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If you are in search of **NCERT solutions for Class 8 Maths Chapter** **9 Mensuration**, then you are at the right place. When it comes to **Mathematics**, it is a subject that holds value for students across various academic streams, be it science, biology, or commerce and having a very good understanding of basic math concepts will help you in achieving your goal and make it much more easier.

Here we provide detailed **NCERT solutions for Class 8 Maths** of all the chapters, **exercise wise** with **updated syllabus by NCERT**. These NCERT solutions for Class 8 Maths have been carefully crafted to provide detailed explanations, ensuring that students can grasp the concepts effectively. We also provide extra practice questions for the same, so feel free to bookmark **ncertforclass8.com** to boost your preparation.

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

**Class 8 Maths Chapter** **9 Mensuration** **Exercise **9.1

**Exercise 9.1**

**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:

ATQ, one parallel side of the trapezium(a) = 1 m

And second parallel side(b) = 1.2 m

Height (h) = 0.8 m

`\[Area\ of\ top\ surface\ of\ the\ table=\frac{1}{2}\times(a+b)\times h\]`

`\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{2}\times\left(1\ +1.2\right)\times0.8\]`

`\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{2}\times\left(2.2\right)\times0.8\ =\ 0.88\ m^2\]`

**2. The area of a trapezium is 34 cm2 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:

Length of one parallel side, a = 10 cm

Let the length of the other parallel side be b.

Height, (h) = 4 cm

Area of a trapezium = 34 cm2

Area of trapezium = (1/2)×(a+b) x h

34 = ½(10+b)×4

34 = 2×(10+b)

34 = 20 + 2b

34-20=2b

14=2b

14/2 = b

b=7 cm

Hence another required parallel side is 7 cm.

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

Perimeter = 120 m

∵ Perimeter of trapezium ABCD

= AB+BC+CD+DA

120 = AB+48+17+40

120 = AB = 105

AB = 120–105 = 15 m

Now, Area of the field= (½)×(BC+AD)×AB=

(½)×(48 +40)×15

= (½)×88×15 = 44 × 15

= 660 m2

Hence, area of the field ABCD is 660 m2 .

**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:

Let h1 = 13 m, h2 = 8 m and AC = 24 m

Area of quadrilateral ABCD = Area of △ ABC+ Area of △ ADC

= ½( b×h1)+ ½(b×h2)

= ½ ×b×(h1+h2)= (½)×24×(13+8)

= (½)×24×21 = 252 m2

Hence, the required area of the field is 252 m2

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

Solution:

Given: d1 = 7.5 cm and d2 = 12 cm

We know that the area of a rhombus = (½ )×d1×d2

= (½)×7.5×12 = 45

∴ Area of the rhombus is 45 cm2 .

**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:

Since a rhombus is also a kind of parallelogram,

The formula for Area of rhombus = Base×Altitude

ATQ, base= 5cm, altitude= 4.8cm and d1=8cm

Area of rhombus = 6×4 = 24

Area of rhombus is 24 cm2

Also, Formula for Area of rhombus = (½)×d1×d2

24 = (½)×8×d2

24= 4 x d2

24/4 = d2

d2 = 6

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

**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 m2 is ₹4.**

Solution:

Length of one diagonal, d1 = 45 cm and d2= 30 cm

∵ Area of one tile = (½)d1d2

= (½)×45×30 = 675 cm2

∴ Area of one tile is 675 cm2

Area of 3000 tiles is = 675×3000 = 2025000 cm2

Since we have to calculate it in m2, therefore dividing 2025000 cm2 by 10000

= 2025000/10000

= 202.50 m2 [∵ 1m2 = 10000 cm2]

Cost of polishing the floor per sq. meter = ₹4

Cost of polishing the floor per 202.50 sq. meter = 4×202.50 = ₹ 810

Hence the total cost of polishing the floor is ₹ 810.

**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 m2 and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river.**

Solution:

Perpendicular distance (h) = 100 m (Given)

Area of the trapezium-shaped field = 10500 m2 (Given)

Let the side along the road be ‘x’ m and the side along the river = 2x

Area of the trapezium field = (½)×(a+b)×h

10500 = (½)×(x+2x)×100

10500 = 3x × 50

10500/50 x 3 = x

10500/150 = x

x= 70

Hence, the side along the river = 2x = 2(70) = 140 m.

**9. 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:

The octagon has eight equal sides, each 5 m. (given)

Construction – Dividing the octagon as shown in the above figure, 2 trapeziums whose parallel and perpendicular sides are 11 m and 4 m respectively and 3rd one is rectangle having length and breadth 11 m and 5 m respectively.

Now, Area of two trapeziums = 2 [(½)×(a+b)×h]

= 2×(½)×(11+5 )×4

= 4×16 = 64

Area of two trapeziums is 64 m2

Also, Area of rectangle = length×breadth

= 11×5 = 55

Area of rectangle is 55 m2

Total area of octagon =Area of two trapeziums + Area of rectangle

=64+55

= 119 m2

**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:

**By Jyoti’s diagram,**

Area of pentagon = Area of trapezium ABCN + Area of trapezium AEDN

= (½)(AN+BC)×CP+(1/2)×(ED+AN)×DN

= (½)(30+15)×CN+(1/2)×(15+30)×DN

= (½)×(30+15)×(CN+DN)

= (½)×45×CD

= (1/2)×45×15

=337.5 m2

Area of pentagon is 337.5 m2

**By Kavita’s diagram,**

Here, a perpendicular AM is drawn to BE.

AM = 30–15 = 15 m

Area of pentagon = Area of triangle ABE+Area of square BCDE (from above figure)

= (½)×15×15+(15×15)

= 112.5+225.0

= 337.5

Hence, the total area of pentagon-shaped park = 337.5 m2

**11. Diagram of the adjacent 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 same.**

Solution:

Here two of the given figures (I) and (II) are similar in dimensions.

And also, figures (III) and (IV) are similar in dimensions.

Area of figure (I) = Area of trapezium

= (½)×(a+b)×h

= (½)×(28+20)×4

= (½)×48×4 = 96

Area of figure (I) = 96 cm2

Also, Area of figure (II) = 96 cm2 [Because area of (I) and (II) are same]

Now, Area of figure (III) = Area of trapezium

= (½)×(a+b)×h

= (½)×(24+16)×4

= (½)×40×4 = 80

Area of figure (III) is 80 cm2

Also, Area of figure (IV) = 80 cm2 [Because area of (III) and (IV) are same]

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