So we get**

**= 2 (31)**

**= 62 cm**

**7. The length and breadth of rectangular piece of land are in the ratio of 5 : 3. If the total cost of fencing it at the rate of ₹48 per metre is ₹19,200, find its length and breadth.**

**Solution:**

**It is given that**

**Length and breadth of rectangular piece of land are in the ratio = 5 : 3**

**Cost of fencing = ₹19,200**

**Rate = ₹48 per metre**

**We know that**

**Perimeter of rectangular piece of land = 19,200/48 = 400 m**

**Consider length = 5x**

**Breadth = 3x**

**So the perimeter = 2 (l + b)**

**Substituting the values**

**400 = 2 (5x + 3x)**

**By further calculation**

**400 = 2 (8x)**

**400 = 16x**

**So we get**

**x = 400/16 = 25**

**Length = 5x = 5 × 25 = 125 m**

**Breadth = 3x = 3 × 25 = 75 m**

**8. A wire is in the shape of square of side 20 cm. If the wire is bent into a rectangle of length 24 cm, find its breadth.**

**Solution:**

**It is given that**

**Side of square = 20 cm**

**So the perimeter of square = 4 × 20 = 80 cm**

**Perimeter of rectangle = 80 cm**

**We know that**

**Length of rectangle = 24 cm**

**So the perimeter of rectangle = 2 (l + b) **

**Substituting the values**

**80 = 2 (24 + b)**

**By further calculation**

**40 = 24 + b**

**b = 40 – 24 = 16 m**

**9. If P = perimeter of a rectangle, l= its length and b = its breadth; find:**

**(i) P, if l = 38 cm and b = 27 cm**

**(ii) b, if P = 88 cm and l = 24 cm**

**(iii) l, if P = 96 m and b = 28 m**

**Solution:**

**(i) It is given that**

**l = 38 cm**

**b = 27 cm**

**We know that**

**Perimeter = 2 (l + b)**

**Substituting the values**

**= 2 (38 + 27)**

**= 2 (65)**

**= 130 cm**

**(ii) It is given that**

**P = 88 cm**

**l = 24 cm**

**Consider b as the breadth**

**We know that**

**P = 2 (l + b)**

**It can be written as**

**b = P/2 – l**

**Substituting the values**

**b = 88/2 – 24 **

**b = 44 – 24**

**b = 20 cm**

**(iii) It is given that**

**P = 96 m**

**B = 28 m**

**Consider l as the length**

**We know that**

**P = 2 (l + b)**

**It can be written as**

**l = P/2 – b**

**Substituting the values**

**l = 96/2 – 28**

**l = 48 – 42**

**l = 20 m**

**10. The cost of fencing a square field at the rate of** **₹75 per meter is** **₹67,500. Find the perimeter and the side of the square field.**

**Solution:**

**Cost of fencing = ₹67,500**

**So the length of fence = 67,500/75 = 900 m**

**We know that the perimeter of square field = length of fence = 900 m**

**Here**

**Perimeter of a square = 4 × Length of its side**

**Substituting the values**

**Length of the side of a square = Perimeter/ 4**

**So we get**

**= 900/4**

**= 225 m**

**11. The length and the breadth of a rectangle are 36 cm and 28 cm. If its perimeter is equal to the perimeter of a square, find the side of the square.**

**Solution:**

**It is given that**

**Length of a rectangle = 36 cm**

**Breadth of a rectangle = 28 cm**

**We know that**

**Perimeter = 2 (l + b)**

**Substituting the values**

**= 2 (36 + 28)**

**= 2 (64)**

**= 128 cm**

**It is given that**

**Perimeter of a square = Perimeter of a rectangle = 128 cm**

**So the side of square = perimeter/4**

**Substituting the value**

**= 128/ 4**

**= 32 cm**

**12. The radius of a circle is 21 cm. Find the circumference (Take π = 3 1/7).**

**Solution:**

**It is given that**

**Radius of a circle = 21 cm**

**We know that π = 22/7**

**So the circumference of a circle = 2 πr**

**Substituting the values**

**= 2 × 22/7 × 21**

**So we get**

**= 2 × 22 × 3**

**= 132 cm**

**13. The circumference of a circle is 440 cm. Find its radius and diameter. (Take π = 22/7).**

**Solution:**

**It is given that**

**Circumference of a circle = 440 cm**

**So the radius = C/ 2π**

**Substituting the values**

**= (440 × 7)/ (2 × 22)**

**So we get**

**= 3088/44**

**= 70 cm**

**Diameter of the circle = 2 × radius**

**So we get**

**= 2 × 70 **

**= 140 cm**

**14. The diameter of a circular field is 56 m. Find its circumference and cost of fencing it at the rate of ₹80 per m. (Take π = 22/7).**

