CBSE Class 7

These NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area InText Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area InText Questions

NCERT In-text Question Page No. 205

Question 1.
What would you need to find, area or
perimeter, to answer the following?
1. How much space does a blackboard occupy?
2. What is the length of a wire required to fence a rectangular flower bed?
3. What distance would you cover by taking two rounds of a triangular park?
4. How much plastic sheet do you need to cover a rectangular swimming pool?
Answer:
1. Area 2.Perimeter 3. Perimeter 4. Area

NCERT In-text Question Page No. 206

Question 1.
Experiment with several such shapes and cutouts. You might find it useful to draw these shapes on squared sheets and compute their areas and perimeters. You have seen that increase in perimeter does not mean that area will also increase.
Answer:
Please do this question yourself with the help of your subject teacher.

Question 2.
Give two examples where the area increases as the perimeter increases.
Answer:
Please do this question yourself with the help of your subject teacher.

Question 3.
Give two examples where the area does not increase when perimeter increases.
Ans: Please do this question yourself with the help of your subject teacher.

NCERT In-text Question Page No. 210

Question 1.
Each of the following rectangles of length 6 cm and breadth 4 cm is composed of congruent polygons. Find the area of each polygon.

Answer:
∴ Length of the rectangle (l) = 6 cm
Breadth of the rectangle (b) = 4 cm
∴ Area of the rectangle = (l x b)
= 6 x 4 cm² = 24 cm²
(i) Here, number of congruent polygons = 6
∴ Area of each polygon = \(\frac{24}{6}\) cm² = 4 cm²

(ii) Here, number of congruent polygons = 4
Area of each polygon = \(\frac{24}{4}\)cm² = 6 cm²

(iii) Here, number of congruent polygons = 2
∴ Area of each polygon = \(\frac{24}{2}\)cm² = 12 cm²

(iv) Number of congruent polygons = 2
∴ Area of each polygon = \(\frac{24}{2}\)cm² = 12 cm²

(v) Number of congruent polygons = 8
∴ Area of each polygon = \(\frac{24}{8}\)cm² = 3 cm²

NCERT In-text Question Page No. 212

Question 1.
Find the area of the following parallelograms:

(iii) In a parallelogram ABCD, AB = 7.2 cm and the perpendicular from C on AB is 4.5 cm.
Answer:
(i) Base = 8 cm, Height = 3.5 cm
∴ Area of the parallelogram = Base x
Height = 8 cm x 3.5 cm = 28 cm²

(ii) Base = 8 cm, Height = 2.5 cm
Area of the parallelogram = Base x Height = 8 cm x 2.5 cm = 20 cm²

(iii) Base of parallelogram ABCD = (AB) = 7.2 cm
Height of parallelogram ABCD = 4.5 cm
Area of parallelogram ABCD
= Base x Height
= 7.2 cm x 4.5 cm = \(\frac { 72 }{ 10 }\)cm x \(\frac { 45 }{ 10 }\)cm
= \(\frac{3240}{100}\) cm² = 32.40 cm²

NCERT In-text Question Page No. 213

Question 1.
Try the activity given on page 213, NCERT Textbook with different types of triangles.
Answer:
Do it yourself.

Question 2.
Take different parallelograms. Divide each of the parallelograms into two triangles by cutting any of its diagonals. Are the triangles congruents.
Answer:
Do it yourself.

NCERT In-text Question Page No. 219

Question 1.
In the adjoining figure,
(a) Which square has the larger perimeter?
(b) Which is larger, perimeter of smaller square or the circumference of the circle?
Answer:
(a) The outer square has the larger perimeter.
(b) The circumference of the circle is larger than the perimeter of the smaller square.

NCERT In-text Question Page No. 222

Question 1.
Draw circles of different radii on a graph paper. Find the drea by counting the number of squares. Also find the area by using the formula. Compare the two answers.
Answer:
Do it yourself.

