AP 8th Class Maths 11th Chapter Mensuration Exercise 11.2 Solutions

Well-designed AP 8th Class Maths Textbook Solutions Chapter 11 Mensuration Exercise 11.2 offers step-by-step explanations to help students understand problem-solving strategies.

Mensuration Class 8 Exercise 11.2 Solutions – 8th Class Maths 11.2 Exercise Solutions

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:
Shape of surface of a table = trapezium in which lengths of parallel sides
a = 1 m, b = 1.2 m
and the perpendicular distance = h = 0.8 m
formula for area of trapezium
= \(\frac{h}{2}\) (a + b) (putting the values)
= \(\frac{0.8}{2}\) (1 + 1.2)
= 0.4 × 2.2 = 0.88 = 0.88 m²

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:
Given area of trapezium = 34 cm²
length of one parallel side = a = 10 cm
length of other parallel side = b = ?
height (h) = 4 cm
Putting the above values in the formula, we get
area = \(\frac{h}{2}\) (a + b)
34= \(\frac{4}{2}\)(10 + b) = 2(10 + b)
⇒ \(\frac{34}{2}\) = 10 + b ⇒ 10 + b = 7
⇒ b = 17 – 10 = 7 cm
∴ Length of other parallel side = 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 AB ⊥ BC.AB ⊥ AD.
So BC, AD are parallel sides
Perimeter of trapezium
= AB + BC + CD + DA = 120
h +48+ 17 +40= 120
⇒ h + 105 = 120
⇒ h = 120 – 105 = 15 cm

then area of trapezium = \(\frac{h}{2}\) (a + b)
= \(\frac{15}{2}\)(AD + BC) = \(\frac{15}{2}\)(40 + 48)
= \(\frac{15}{2}\) × 88 = 15 × 44 = 660 m²

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:
Diagonal (d) = 24 m
two perpendicular dropped on it are
h₁ = 8 m , and h₂ = 13 m
∴ Formula for area of quadrilateral
= \(\frac{1}{2}\) d(h₁ + h₂)= \(\frac{1}{2}\) × 24 × (8 + 13)
= \(\frac{1}{2}\) × 24 × 21 = 12 × 21 = 252 m²
Area of the field = 252 m²

Question 5.
The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.
Solution:
Two diagonals of Rhombus d₁ = 7.5 cm ; d₂ = 12 cm

then formula for area of rhombus
= \(\frac{1}{2}\)d₁d₂ = \(\frac{1}{2}\) × 7.5 × 12 = 7.5 × 6 = 45 cm²

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 Rhombus = 5 cm
Altitude = shortest perpendicular
distance from its opposite side h = 4.8 cm
then area of Rhombus = bh
= 4.8 x 5 = 24 cm² ………. (1)

Let the first diagonal d₁ = 8 cm
then the second diagonal = d₂
So from the formula for area of rhombus
\(\frac{1}{2}\)d₁ d₂ = Area
\(\frac{1}{2}\) × 8 × d₂ = 24 cm² => 4 × d₂ = 24 cm
d₂ = \(\frac{24}{4}\) = 6 cm
So the length of second diagonal = 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:
Diagonals (d₁, d₂) of each
Rhombus shaped tile = 45 cm, 30 cm,
d₁ = 45 cm ; d₂ = 30 cm
then area of each tile = \(\frac{\mathrm{d}_1 \mathrm{~d}_2}{2}\)
= \(\frac{45 \times 30}{2}\) = 45 × 15 = 675 cm²
total no.of tiles used = 3000 tiles
∴ total floor area = area of 3000 tiles = 3000 × 675
(∵ m² = m × m = 100 × 100 cm²)
Converting into square meter

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

Solution:
Shape of the field = trapezium
let the length of Road side = x m
then the length of River side = 2x m
distance between those two parallel side = h = 100 m
∴ area of trapezium = 10500 m
⇒ \(\frac{h}{2}\)(a + b) = 10500 m²
⇒ \(\frac{100}{2}\)(x + 2x) = 10500 m²
⇒ 3x = \(\frac{10500}{50}\) = 210 m
⇒ x = \(\frac{210}{3}\) = 70 m
∴ length of the side along the river = 2x = 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.

As it is a regular octagon
then AB = BC = CD = DE = EF = FG = GH = HA = 5 cm

then area of BCIJ = L × B
= 5 × 4 = 20 m² …………..(1)
∴ area of GFLK = 20 m² ……..(2)
BC = IJ = 5 m
∴ AI + IJ + JD = 11
⇒ AI + JD = 11 – 5 = 6m
AI = \(\frac{6}{2}\) = 3 m
then area of ∆AIB = \(\frac{1}{2}\) × AI × BI
= \(\frac{1}{2}\) × 3 × 4 = \(\frac{12}{2}\) = 6 m²
∴ area of CJD = ∆LEF = ∆KGH = 6 m2 each now area of ADEH = AD × DE
= 11 × 5 = 55 m2 (v rectangle)
Area of octagon
= Area of ∆AIB + ∆CJD + ∆LED + ∆KGH + Area of rectangle BCJI + rectangle KLFG + rectangle ADEH
= 6 + 6 + 6 + 6 + 20 + 20 + 55
Total area of octagon = 119 m²