class 8 maths chapter 9 exercise 9.1
Keeping the examination point of view in mind the mathematicsandinformationtechnology.com team has prepared NCERT Solutions for Class 8 Maths Chapter 9 Mensuration Exercise 9.1. The NCERT Solutions for chapter Mensuration solutions explains the easy and simple way to solve the problems. By understanding these ways in NCERT Solutions for Class 8, students will be confident while solving such problems found in Chapter 9 Mensuration Exercise 9.1.
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:
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 = \(({1\over 2})\times(a+b)h\)
\(= {1\over 2}\times(1+1.2)0.8\)
\(= {1\over 2}\times 2.2\times0.8 \)
\(= 0.88\)
Therefore, the area of the top surface of the table is \(0.88 \text{ m}^2\).
Question 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:
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 \(= 34 \text{cm}^2\)
The area of a trapezium = \(({1\over 2})\times(a+b)h\)
\(34 = {1\over 2}(10+b)\times 4\)
\(34 = 2×(10+b)\)
\(10+b={34 \over 2}\)
\(10+b=17\)
\(b=17-10\)
b = 7
Hence, the length of the other parallel side is 7 cm.
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: 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
Area of the field = \(({1\over 2})\times(BC+AD)AB\)
\(= {1\over 2}×(48 +40)×15\)
\(={1\over 2}×88×15\)
\(= 660\)
Hence, the area of the field ABCD is \(660 \text{ m}^2\).
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.
Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ADC
\(= {1 \over 2}( b\times h_1)+ {1 \over 2}(b \times h_2)\)
\(= {1 \over 2} \times b(h_1+h_2)\)
\(= {1 \over 2} \times 24 \times (13+8)\)
\(= {1 \over 2} \times 24 \times 21 = 252\)
Hence, the required area of the field is \(252 \text{ m}^2\).
Question 5:
The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.
Solution:
Given: \(d_1 = 7.5 \text{ cm and } d_2 = 12 \) cm
Area of the rhombus = \({1 \over 2} \times d_1 \times d_2\)
= \({1 \over 2} \times 7.5 \times 12 = 45\)
Therefore, the area of the rhombus is \(45 \text{ cm}^2\).
Question 6:
Find the area of a rhombus whose side is 6 cm and whose altitude is 4 cm. If one of its diagonals is 8 cm long, find the length of the other diagonal.
Hence, the length of the other diagonal is 6 cm.
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 m2 is ₹4.
Solution:
Length of one diagonal, \(d_1 = 45 \text{cm and } d_2= 30\) cm
∵ Area of one tile = \({1 \over 2}d_1\times d_2\)
=\({1 \over 2}\times 45 \times 30 = 675\)
The area of one tile is \(675 \text{cm}^2\)
So, the area of 3000 tiles is
\(= 675×3000 = 2025000 \text{cm}^2\)
\(= {2025000 \over 10000}\)
\(= 202.50 \text{m}^2 [∵ 1 \text{m}^2 = 10000 \text{cm}^2]\)
Cost of polishing the floor per sq. meter = ₹ 4
Cost of polishing the floor per 202.50 sq. meter \(= 4\times 202.50 = \)₹ 810
Hence, the total cost of polishing the floor is ₹ 810.
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. It 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 \(= 10500 \text{m}^2\) (Given)
Let the side along the road be ‘x’ m,
Side along the river = 2x m
Area of the trapezium field \(= {1 \times 2} \times(a+b)\times h\)
\(10500 = {1 \over 2} \times (x+2x)×100\)
\(10500 = {3x \times 50}\)
\(x= {10500 \over {50 \times 3}}\)
x = 70,
The side along the river is 70 m
Hence, the side along the river \(= 2 \times x = 2( 70) = 140\) m.
Question 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.
Area of two trapeziums \(= 2 [({1 \over 2})×(a+b)\times h]\)
\(= 2×({1 \over 2})×(11+5 )×4\)
\(= 4×16 = 64\)
Area of two trapeziums is 64 \(\text{m}^2\)
Also, the area of the rectangle = length×breadth \(= 11×5 = 55\)
The area of the rectangle is 55 \(\text{m}^2\)
The total area of the octagon = 64+55 = 119 \(\text{m}^2\)
Therefore, the area of the octagonal surface is 119 \(\text{m}^2\).
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?
First way: By Jyoti’s diagram,
Area of pentagon \(= ({1 \over 2})(AP+BC)×CP+({1 \over 2})×(ED+AP)×DP \)
\(= ({1 \over 2})(30+15)×CP+(1/2)×(15+30)×DP \)
\(= ({1 \over 2})×(30+15)×(CP+DP) \)
\(= ({1 \over 2})×45×CD \)
\(= ({1 \over 2})×45×15 \)
\(= 337.5 \text{m}^2\)
Therefore, the area of the pentagon is \(= 337.5 \text{m}^2\).
Second way: By Kavita’s diagram
side = 30–15 = 15 m
Area of pentagon \(= {1 \over 2}×15×15+(15×15)\)
\(= 112.5+225.0\)
\(= 337.5\)
Hence, the total area of the pentagon-shaped park \(= 337.5 \text{m}^2\).
Question 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.
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