Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 2 Measurements Ex 2.2 Text Book Back Questions and Answers, Notes.

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

Question 1.

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

(i)

Answer:

Length of the arc of the semicircle = \(\frac{1}{2}\) × 2πr units

= \(\frac{22}{7} \times \frac{7}{2}\) m =11 m

∴ Perimeter = Sum of all lengths of sides that form the closed boundary

P = 11 + 10 + 7 + 10 m

Perimeter = 38 m

Area = Area of the rectangle – Area of semicircle

= (l × b) – \(\frac{1}{2}\) πr^{2} sq. units

= 50.75 m^{2} (approx)

Area of the figure = 50.75 m^{2} approx.

(ii)

Answer:

Perimeter = sum of outside lengths

Length of the arc of quadrant circle = \(\frac{1}{4}\) × 2πr units

= \(\frac{1}{2} \times \frac{22}{7}\) × 3.5cm

= 11 × 0.5 cm = 5.5cm

∴ Length of arc of 2 sectors 2 × 5.5 cm

= 11 cm

∴ Perimeter P = 11 + 6 + 3.5 + 6 + 35 cm

P = 30 cm

Area Area of 2 quadrant circle + Area of a rectangle.

= 2 × \(\frac{1}{4}\)πr^{2} + lb sq. units

= (\(\frac{1}{2} \times \frac{22}{7}\) × 3.5 × 3.5) + (6 × 3.5) cm^{2}

= (11 × 3.5 × 0.5) + 21 cm^{2}

= (19.25 + 21) cm^{2} = 40.25 cm^{2}

∴ Area = 40.25 cm^{2} approx

Question 2.

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

(i)

Answer:

Area of the shaded part = Area of 4 quadrant circles of radius \(\frac { 10 }{ 2 }\) cm

= 4 × \(\frac { 1 }{ 4 }\) × πr^{2} = 3.14 × \(\frac { 10 }{ 2 }\) × \(\frac { 10 }{ 2 }\) cm^{2}

= \(\frac { 314 }{ 4 }\) cm^{2} = 78.5 cm^{2}

Area of the shaded part = 78.5 cm^{2}

Area of the unshaded part = Area of the square – Area of shaded part

= a^{2} – 78.5 cm^{2} = (10 × 10) – 78.5 cm^{2}

= 100 – 78.5cm^{2} = 21.5cm^{2}

Area of the unshaded part = 21.5 cm^{2} (approximately)

(ii)

Answer:

Area of the shaded part = Area of semicircle – Area of the triangle

= \(\frac { 1 }{ 2 }\) × 3.14 × 7 × 7 – \(\frac { 1 }{ 2 }\) × 14 × 7cm^{2}

= \(\frac{153.86}{2}\) – 49 cm^{2} = 76.93 – 49 cm^{2}

= 27.3 cm^{2}

∴ Area of the shaded part = 27.93 cm^{2} (approximately)

Question 3.

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

Answer:

Area of the figure = Area of 2 parallelograms with base 8 cm and height 3 cm

= 2 × (bh) sq. units

= 2 × 8 × 3cm^{2} = 48 cm^{2}

∴ Area of the given figure = 48 cm^{2}

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)

Answer:

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

= \(\left(\frac{1}{2} \pi r^{2}\right)+\left(2 \times \frac{1}{2} b h\right)\) sq. units

= \(\frac { 1 }{ 2 }\) × 3.14 × 3 × 3 + (2 × \(\frac { 1 }{ 2 }\) × 9 × 3)cm^{2}

= \(\frac{28.26}{2}\) + 27 cm^{2} = 14.13 + 27 cm^{2} = 41.13 cm^{2}

∴ Area of the figure = 41.13 cm^{2} (approximately)

Question 5.

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

Answer:

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)cm^{2}

= 35 × 160 cm^{2} = 5600 cm^{2}

∴ Area of the door mat = 5600 cm^{2}

Question 6.

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

Answer:

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 = 30cm

∴ 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) cm^{2}

= 2400 + 300 + 800 cm^{2} = 3500 cm^{2}

Area of the figure = 3500 cm^{2}

Question 7.

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

Answer:

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

Area of the triangle = \(\frac { 1 }{ 2 }\) bhsq.units

Area of the trapezium = \(\frac { 1 }{ 2 }\) × h × (a + b) sq.units

Area of the trapezium FBCH = \(\frac { 1 }{ 2 }\) × (10 + 8) × (8 + 3)m^{2} = 9 × 11 = 99 m^{2} ….. (1)

Area of the ∆DHC = \(\frac { 1 }{ 2 }\) × 8 × 5 m^{2} = 20 m^{2} ….. (2)

Area of ∆EGD = \(\frac { 1 }{ 2 }\) × 8 × 15m^{2} = 60 m^{2} …….. (3)

Area of ∆EGA = \(\frac { 1 }{ 2 }\) × 8 × (8 + 6)m^{2} = 4 × 14 m^{2}

= 56m^{2}

Area of ∆BFA = \(\frac { 1 }{ 2 }\) × 3 × 6m^{2} = 9 m^{2}

∴ Area of the field = 99 + 20 + 60 + 56 + 9 m^{2}

= 244 m^{2}

Area of the field = 244 m^{2}