**Solution:**

**It is given that**

**Diameter of a circular field = 56 m**

**So the radius = 56/2 = 28 m**

**We know that**

**Circumference of the circle = 2 πr**

**Substituting the values**

**= 2 × 22/7 × 28**

**So we get**

**= 2 × 22 × 4**

**= 176 m**

**Here the cost of fencing 176 m = 176 × 80 = ₹ 14,080**

**15. The radii of two circles are 20 cm and 13 cm. Find the difference between their circumferences. (Take π = 22/7).**

**Solution:**

**It is given that**

**Radius of first circle = 20 cm**

**We know that**

**Circumference of the circle = 2 πr**

**Substituting the values**

**= 2 × 22/7 × 20**

**So we get**

**= 880/7**

**= 122.8 cm**

**Similarly **

**Radius of the second circle = 13 cm**

**We know that**

**Circumference of the circle = 2 πr**

**Substituting the values**

**= 2 × 22/7 × 13**

**So we get**

**= 572/7**

**= 81.7**

**So the difference of circumference of two circles = 122.8 – 81.7 = 41.1 cm**

#### Exercise 20B page: 228

**1. Find the area of a rectangle whose length and breadth are 25 cm and 16 cm.**

**Solution:**

It is given that

Length of a rectangle = 25 cm

Breadth of a rectangle = 16 cm

We know that

Area of a rectangle = l × b

Substituting the values

= 25 × 16

= 400 cm^{2}

**2. The diagonal of a rectangular board is 1 m and its length is 96 cm. Find the area of the board.**

**Solution:**

**It is given that**

**Length of rectangular board = 96 cm**

**Diagonal of rectangular board = 1 m = 100 cm**

**Using the Pythagoras theorem in right angled triangle ABC**

**AC ^{2} = AB^{2} + BC^{2}**

**Substituting the values**

**100 ^{2} = 96^{2} + BC^{2}**

**By further calculation**

**10000 = 9216 + BC ^{2}**

**So we get**

**BC ^{2} = 10000 – 9216 = 784**

**BC = **√**784 = 28 cm**

**Here the area of rectangular board = l × b**

**Substituting the values**

**= 96 × 28**

**= 2688 cm ^{2}**

**3. The sides of a rectangular park are in the ratio 4 : 3. If its area is 1728 m ^{2}, find:**

**(i) its perimeter**

**(ii) cost of fencing it at the rate of ₹40 per meter.**

**Solution:**

**It is given that**

**Ratio in the sides of a rectangular park = 4 : 3**

**Area = 1728 m ^{2}**

**Consider 4x as the length and 3x as the breadth**

**We know that**

**Area = l × b**

**Substituting the values**

**1728 = 4x × 3x**

**By further calculation**

**12x ^{2} = 1728**

**x ^{2}**= 1728/12 = 144

**x = **√144 = 12

Here we get

Length = 4x = 4 × 12 = 48 m

Breadth = 3x = 3 × 12 = 36 m

(i) We know that

**Perimeter = 2 (l + b)**

**Substituting the values**

**= 2 (48 + 36)**

**So we get**

**= 2 (84)**

**= 168 m**

**(ii) It is given that**

**Rate of fencing = ₹40 per meter**

**So the total cost of fencing = 168 **× 40 = **₹ 6720**

**4. A floor is 40 m long and 15 m broad. It is covered with tiles, each measuring 60 cm by 50 cm. Find the number of tiles required to cover the floor.**

