Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.2

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Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.2

Question 1.
Find the perimeter and area of the combined figures given below, (π = \(\frac{22}{7}\))

Solution:

∴ Perimeter = Sum of all lengths of sides that form the closed boundary
P = 11 + 10 + 7 + 10m
Perimeter = 38 m
Area = Area of the rectangle – Area of semicircle

= 50.75 m² (approx)
Area of the figure = 50.75m² approx.

(ii) Perimeter = sum of outside lengths

Question 2.
Find the area of the shaded part of the following figures. (π = 3.14 )

Solution:
Area of the shaded part = Area of 4 quadrant circles of radius \(\frac{10}{2}\) cm
= 4 × \(\frac{1}{4}\) × πr² = 3.14 × \(\frac{10}{2} \times \frac{10}{2}\) cm²
= \(\frac{314}{4}\) cm² = 78.5cm²
Area of the shaded part = 78.5 cm²
Area of the unshaded part = Area of the square – Area of shaded part
= a² – 78.5 cm² = (10 × 10) – 78.5 cm²
= 100 – 78.5 cm² = 21.5 cm²
Area of the unshaded part = 21.5 cm² (approximately)

(ii) Area of the shaded part = Area of semicircle – Area of the triangle

∴ Area of the shaded part = 27.93 cm² (approximately)

Question 3.
Find the area of the combined figure given which is got by joining of two parallelograms.

Solution:
Area of the figure = Area of 2 parallelograms with base 8 cm and height 3 cm
= 2 × (bh) sq. units
= 2 × 8 × 3 cm² = 48 cm²
∴ Area of the given figure = 48 cm²

Question 4.
Find the area of the combined figure given, formed by joining a semicircle of diameter 6 cm with a triangle of base 6 cm and height 9 cm. (π = 3.14 )

Solution:
Area of the figure = Area of the semicircle of radius 3 cm + 2 (Area of triangle with b = 9 cm and h = 3 cm)

= 14.13 + 27 cm² = 41.13 cm²
∴ Area of the figure = 41.13 cm² (approximately)
The door mat which is in a hexagonal shape

Question 5.
The door mat which is in a hexagonal shape has the following measures as given in the figure. Find its area.

Solution:
Area of the doormat = Area of 2 trapezium
Height of the trapezium h = \(\frac{70}{2}\) cm;
a = 90 cm; b = 70 cm
∴ Area of the trapezium
= \(\frac{1}{2}\)h (a + b) sq. units
Area of the door mat
= 2 × \(\frac{1}{2}\) × 35 (90 + 70) cm²
= 35 × 160 cm² = 5600 cm²
∴ Area of the door mat = 5600 cm²

Question 6.
Find the area of an invitation card which has two semicircles attached to a rectangle as in the figure given. (π = \(\frac{22}{7}\))

Solution:
Area of the card = Area of the rectangle + area of 2 semicircles
Length of the rectangle l = 30 cm
Breadth b = 21 cm
Radius of the semicircle = \(\frac{21}{27}\)
∴ Area of the card
= (l × b) + (2 × \(\frac{1}{2}\) πr²) sq. units
= 30 × 21 + \(\frac{22}{7} \times \frac{21}{2} \times \frac{21}{2}\) cm² = 630 + 346.5
= 976.5 cm² (approximately)
∴ Area of the Invitation card = 976.5 cm²

Question 7.
Find the area of the combined figure given, which has two triangles attached to a rectangle.

Solution:
Area of the combined shape = Area of the rectangle + Area of 2 triangles
Length of the rectangle l = 10 cm
Breadth b = 8 cm
Base of the triangle base = 8 cm
Height h = 6 cm
∴ Area of the shape = (l × b) + (2 × \(\frac{1}{2}\) × base × h) cm²
= (10 × 8) + (8 × 6) cm² = 80 + 48 cm² = 128 cm²
Area of the given shape = 128 cm²

Question 8.
A rocket drawing has the measures as given in the figure. Find its area.

Solution:
Area = Area of a rectangle + Area of a triangle + Area of a trapezium
For rectangle length l = 120 – 20 – 20 cm = 80 cm
Breadth b = 30 cm
For the triangle base = 30 cm
Height = 20 cm
For the trapezium height h = 20 cm
Parallel sided a = 50 cm
b = 30 cm
∴ Area of the figure = (l × b) + (\(\frac{1}{2}\) × base × height) + \(\frac{1}{2}\) × h × (a + b) sq. units
= (80 × 30) + ( \(\frac{1}{2}\) × 30 × 20) + \(\frac{1}{2}\) × 20 × (50 + 30) cm²
= 2400 + 300 + 800 cm² = 3500 cm²
Area of the figure = 3500 cm²

Question 9.
Find the area of the glass painting which has a triangle on a square as given in the figure.

Solution:
Area of the glass painting = Area of the square + Area of the triangle
Side of the square a = 30 cm
Base of the triangle b = 30 cm
Height of the triangle h = 8 cm
∴ Area of the painting = (a)² + (\(\frac{1}{2}\) bh) sq. units
= (30 × 30) + (\(\frac{1}{2}\) × 30 × 8) cm² = 900 + 120 cm²
= 1020 cm²
Area of the glass painting = 1020 cm²

Question 10.
Find the area of the irregular polygon shaped fields given below.

Solution:
(i) Area of the irregular field = Area of ∆AHF + Area of trapezium FHIE + Area of triangle EID + Area of ∆JDC + Area of rectangle BGJC + Area of ∆AGB

(ii) Area of the field = Area of trapezium FBCH + Area of ∆DHC + Area of ∆EGD + Area of ∆EGA + Area of ∆BFA