7th Maths : Term 2 Unit 2 : Measurements : Exercise 2.4 : Miscellaneous Practice problems, Challenge Problems, Text Book Back Exercises Questions with Answers, Solution

**Exercise
2.4**

__Miscellaneous
Practice problems__

** **

**1. A wheel of a car covers a distance
of 3520 cm in 20 rotations. Find the radius of the wheel?**

In 20 rotations the wheel of a car covers the distance = 3520 cm

In one rotation the wheel covers the distance = 3520 / 20 = 176
cm

Circumference ofthe circle 2πr = 176

2 × 22/7 ⊂ r = 176

r = 176×7 / 2×22 = 28 cm

The radius of the wheel = 28 cm

** **

**2. The cost of fencing a circular race
course at the rate of ₹8 per metre is ₹2112. Find the diameter of the race course.**

The distance cover on fencing for ₹ 8 = 1 metre

The distance cover on fencing for ₹ 2112 = 2112 / 8 m

= 264 m

The circumference of the race course 2πr = 264 m

2
× 22/7 ⊂ r = 264

r = 264×7 / 2×22 = 42 m

Diameter d = 2r = 2 × 4 = 84 m

The diameter of the race course = 84 m

** **

**3. A path 2 m long and 1 m
broad is constructed around a rectangular ground of dimensions 120 m and
90 m respectively. Find the area of the path.**

Inner area of the rectangular ground = 120 × 90 = 10800 m^{2 }

Outer area = (120 + 2×2) (90 + 2×1) m^{2}

= (120 + 4) (90 + 2) m^{2}

= 124 × 92 m^{2 }=
11408 m^{2}

The area of the path = Outer area – Inner area

= (11408 – 10800) m^{2}

The area of the Path = 608 m^{2}

** **

**4. The cost of decorating the circumference
of a circular lawn of a house at the rate of ₹55 per metre is ₹16940. What is the
radius of the lawn?**

The distance decorated for ₹ 55 = 1 metre

The distance decorated for ₹ 16940 =16940 / 55 = 308 m

Circumference of the circle 2πr = 308 m

2 × 22/7 ⊂r = 308 m

r = 308×7 / 2×22 = 49 m

The radius of the lawn r = 49 m.

** **

**5.
Four circles are drawn side by side in a line and enclosed by a rectangle as shown
below. If the radius of each of the circles is 3 cm, then calculate:**

**(i)
The area of the rectangle.**

**(ii)
The area of each circle.**

**(iii)
The shaded area inside the rectangle.**

The radius of the circle = 3 cm

Length of the rectangle = 8 × radius

= 8 × 3 = 24 cm

Breath = 2 × radius

= 2 × 3 = 6 cm

i) The area of the rectangle = *l* ×* b*

= 24 × 6 = 144 cm^{2}

ii) Radius of a circle r = 3 cm

Area = πr^{2}

The area of each circle = 22/7 × 3 × 3 = 198/7 = 28.28 cm^{2}

iii) Area of a circle = 28.28 cm^{2}

Area of 4 circles = 4 × 28.28 cm^{2}

^{ }= 113.12 cm^{2}

The shaded area inside the rectangle = Area ofthe rectangle – Area
of 4 circles.

= (144 – 113.12) cm^{2}

= 30.88 cm^{2}

** **

__Challenge
Problems__

** **

**6.
A circular path has to be constructed around a circular lawn. If the outer and inner
circumferences of the path are 88 cm and 44 cm respectively, find
the width and area of the path.**

Outer circumference of the circle = 88 cm

2 πR = 88 cm

R = [ 88 × 7 ] / [ 2 × 22 ] = 14 cm

Inner circumference = 44 cm

2 πr = 88 cm

R = 44×7 / 2×22 = 7 cm

The width of the path = R – r

= 14 – 7 = 7 cm

The area of the path = π (R^{2} – r^{2})

= 22/7 (14^{2} –
7^{2})

= 22/7 (196 – 49)

= 22/7 × 147 cm^{2}

^{ }= 462 cm^{2}

The area of the path = 462 cm^{2}

** **

**7. A cow is tethered with a rope of length
35 m at the centre of the rectangular field of length 76 m and breadth
60 m. Find the area of the land that the cow cannot graze?**

The area of the rectangle = 76 × 60 m^{2}

= 4560 m^{2}

The area of the land that cow can graze = Area of the circle of
radius

r = 35 m

Area of the land that cow can graze = πr^{2}

= 22/7 × 35 × 35 m^{2
}

= 3850 m^{2}

The area of the land that cow can not graze = Area of the
rectangle – Area of the land that cow can graze.

= (4560 – 3850) m^{2}

The area of the land that cow can not graze = 710 m^{2}.