**Solution:**

**Given below are the dimensions of floor**

**Length = 40 m**

**Breadth = 15 m**

**We know that**

**Area = l **× b = 40 × 15 = 600 m^{2}

Here the length of one tile = 60 cm = 6/10 m

Breadth of one tile = 50 cm = 5/10 m

So the area of one tile = 6/10 × 5/10 = 30/100 = 3/10 m^{2}

We know that

Number of tiles = total area of floor/ area of one tile

Substituting the values

= 600/ 3/10

= (600 × 10)/ 3

= 2000

**5. The length and breadth of a rectangular piece of land are in the ratio 5 : 3. If the total cost of fencing it at the rate of ₹24 per meter is ₹9600, find its:**

**(i) length and breadth**

**(ii) area**

**(iii) cost of levelling at the rate of ₹60 per m ^{2}.**

**Solution:**

**It is given that**

**Ratio in length and breadth of a rectangular piece of land = 5 : 3**

**Cost of fencing = ₹9600**

**Rate = ₹24 per meter**

**We know that**

**Perimeter = total cost of fencing/ rate per meter **

**Substituting the values**

**= 9600 /24 **

**= 400 m**

**Consider 5x as the length and 3x as the breadth**

**Here perimeter = 2 (l + b)**

**Substituting the values**

**400 = 2 (5x + 3x)**

**By further calculation**

**400 = 2 (8x)**

**So we get**

**400 = 16x**

**x = 400/16 = 25**

**(i) Length of land = 5x = 5 × 25 = 125 m**

**Breadth of land = 3x = 3 × 25 = 75 m**

**(ii) Area = l × b**

**Substituting the values**

**= 125 × 75 **

**= 9375 m ^{2}**

**(iii) Cost of levelling at the rate of ₹60 per m ^{2} = 60 × 9375 = ₹ 5,62,500**

**6. Find the area of the square whose perimeter is 56 cm.**

**Solution:**

**It is given that**

**Perimeter of the square = 56 cm**

**We know that**

**4 × side = 56 cm**

**So we get**

**Side = 56/4 = 14 cm**

**Here**

**Area of the square = side ^{2}**

**Substituting the values**

**Area of the square = 14 ^{2} = 196 cm^{2}**

**7. A square lawn is surrounded by a path 2.5 m wide. If the area of the path is 165 m ^{2}, find the area of the lawn.**