NCERT In-text Question Page No. 225

Question 1.
Convert the following:
(i) 50 cm² in mm²
(ii) 2 ha in m²
(iii) 10 m2 in cm²
(iv) 1000 cm² in m²
Answer:
(i) 50 cm² in mm²
∵ 1 cm² = 100 mm²
∴ 50 cm² = 50 x 100 mm² = 5000 mm²

(ii) 2 ha in m²
1 ha = 1000 m²
∴ 2 ha = 2 x 1000 m2 = 20000 m²

(iii) 10 m² in cm²
∵ 1 cm² = 10000 cm²
∴ 10 m²= 10 x 10000 cm² = 100000 cm²

(iv) 1000 cm² in m²
∵ 10000 cm² = 1 m²
∴ 1 cm² = \(\frac{1}{10000}\) m²
So, 1000 cm² = \(\frac{1}{10000}\) x 1000 m²
= \(\frac{1}{10}\) m² = 0.1m²

These NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Exercise 11.4

Question 1.
A garden is 90 m long and 75 m broad. A path 5 m wide is to be built outside and around it. Find the area of the path. Also, find the area of the garden in hectare.
Answer:
Length of the garden (1) = 90 m
Breadth of the garden (b) = 75 m
Area of the garden = 1 x b sq units = 90 m x 75 m = 6750 m²

Length of the outer rectangle (L)
= 90 + 5 + 5 m
= 100 m
Breadth of the outer rectangle (B)
= 75 + 5 + 5 m
= 85 m
Area of the outer rectangle
= L x B = 100 x 85 m²
= 8500 m²
Area of the pathway = Area of the outer rectangle- Area of the inner rectangle
= (8500 – 6750)m²
= 1750m²
(i) Area of the garden = 6750 m²
= \(\frac{6750}{10000}\) ha
= 0.675 ha (1 m² = \(\frac{1}{10000}\) ha)
(ii) Area of the path = 1750 m²

Question 2.
A 3 m wide path runs outside and around a rectangular park of length 125 m and breadth 65 m. Find the area of the path.
Answer:
Length of the park (1) = 125 m
breadth of the park (b) = 65 m
Area of the park = l x b
= 125 x 65 m²
= 8125m²

Length of the outer region (L)
= (125 + 3 + 3) m
= 131 m
Breadth of the outer region (B)
= (65 + 3+ 3) = 71 m
Area of the outer region
= L x B m²
= (131 x 71) m²
= 9301 m²
Area of the path = Area of the outer region – Area of the park
= 9301 m² – 8125 m²
= 1176 m²

Question 3.
A picture is painted on a cardboard 8 cm long and 5 cm wide such that there is a margin of 1.5 cm along each of its sides. Find the total area of the margin.
Answer:
Length of the cardboard (l) = 8 cm
Width of the cardboard (b) = 5 cm
Area of the card board = 1 x b sq.
= 8 x 5 = 40 cm²

Width of the margin = 1.5 cm
Length of the inner rectangle
= 8 – (1.5 + 1.5) cm
= 5 cm
Breadth of the inner rectangle
= 5 – (1.5 + 1.5)cm
= 2 cm
Area of the inner rectangle
= 5 x 2 = 10 cm²
Area of the margin = Area of the cardboard – Area of the inner rectangle
= 40 cm² – 10 cm²
= 30 cm²

Question 4.
A verandah of width 2.25 m is constructed all along outside a room which is 5.5 m long and 4 m wide. Find:
(i) the area of the verandah.
(ii) the cost of cementing the floor of the verandah at the rate of ₹ 200 per m² .
Answer:
Length of the room= 5.5 m
Breadth of the room = 4 m
Area of the room = 5.5 m x 4 m = 22 m²
Width of the verandah = 2.25 m
Length of the verandah
= 5.5 + (2.25 +2.25) m
= 10 m

Breadth of the Verandah
= 4 + (2.25 + 2.25) m
= 8.5 m
Area of the outer rectangle
= 10 m x 8.5 m ‘
= 85 m²
Area of the verandah = Area of the outer rectangle – Area of the inner rectangle
= 85 m² – 22 m²
= 63 m²
Cost of cementing the verandah
= ₹ 200 x 63
= ₹ 12600

Question 5.
A path 1 m wide is built along the border and inside a square garden of side 30 m. Find:
(i) the area of the path
(ii) the cost of planting grass in the remaining portion of the garden at the rate of? 40 per m².
Answer:
(i) Length of the outer square = 30 m
Area of the outer square = side x side
= 30 x 30 = 900 m²

Let width of the path = 1 m
Side of the inner square
= [30 – (1 + 1)3 m
= 30m-2m = 28m
Area of the inner square
= (28 x 28) m²
= 784 m²
Area of the path = Area of the outer square – Area of the inner square
= (900 – 784) m²
Area of the path = 116 m²

