Class 8 - Mathematics
Algebraic Expressions and Identities - Exercise 11.2
Top Block 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.
Answer :
Here one parallel side of the trapezium a = 1 m
And second side b = 1.2 m and height h = 0.8 m
∴ Area of top surface of the table
= ½(𝒶 + 𝒷) x 𝒽 = ½(1 + 1.2) x 0.8
= ½(2.2) x 0.8 = 0.88 m²
Hence surface area of the table is 0.88 m²
Question :2. The area of a trapezium is 34cm² 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.
Mddle block 1
Answer :
Let the length of the other parallel side be 𝒷
Length of one parallel side 𝒶 = 10 am and height 𝒽 = 4 cm
Area of trapezium = ½(𝒶 + 𝒷) x 𝒽
⇒ 34 = ½(10 + 𝒷) x 4
⇒ 34 = (10 + 𝒷) x 2
⇒ 34 = 20 + 2𝒷
⇒ 34 – 20 = 2𝒷
⇒ 14 = 2𝒷
⇒ 𝒷 = 7
Hence another required 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.
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.
Answer :
Here 𝒽1 = 13 m, 𝒽2 = 8 m and
AC = 24 m
Area of quadrilateral ABCD
= Area of ΔABC + ΔArea of ADC
= ½𝒷 x 𝒽1 + ½𝒷 x 𝒽2
= ½𝒷 x (𝒽1 + 2)
= ½ x 24 x (13 + 8)
= ½ x 24 x 21
= 12 x 21
=252 m²
Hence required area of the field is 252 m²
Question :5. The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.
Answer :
Given: 𝒹1 =7.5 cm and 𝒹2 = 12 cm
We know that,
Area of rhombus = ½ x 𝒹1 x 𝒹2
= ½ x 7.5 x 12
= 45 45 cm²
Hence area of rhombus is 45cm².
Question :6. Find the area of a rhombus whose side is 6 cm and whose altitude is 4 cm. If one of the diagonals is 8 cm long, find the length of the other diagonal.
Answer :
Since rhombus is also a kind of parallelogram.
∴ Area of rhombus
= Base x Altitude
= 6 x 4 = 24cm²
Area of rhombus = ½ x 𝒹1 x 𝒹2
⇒ 24 = ½ x 8 x 𝒹2
⇒ 24 = 4𝒹2
⇒ 𝒹2 =6
= 6 cm
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 荷4.
Answer :
Here, 𝒹1 = 45 cm and 𝒹2 = 30 cm
∴ Area of one tile = ½ x 𝒹1 x 𝒹2
= ½ x 45 x 30
= 675 cm²
∴ Area of 3000 tiles
= 675 3000 = 2025000 cm²
= 2025000⁄10000
= 202.50 m² [∵ 1m² = 10000cm²]
∴ Cost of polishing the floor per
sq. meter = 4
∴ Cost of polishing the floor per 202.50 sq. meter = 4 x 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 sidealong 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.
Answer :
Given: Perpendicular distance
= 100 m
Area of the trapezium shaped field
= 10500 m²
Let side along the road be 𝓍 m and side along the river = 2𝓍m
Area of the trapezium field
= ½(a + b) x h
10500 = ½(𝓍 + 2𝓍) x 100
10500 = 3𝓍 X 50
3𝓍 = 10500⁄50
70 m
Hence the side along the river = 2𝓍
= 2 x 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.
Answer :
Given: Octagon having eight equal sides, each 5 m.
Construction: Divided the octagon in 3 figures, two trapeziums whose parallel and perpendicular sides are 11 m and 4 m respectively and third figure is rectangle having length and breadth 11 m and 5 m respectively.
= 2 x ½(11 + 5) x 4 = 4 x 16 = 64m²
And Area of rectangle = length breadth
= 11 x 5 = 55
∴ Total area of octagon = 64 + 55
= 119 m²
Question :10. There is a pentagonal shaped park as shown in the figure.
For finding its are a 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?
Answer :
First way: By Jyoti’s diagram,
Area of pentagon = Area of trapezium ABCP + Area of trapezium AEDP
= ½(AP + BC) x CP + ½(ED + AP) DP = (30 + 15 ) x CP + (15 + 30) DP
= ½(30 + 15) (CP + DP)
= ½ x 45 CD
= 337.5 m²
Second way:
By Kavita’s diagram
Here, a perpendicular AM drawn to BE.
AM = 30 – 15 = 15 m
Area of pentagon
= Area of ΔABE + Area of square BCDE
= 112.5 + 225.0
= 337.5 m²
Hence total area of pentagon shaped
park = 337.5 m².
Question :11. Diagram of the adjacent picture frame has outer dimensions = 24 cm x 28 cm and inner dimensions 16 cm x 20 cm. Find the area of each section of theframe, if the width of each section is same.
Answer :
Here two of 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) x 𝒽
= ½(28 + 20) x 4
= ½ x 48 x 4
= 96
Also Area of figure (II) = 96 cm²
Now Area of figure (III)
= Area of trapezium = ½(𝒶 + 𝒷) x 𝒽
= ½(24 + 16) x 4 = ½ x 40 x 4 = 80 cm²
Also Area of figure (IV) = 80 cm²