**Solution:**

**It is given that**

**Area of the path = 165 m ^{2}**

**Width of the path = 2.5 m**

**Consider x m as the side of square lawn **

**So the outer side = x + 2 × 2.5**

**We get**

**Outer side = (x + 5) m**

**Here the area of path = (x + 5) ^{2} – x^{2}**

**Substituting the values**

**x ^{2} + 10x + 25 – x^{2} = 165**

**By further calculation**

**10x = 165 – 25 = 140**

**So we get**

**x = 140/10 = 14 m**

**Here the side of lawn = 14 m**

**Area of the lawn = 14 ^{2} = 196 m^{2}**

**8. For each figure, given below, find the area of shaded region: (All measurements are in cm)**

**Solution:**

**(i) It is given that**

**Outer length = 20 cm**

**Outer breadth = 16 cm**

**Outer area = l × b**

**Substituting the values**

**= 20 × 16 **

**= 320 cm ^{2}**

**Similarly**

**Inner length = 15 cm**

**Inner breadth = 10 cm**

**So the inner area = l × b**

**Substituting the values**

**= 15 × 10 **

**= 150 cm ^{2}**

**Here the area of shaded region = area of whole region – area of unshaded region**

**Substituting the values**

**= 320 – 150**

**= 170 cm ^{2}**

**(ii) It is given that**

**Outer length = 30 cm**

**Outer breadth = 20 cm**

**Outer area = l × b**

**Substituting the values**

**= 30 × 20 **

**= 600 cm ^{2}**

**Similarly**

**Inner length = 12 cm**

**Inner breadth = 12 cm**

**So the inner area = l × b**

**Substituting the values**

**= 12 × 12 **

**= 144 cm ^{2}**

**Here the area of shaded region = area of outer figure – area of inner figure**

**Substituting the values**

**= 600 – 144**

**= 456 cm ^{2}**

**(iii) Here the area shaded portion = area of outer region – area of unshaded region**

**Substituting the values**

**= 40 × 40 – 32 × 15**

**So we get**

**= 1600 – 480**

**= 1120 cm ^{2}**

**(iv) Here the area of shaded region = area of outer region – area of inner region**

**Substituting the values**

**= 40 × 40 – 15 × 15**

**So we get**

**= 1600 – 225**

**= 1375 cm ^{2}**

**(v) Area of shaded portion = 2 × 20 + 2 × 8 + 2 × (12 + 2)**

**By further calculation**

**= 40 + 16 + 28**

**= 84 cm ^{2}**

**9. One side of a parallelogram is 20 cm and its distance from the opposite side is 16 cm. Find the area of the parallelogram.**

**Solution:**

**We know that**

**Area of parallelogram = base × height**

**Here **

**Area of parallelogram = AB × DE**

**Substituting the values**

**= 20 × 16**

**= 320 cm ^{2}**

**10. The base of a parallelogram is thrice it height. If its area is 768 cm ^{2}, find the base and the height of the parallelogram.**

**Solution:**

**It is given that**

**Area of parallelogram = 768 cm ^{2}**

**Consider x as the height and 3x as the base of parallelogram**

**So we get**

**Area = base × height**

**Substituting the values**

**768 = 3x × x**

**By further calculation**

**768 = 3x ^{2}**

**x ^{2} = 768/3 = 256 cm**

**x = **√ 256 = 16 cm

Height = x = 16 cm

Base = 3x = 3 × 16 = 48 cm

**11. Find the area of the rhombus, if its diagonals are 30 cm and 24 cm.**

**Solution:**

**It is given that**

**Diagonal d _{1} = 30 cm**

**Other diagonal d _{2} = 24 cm**

**Consider AC and BD as the diagonals of rhombus**

**We know that**

**Area = ½ **× product of diagonals

So we get

= ½ × AC × BD

= ½ × **d _{1} **×

**d**

_{2}**Substituting the values**

**= ½ **× 30 × 24

= 15 × 24

= 360 cm^{2}

Hence, the area of the rhombus is 360 cm^{2}.

**12. If the area of a rhombus is 112 cm ^{2} and one of its diagonals is 14 cm, find its other diagonal.**

**Solution:**

**It is given that**

**Area of a rhombus = 112 cm ^{2}**

**One diagonal = 14 cm**

**Consider x cm as the second diagonal**

**We know that**

**Area = product of diagonal/2**

**Substituting the values**

**112 = (14 **× x)/ 2

By further calculation

x = (112 × 2)/ 14

So we get

x = 224/14 = 16 cm

Hence, the other diagonal of the rhombus is 16 cm.

**13. One side of a parallelogram is 18 cm and its area is 153 cm ^{2}. Find the distance of the given side from its opposite side.**

**Solution:**

**It is given that**

**Area of parallelogram = 153 cm ^{2}**

**One side = 18 cm**

**So the distance between AB and DC = area/ base**

**Substituting the values**

**= 153/ 18**

**= 17/2**

**= 8.5 cm**

**14. The adjacent sides of a parallelogram are 15 cm and 10 cm. If the distance between the longer sides is 6 cm, find the distance between the shorter sides.**

**Solution:**

**It is given that**

**AB = DC = 15 cm**

**BC = AD = 10 cm**

**Here the distance between longer sides AB and DC = 6 cm**

**So perpendicular DL = 6 cm**

**DM is perpendicular to BC**

**We know that**

**Area of parallelogram = base **× altitude

So we get

= AB × DL

Substituting the values

= 15 × 6

= 90 cm^{2}

**Consider DM = x cm**

**Similarly **

**Area of parallelogram ABCD = BC **× DM

Substituting the values

= 10 × x

= 10 x cm^{2}

By equating both

10 x = 90

x = 90/10 = 9 cm

**15. The area of a rhombus is 84 cm ^{2} and its perimeter is 56 cm. Find its height.**

**Solution:**

**It is given that**

**Area of a rhombus = 84 cm ^{2}**

**Perimeter of a rhombus = 56 cm**

**So the side of a rhombus = 56/4 = 14 cm**

**Here the height = area/ base **

**Substituting the values**

**= 84/14 **

**= 6 cm**