7th Maths : Term 2 Unit 2 : Measurements : Area of Pathways : Text Book Back Exercises Questions with Answers, Solution

**Exercise
2.3**

**1. Find the area of a circular pathway
whose outer radius is 32 cm and inner radius is 18 cm.**

Outer radius R = 32 cm

Inner radius r = 18 cm

Area of the circular pathway = π (R^{2} – r^{2})

= 22/7 (32^{2}–18^{2}) cm^{2}

= 22/7 (32 +18)(32 –18) cm^{2}

= 22/7 × 50 × 14 cm^{2}

= 2200 cm^{2}

Area of the Circular pathway = 2200 cm^{2}.

** **

**2. There is a circular lawn of radius
28 m. A path of 7 m width is laid around the lawn. What will be the
area of the path?**

Inner radius of the circle r = 28 m

Width of the path w = 7m

Outer radius of the circle R = r + w

= 28 + 7 = 35 m

Area of the pathway = π (R^{2} – r^{2})

= π (R+ r)(R+ r)

= 22/7 (35 + 28)(35 – 28) m^{2}

= 22/7 × 63 ⊂ 7m^{2}

= 1386 m^{2}

Area of the pathway = 1386 m^{2}.

** **

**3.
A circular carpet whose radius is 106 cm is laid on a circular hall of radius
120 cm. Find the area of the hall uncovered by the carpet.**

Radius of the circular Hall R = 120 cm

Radius of the circular carpet r = 106 cm

The area of the hall uncovered by the carpet = πR^{2 }– πr^{2}

= π (R^{2 }– r^{2})

= π (R+ r)(R – r)

= 22/7 (120 +106)(120 –106) cm^{2}

= 22/7 × 226 × 14 cm^{2}

= 9944 cm^{2}

The area of the hall uncovered by the Carpet = 9944 cm^{2}.

** **

**4.
A school ground is in the shape of a circle with radius 103 m. Four tracks
each of 3 m wide has to be constructed inside the ground for the purpose
of track events. Find the cost of constructing the track at the rate of ₹50 per
sq.m.**

Outer radius of the circular ground R = 103 m

Width of the track w = 3m

Inner radius of the ground r = R – w

= (103 –3) m = 100 m

Area of the track = π (R^{2 }– r^{2})

= 22/7 (103^{2} – 100^{2}) m^{2}

= 22/7 (10609 – 10000) m^{2}

= 22/7 × 609 = 1914 m^{2}

Area of the track = 1914 m^{2}

The cost of constructing the track for 1 sq.m = ₹ 50

The cost of constructiog 1914 squ.m = ₹ 1914 × 50

= ₹ 95700

The cost of constructing the track = ₹ 95700.

** **

**5. The figure shown is the aerial view
of the pathway.Find the area of the pathway.**

Area of the outer rectangle = 80 × 50 m^{2}

= 4000 m^{2}

Area of the inner rectangle = 70 × 40 = 2800 m^{2}

The area of the pathway = Area of the outer rectangle – Area of
the inner rectangle.

= 4000m^{2} – 2800m^{2} = 1200 m^{2}

** **

**6.
A rectangular garden has dimensions 11 m **

Area of the outer rectangular garden = 11 × 8 = 88 m^{2}

Width of the path = 2 m

Area of the inner rectangular garden = (11 – 4) (8 – 4) m^{2}

= 7 × 4 = 28 m^{2}

Area of the path = Outer Area – Inner Area

= (88 – 28)m^{2} =
60 m^{2}

Area of the path = 60 m^{2}.

** **

**7. A picture is painted on a ceiling
of a marriage hall whose length and breadth are 18 m and 7 m respectively.
There is a border of 10 cm along each of its sides. Find the area of the
border.**

Area of the ceiling = 18 × 7 = 126 m^{2}

Width ofthe border w = 10
cm = 0.1m

Inner area of the picture = (18 – 0.2) (7 – 0.2)m^{2}

= 17.8 × 6.8 = 121.04 m^{2}

Area ofthe border = Area ofthe Ceiling – Area ofthe Picture

= (126 – 121.04)m^{2} = 4.96 m^{2}

The area of the border = 4.96 m^{2}.

** **

**8.
A canal of width 1 m is constructed all along inside the field which is 24
m long and 15 m wide. Find (i) the area of the canal (ii) the cost
of constructing the canal at the rate of ₹12 per sq.m.**

Outer area of the field = 24 × 15 m^{2} = 360 m^{2}

Width of the Canal = 1 m

Inner area of the field = (24 – 2) (15 – 2) m^{2}

= 22 × 13m^{2} =
286m^{2}

i) Area of the canal = Outer area of the field – Inner area of
the field.

= (360 – 286) m^{2} = 74m^{2}

ii) The cost of constructing the canal for 1 sq.m = ₹ 12

The cost of constructing the canal for 74 m^{2} = ₹ 12 ×
74

= ₹ 888

The cost of constructing the canal = ₹ 888

** **

__Objective type questions__

**9.
The formula to find the area of the circular path is**

(i) π(*R* ^{2} −
*r*^{2} ) *sq. units*

(ii) πr^{2} *sq. units*

(iii) 2πr^{2 }*sq. units*^{}

(iv) πr^{2} +
2*r sq. units*

**Answer: **(i)
π (R^{2 }– r^{2}) sq. units

**10.
The formula used to find the area of the rectangular path is**

(i) π(*R*^{2} −
*r*^{2} ) *sq. units *

(ii) (*L* ×
*B*) − (*l* × *b*) *sq. units*

(iii) *LB sq. units*

(iv) *lb sq. units*

**Answer: **(ii)
(L × B) – (l × b) sq. units

**11.
The formula to find the width of the circular path is**

(i) ( *L* −
*l* ) *units*

(ii) ( *B* −
*b* ) *units*

(iii) ( *R* −
*r* ) *units*

(iv) ( *r* −
*R* ) *units*

**Answer: **(iii)
(R – r) units

* *

__ANSWERS:__

**Exercise 2.3**

1. 2200 *cm*^{2}

2. 1386 *m*^{2}

3. 9944 *cm*^{2}

4. ₹95700

5. 1200 *m*^{2}

6. 60 *m*^{2}

7. 4.96 *m*^{2}

8. (i) 74 *m*^{2} (ii) ₹888

**Objective type questions **

9. (i) π( R^{2}
− *r*^{2} ) *sq.units*

10. (ii) (L × B) − (l
× b) *sq.units*

11. (iii) R − r

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

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7th Maths : Term 2 Unit 2 : Measurements : Exercise 2.3 (Area of Pathways) | Questions with Answers, Solution | Measurements | Term 2 Chapter 2 | 7th Maths

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