The Class 8 Mathematics NCERT questions, provided with appropriate solutions, will come in handy in revising the Class 8 Maths CBSE syllabus. The textbook provides a wide range of exercises, including hundreds of questions based on CBSE guidelines. NCERT Solutions for Class 8 Maths Chapter 11, Exercise 11.2 is a set of questions and answers according to the textbook exercise questions. All the solutions are prepared by the subject-matter experts at BYJU’S to help students ace the exam confidently. Download the free NCERT Solutions for Maths Chapter 11 – Mensuration and score well in the annual exams.
NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Exercise 11.2
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Access Other Exercise Solutions of Class 8 Maths Chapter 11 Mensuration
Exercise 11.1 Solutions: 5 Questions (Long Answers)
Exercise 11.3 Solutions: 10 Questions (2 Short Answers, 8 Long Answers)
Exercise 11.4 Solutions: 8 Questions (2 Short Answers, 6 Long Answers)
Access Answers to NCERT Class 8 Maths Chapter 11 Mensuration Exercise 11.2 Page Number 177
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 the perpendicular distance between them is 0.8 m.
Solution:
One parallel side of the trapezium (a) = 1 m
The second side (b) = 1.2 m
height (h) = 0.8 m
Area of top surface of the table = (½)×(a+b)h
= (½)×(1+1.2)0.8
= (½)×2.2×0.8 = 0.88
Therefore, the area of the top surface of the table is 0.88 m2.
2. The area of a trapezium is 34 cm2, 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:
Let the length of the other parallel side be b.
Length of one parallel side, a = 10 cm
height, (h) = 4 cm
Area of a trapezium is 34 cm2
Formula for the area of a trapezium = (1/2)×(a+b)h
34 = ½(10+b)×4
34 = 2×(10+b)
After simplifying, b = 7
Hence, the length of the other 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,
AD = 40 m and 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
= 660
Hence, the area of the field ABCD is 660m2.
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:
Consider, h1 = 13 m, h2 = 8 m and AC = 24 m
Area of quadrilateral ABCD = Area of triangle ABC+Area of triangle ADC
= ½( bh1)+ ½(bh2)
= ½ ×b(h1+h2)= (½)×24×(13+8)
= (½)×24×21 = 252
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 the rhombus = (½ )×d1×d2
= (½)×7.5×12 = 45
Therefore, the 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 the 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 the area of a rhombus = Base×Altitude
Putting values, we have
Area of rhombus = 5×4.8 = 24
The area of the rhombus is 24 cm2
Also, the formula for the area of the rhombus = (½)×d1d2
After substituting the values, we get
24 = (½)×8×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 is 45 cm and 30 cm in length. Find the total cost of polishing the floor, if the cost per m2 is Rs. 4.
Solution: Length of one diagonal, d1 = 45 cm and d2= 30 cm
∵ Area of one tile = (½)d1d2
= (½)×45×30 = 675
The area of one tile is 675 cm2
So, the area of 3000 tiles is
= 675×3000 = 2025000 cm2
= 2025000/10000
= 202.50 m2 [∵ 1 m2 = 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 Rs. 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 m2Â 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 m
Area of the trapezium field = (½)×(a+b)×h
10500 = (½)×(x+2x)×100
10500 = 3x×50
After simplifying, we have x = 70, which means the side along the river is 70 m
Hence, the side along the river =Â 2x = 2(Â 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)
Divide the octagon as shown in the below figure, 2 trapeziums whose parallel and perpendicular sides are 11 m and 4 m, respectively, and the 3rd one is a rectangle having length and breadth of 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 64Â m2
Also, the area of the rectangle = length×breadth
= 11×5 = 55
The area of the rectangle is 55 m2
The total area of the octagon = 64+55
= 119Â m2
Therefore, the area of the octagonal surface is 119Â m2.
10. There is a pentagonal-shaped park, as shown in the figure.
To find its area, Jyoti and Kavita divided it into two different ways.
Find the area of this park using both ways. Can you suggest some other way of finding its area?
Solution:
First way: By Jyoti’s diagram,
Area of pentagon = Area of trapezium ABCP + Area of trapezium AEDP
= (½)(AP+BC)×CP+(1/2)×(ED+AP)×DP
= (½)(30+15)×CP+(1/2)×(15+30)×DP
= (½)×(30+15)×(CP+DP)
= (½)×45×CD
= (1/2)×45×15
= 337.5 m2
Therefore, the area of the pentagon is 337.5 m2.
Second way: 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 the above figure)
= (½)×15×15+(15×15)
= 112.5+225.0
= 337.5
Hence, the total area of the 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 the same.
Solution:
Divide the given figure into 4 parts, as shown below:
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
The area of figure (I) = 96 cm2
Also, the area of figure (II) = 96 cm2
Now, the area of figure (III) = Area of trapezium
= (½)×(a+b)×h
= (½)×(24+16)4
= (½)×40×4 = 80
Area of figure (III) is 80Â cm2
Also, the area of figure (IV) = 80 cm2.
Chapter 11 Mensuration, Exercise 11.2 is about the area of a trapezium, general quadrilateral and polygon and their real-life examples. Download and practise NCERT Class 8 Maths Solutions and improve your Maths skills and score good marks in the annual exam.
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