**.**

**NCERT Solutions for Class 8 Maths Chapter 9 ****Mensuration**** Ex 9.1**

**Ex 9.1 Class 8 Maths Question 1.**

### The shape of the top surface of a table is a trapezium. Find its area if its parallel sides are 1 m and 1.2 m and perpendicular distance between them is 0.8 m.

**Solution:**Area of a trapezium = ½ × (a + b) × h

= ½ × (1.2 + 1) × 0.8

= ½ × 2.2 × 0.8

= 0.88 m

^{2}

Hence, the required area of the table top is 0.88 m

^{2}.

^{}

**Ex ****9.1**** Class 8 Maths Question 2.**

### The area of a trapezium is 34 cm^{2} and the length of one of the parallel sides is 10 cm and its height is 4 cm. Find the length of the other parallel side.

**Solution:**Given: The area of trapezium = 34 cm

^{2}

Length of one of the parallel sides (a) = 10 cm

Height (h) = 4 cm

Area of a trapezium = ½ × (a + b) × h

34 = ½ × (10 + b) × 4

⇒ 34 = (10 + b) × 2

⇒ 17 = 10 + b

⇒ b = 17 – 10 = 7 cm

Hence, the other parallel side is 7 cm.

**Ex ****9.1**** Class 8 Maths Question 3.**

### Length of the fence of a trapezium-shaped field ABCD is 120 m. If BC = 48 m, CD = 17 m and AD = 40 m, find the area of this field. Side AB is perpendicular to the parallel sides AD and BC.

**Solution:**

Given: AB + BC + CD + DA = 120 m

BC = 48 m, CD = 17 m and AD = 40 m

AB = 120 m – (48 m + 17 m + 40 m) = 120 – 105 m = 15 m

Area of the trapezium ABCD = ½ × (BC + AD) × AB

= ½ × (48 + 40) × 15

= ½ × 88 × 15

= 44 × 15 = 660 m^{2}.

Hence, the area of the field is 660 m^{2}.^{}

**Ex ****9.1**** Class 8 Maths Question 4.**

### The diagonal of a quadrilateral shaped field is 24 m and the perpendiculars dropped on it from the remaining opposite vertices are 8 m and 13 m. Find the area of the field.

**Solution:**Area of the field = area of ∆ABD + area of ∆BCD

= ½ × b × h + ½ × b × h

= ½ × 24 × 13 + ½ × 24 × 8

= ½ × 24 (13 + 8)

= 12 × 21

= 252 m

^{2}

Hence, the required area of the field is 252 m

^{2}.

**Ex ****9.1**** Class 8 Maths Question 5.**

### The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.

**Solution:**Here, d

_{1}= 7.5 cm, d

_{2}= 12 cm

Area of a rhombus = ½ × d

_{1}× d

_{2}

= ½ × 7.5 × 12

= 7.5 × 6

= 45 cm

^{2}

Hence, the area of the rhombus is 45 cm

^{2}.

**Ex ****9.1**** Class 8 Maths Question 6.**

### Find the area of a rhombus whose side is 5 cm and whose altitude is 4.8 cm. If one of its diagonals is 8 cm long, find the length of the other diagonal.

**Solution:**Given: Side of the rhombus = 5 cm

Altitude = 4.8 cm

Length of one diagonal = 8 cm

Area of a rhombus = Side × Altitude = 5 × 4.8 = 24 cm

^{2}

Area of a rhombus = ½ × d

_{1}× d

_{2}

24 = ½ × d

_{1}× d

_{2}

24 = ½ × 8 × d

_{2}

24 = 4d_{2}

d_{2} = 6 cm

Hence, the length of the other diagonal is 6 cm.

**Ex ****9.1**** Class 8 Maths Question 7.**

### The floor of a building consists of 3000 tiles which are rhombus shaped and each of its diagonals are 45 cm and 30 cm in length. Find the total cost of polishing the floor, if the cost per m^{2} is ₹ 4.

**Solution:**Given: Number of tiles = 3000

Length of the two diagonals of a tile = 45 cm and 30 cm

Area of one tile = ½ × d

_{1}× d

_{2}

= ½ × 45 × 30

= 45 × 15

= 675 cm

^{2}

Area covered by 3000 tiles = 3000 × 675 cm

^{2}= 2025000 cm

^{2}= 202.5 m

^{2}

The cost of polishing the floor = 202.5 × ₹ 4 = ₹ 810

Hence, the required cost of polishing the floor is ₹ 810.

**Ex ****9.1**** Class 8 Maths Question 8.**

### Mohan wants to buy a trapezium-shaped field. Its side along the river is parallel to and twice the side along the road. If the area of this field is 10500 m^{2} and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river.

Solution:

Let the side of the trapezium (roadside) be x cm.

The opposite parallel side = 2x m

h = 100 m

Area of the field = 10500 m^{2}

Area of a trapezium = ½ (a + b) × h

10500 = ½ (2x + x) × 100

2 × 10500 = 3x × 100

21000 = 300x

x = 70 m

So, AB = 2x = 2 × 70 = 140 m

Hence, the length of the side along the river is 140 m.

**Ex ****9.1**** Class 8 Maths Question 9.**

### The top surface of a raised platform is in the shape of a regular octagon as shown in the figure. Find the area of the octagonal surface.

Solution:

Area of the octagonal surface = area of trapezium ABCH + area of rectangle HCDG + area of trapezium GDEF

Area of trapezium ABCH = Area of trapezium GDEF

= ½ (a + b) × h

= ½ (11 + 5) × 4

= ½ × 16 × 4

= 32 m^{2}

Area of rectangle HCDG = l × b = 11 m × 5 m = 55 m^{2}

Area of the octagonal surface = 32 m^{2} + 55 m^{2} + 32 m^{2} = 119 m^{2}

Hence, the area of the octagonal surface is 119 m^{2}.

**Ex ****9.1**** Class 8 Maths Question 10.**

### There is a pentagonal shaped park as shown in the figure. For finding its area Jyoti and Kavita divided it in two different ways.

Find the area of this park using both ways. Can you suggest some other way of finding its area?

**Solution:**(i) From Jyoti’s diagram:

= 2 × Area of trapezium ABCD

= 2 × ½ (a + b) × h

= (15 + 30) × 7.5

= 45 × 7.5

= 337.5 m

^{2}

(ii) From Kavita’s diagram:

Area of the pentagonal shape = Area of ∆ABE + Area of square BCDE

= ½ × b × h + 15 × 15

= ½ × 15 × 15 + 225

= 112.5 + 225

= 337.5 m

^{2}

Yes, we can also find the other way to calculate the area of the given pentagonal shape.

Join CE to divide the figure into two parts, i.e., trapezium ABCE and right-angled triangle EDC.

Area of ABCDE = Area of ∆EDC + Area of trapezium ABCE

**Ex ****9.1**** Class 8 Maths Question 11.**

### Diagram of the picture frame has outer dimensions = 24 cm × 28 cm and inner dimensions 16 cm × 20 cm. Find the area of each section of the frame, if the width of each section is the same.

**Solution:**

^{2}, 96 cm

^{2}, 80 cm

^{2 }and 96 cm

^{2}respectively.

**Related Links:**

**NCERT Solutions for Maths Class 9**

**NCERT Solutions for Maths Class 10**

**NCERT Solutions for Maths Class 11**