** **

**8. A path 5 m wide runs along
the inside of the rectangular field. The length of the rectangular field is three
times the breadth of the field. If the area of the path is 500 m^{2}
then find the length and breadth of the field.**

Length of outer rectangle =*
l* m

Outer breadth = *b* m

Width of the path = 5 m

Length of inner rectangle = *l*
– 2w = *l* – 10 m

Inner breadth = *b *– 2w
= *b *– 10m

Area of the outer rectangle = *l *x* b* m^{2}

Area ofthe inner rectangle = (*l* – 10) (*b *– 10) m^{2}

= (*l* *b *– 10* l* –10* b* + 100)m^{2}

^{}

Area of the path = Area of the outer rectangle – Area of the
inner rectangle

= (*l* *b *–*
lb* + 10*l* + 10*b* – 100 = 500

= 10 (*l + b*) = 500 +
100 = 600 m^{2}

= *l + b* = 600 / 10 =
60

Length of the rectangle = 3 times its breadth.

*l *= 3*b*

*l + b* = 60

3*b* + *b* = 60

4*b* = 60

*b *= 60 / ~~4~~ = 15 m

*l *= 3*b* = 3 × 15 = 45m

The length of the field = 45 m

Breadth = l5m

** **

**9. A circular path has to be constructed
around a circular ground. If the areas of the outer and inner circles are 1386 m^{2}
and 616 m^{2} respectively, find the width and area of the path.**

Area of the outer circle = 1386 m^{2}

πR^{2} = 1386

R^{2} = 1386/22 ×
7 = 9 × 7 × 7

R = 3 × 7 = 21m

Area of the inner circle πr^{2} = 616

r^{2} = 616/22 × 7 = 4 × 7 × 7

r = 2 × 7 = 14 m

Width of the circle w = R – r

= 21 – 14 = 7 m

Area of the Path = Area ofthe outer circle – Area of the inner
circle.

= (1386– 616) m^{2}

Area of the Pathe = 770 m^{2}.

** **

**10. A goat is tethered with a rope of
length 45 m at the centre of the circular grass land whose radius is 52 m.
Find the area of the grass land that the goat cannot graze.**

Radius ofthe circular grass land r = 52 m

Area of the grass land = ( 22/7) × 52 × 52 m^{2}

Area of the grass land = 59488 / 7 = 8498.3 m^{2}

The length ofthe tethered rope = radius r = 45 m

Area of the grass land that the goat can graze πr^{2} =
22/7 × 45 × 45m^{2}

= 44550 / 7 = 6364.3 m^{2}

Area of th grass land that the goat can not graze = Area of the
circle – Area of the grass land that the goat can graze.

= (8498.3 – 6364.3)m^{2}

= 2134 m^{2}

The area ofthe grass land that the goat cannot graze. = 2134m^{2}

** **

**11. A strip of 4 cm wide is cut
and removed from all the sides of the rectangular cardboard with dimensions 30 cm
**

Area of the rectangular cardboard = 30 × 20 cm^{2}

= 600 cm^{2}

Width of the cutting portion w = 4 cm

Area ofthe inner rectanglre = (30 – 8) (20 – 8) cm^{2}

= 22 × 12
cm^{2}

Area of the cutting portion = 264 cm^{2}

Area of the removed portion = Area of the rectangle – Area of
the cutting portion.

= (600 – 264) cm^{2}

= 336 cm^{2}

Area of the cutting portion = 264 cm^{2}

Area of the removed portion = 336 cm^{2}

** **

**12.
A rectangular field is of dimension 20 m **

Area of the longer path = 20 × 2 = 40 m^{2}

Area of the shorter path = 15 × 1 = 15 m^{2}

Area of the centre rectangle = 2 × 1 = 2 m^{2}

Area of the path = (40 + 15 – 2) m^{2}

= 53 m^{2}

^{}

i) Area of the pathe = 53 m^{2}

Area of the field = 20 × 15= 300m^{2}

ii) Area of the remaining portion = Area of the field – Area of
the path.

= (300 – 53) m^{2}
= 247 m^{2}

iii) The cost of constructing the road of 1 sq.m = ₹ 10

The cost of constructing the road for 53 sq.m = ₹ 53 × 10

= ₹ 530

The cost of constructing the road = ₹ 530.

__ANSWERS:__

**Exercise 2.4**

1. 28* cm *

2. 84* m *

3.668 *m*^{2} * *

4. 49* m *

5. (i) 196 *cm*^{2} (ii) 38.5 *cm*^{2} (iii) 42 *cm*^{2}

**Challenge problems **

6. 7 *cm*; 462 *cm*^{2}

7. 710 *m*^{2}

8. 30 *cm*; 10* cm *

9. 7 m; 770 *m*^{2}

10. 2134 *m*^{2}

11. 264 *cm*^{2} ; 336 *cm*^{2}

12. (i) 53 *m*^{2} (ii) 247 *m*^{2} (iii) ₹530

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7th Maths : Term 2 Unit 2 : Measurements : Exercise 2.4 | Questions with Answers, Solution | Measurements | Term 2 Chapter 2 | 7th Maths

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