8th Maths : Chapter 2 : Measurements : Perimeter and Area of Combined shapes : Exercise 2.2 : Text Book Back Exercises Questions with Answers, Solution

**Exercise 2.2**

** **

**1. Find the perimeter and area of the
figures given below. (***π =*** 22/7)**

**Solution: **

**(i)**

Length of the arc of the semicircle = (1/2) × 2π*r* units

= (22/7) × (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*) – (1/2) π*r*^{2} sq. units

= (10 × 7) – (1/2) × (22/7)
× (7/2) × (7/2)

= 70 – [(11 × 7) / (2 × 2)] = [280 – 77] / 4 = 203/4

= 50.75 m^{2} (approx)

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

**(ii)** Perimeter = sum of outside lengths

Length of the arc of quadrant circle = (1/4) × 2π*r* units

= (1/2) × (22/7) × 3.5 cm

= 11 × 0.5 cm = 5.5 cm

∴ Length of arc of 2
sectors = 2 × 5.5 cm

= 11 cm

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

P = 30 cm

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

= 2 × (1/4)π*r*^{2} +* lb* sq. units

= (1/2 × 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

** **

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

**Solution:**

**(i) **

Area of the shaded part = Area of 4 quadrant circles of radius 10/2
cm

= 4 × 1/4 × π*r*^{2} = 3.14 × 10/2 × 10/2 cm^{2}

= 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 − 78.5 cm^{2} = 21.5 cm^{2}

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

**(ii) **

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

= ( 1/2 π*r*^{2}) – (1/2 *bh*) cm^{2}

^{}

= (½) × 3.14 × 7 × 7 – (½) × 14 × 7 cm^{2}

= [ 153.86 / 2 ] – 49 cm^{2} = 76.93 – 49 cm^{2}

= 27.93 cm^{2}

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

** **

**3. Find the area of the combined figure
given which is got by the 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 cm^{2} = 48 cm^{2}

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

** **

**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. ( **

**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)

= ( (1/2) π*r*^{2} ) + (2 × [ 1/2 *bh* ] ) sq.
units

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

= [28.26 / 2] + 27 cm^{2} = 14.13 + 27 cm^{2} =
41.13 cm^{2}

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

** **

**5. The door mat which is hexagonal in
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* = 70/2 cm; *a* = 90 cm; *b*
= 70 cm

∴ Area of the trapezium = 1/2 [ *h* (*a + b*) ] sq.
units

Area of the doormat = 2 ×
(1/2) × 35 (90 + 70 ) cm^{2}

= 35 × 160 cm^{2} = 5600 cm^{2}

∴ Area of the door mat = 5600 cm^{2}

** **

**6. 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*) + (1/2 × base × height)
+ 1/2 × *h* × (*a + b*) sq. units

= (80 × 30) + (1/2 × 30 × 20) + 1/2 × 20 × (50 + 30) cm^{2}

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

Area of the figure = 3500 cm^{2}

** **

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

**Solution:**

Area of the field = Area of trapezium FBCH + Area of ΔDHC + Area
of ΔEGD + Area of ΔEGA + Area of ΔBFA

Area of the triangle = 1/2 *bh* sq. units

Area of the trapezium = 1/2 × *h* × (*a + b*) sq. units

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

Area of the ΔDHC = 1/2 × 8 × 5 m^{2} = 20 m^{2} ...(2)

Area of ΔEGD = 1/2 × 8 × 15 m^{2} = 60 m^{2} ...(3)

Area of ΔEGA = 1/2 × 8 × (8 + 6) m^{2 }= 4 × 14 m^{2}

= 56 m^{2} ...(4)

Area of ΔBFA = 1/2 × 3 × 6 m^{2 }= 9 m^{2 }...(4)

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

= 244 m^{2}

Area of the field = 244 m^{2}

**Answer:**

**Exercise
2.2 **

**1. 38 m, 50.75 m^{2} (approximately) (ii) 30 cm,40.25 cm^{2} (approximately)**

**2 (i) 21.5 cm^{2} (approximately) (ii) 27.93 cm^{2}
(approximately)**

**3. 48 cm^{2}**

**4. 41.13 cm^{2} (approximately)**

**5. 5600 cm^{2}**

**6. 3500 cm^{2}**

**7. 244 m^{2} **

^{}

Tags : Questions with Answers, Solution | Measurements | Chapter 2 | 8th Maths , 8th Maths : Chapter 2 : Measurements

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8th Maths : Chapter 2 : Measurements : Exercise 2.2 (Perimeter and Area of Combined shapes) | Questions with Answers, Solution | Measurements | Chapter 2 | 8th Maths

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