(ii) Rate of planting grass
= ₹ 40 per m²
Cost of planting grass = ₹ 40 x 784
= ₹ 31,360

Question 6.
Two cross roads, each of width 10 m, cut at right angles through the centre of a rectangular park of length 700 m and breadth 300 m and parallel to its sides. Find the area of the roads. Also find the area of the park excluding cross roads. Give the answer in hectares.
Answer:
Length of the rectangular park (l) = 700 m
Breadth of the rectangular Park (b) = 300 m

Area of the park = l x b sqm
= 700 x 300 m²
= 210000 m²
Area of the road HEFG (length wise)
= 700 m x 10 m
= 7000 m²
Area of the road PQRS (breadth wise)
= 300 x 10 = 3000 m²
Area of KLMN = 10 x 10 = 100 m²
Area of the roads = Area of the road HEFG + Area of the Road PQRS – Area of KLMN (which is repeated two times.)
= (7000 +3000-100) m²
= 9900 m²
Area of the park excluding cross roads = Area of the park – Area of the cross roads
= 21,0000 m² – 9900 m² = 200100 m²
= \(\frac{200100}{10000}\) ha
∴ Area of the park = 20.01ha

Question 7.
Through a rectangular field of length 90 m and breadth 60 m, two roads are constructed which are parallel to the sides and cut each other at right angles through the centre of the fields. If the width of each road is 3 m, find
(i) the area covered by the roads.
(ii) the cost of constructing the roads at the rate of? 110 per m2.
Answer:
Length of the rectangular field = 90 m
Breadth of the rectangular field = 60 m
Width of each road = 3m.

Area of the road ABCD = 90 x 3 m²
= 270 m²
Area of the road EFGH = 60 x 3
= 180 m²
Area of PQRS = 3 x 3 = 9 m²
(i) Area covered by the roads = Area of the road ABCD + Area of the road EFGH – Area of PQRS (which is repeated two times)
= (270 + 180 – 9) m²
= 450 m² – 9 m²
= 441 m²
(ii) Rate of construction of roads = ₹ 110/m²
Cost of construction of roads
= ₹ 110 x 441
= ₹ 48510

Question 8.
Pragya wrapped a cord around a circular pipe of radius 4 cm (figure given below) and cut off the length required of the cord. Then she wrapped it around a square box of side 4 cm (also shown). Did she have any cord left? (π = 3.14)
Answer:
Radius of the circular pipe (r) = 4 cm
Circumference of the pipe = 2πr.
= 2 x 3.14 x 4 cm = 25.12 cm

Side of the square box = 4 cm
Perimeter of the square box
= 4 x 4 cm
= 16 cm
Since, 25.12 cm > 16 cm
Difference in length
= 25.12 cm – 16 cm
= 9.12 cm.
Yes, she has 9.12 cm of length cord left.

Question 9.
The adjoining figure represents a rectangular lawn with a circular flower bed in the middle. Find: (Take π = 3.14)
(i) the area of the whole land
(ii) the area of the flower bed
(iii) the area of the lawn excluding the area of the flower bed
(iv) the circumference of the flower bed.
In the following figure, find the area of the shaded portions:

Answer:
Length of the land = 10 m
Breadth of the land = 5 m
(i) Area of the whole land
= (10 x 5) m²
= 50 m²
(ii) Radius of the flower bed (r)
= 2 m.
Area of the flower bed = πr² sq.m
= 3.14 x 2 x 2 m² = 12.56 m²

(iii) Area of the lawn excluding the area of the flower bed = 50 m² – 12.56 m²
= 37.44 m²

(iv) Circumference of the flower bed
= 2πr.
= 2 x 3.14 x 2 m
= 12.56 m

Question 10.
In the following figures, find the area of the shaded portions.

Answer:
(i) Area of the whole rectangle ABCD
= 18 x 10 cm² = 180 cm²
Area of the right ΔAEF
= \(\frac { 1 }{ 2 }\) x base x height
= \(\frac { 1 }{ 2 }\) x 6 x 10 cm²
= 30 cm²
Area of the right ΔCBE
= \(\frac { 1 }{ 2 }\) x base x height
= \(\frac { 1 }{ 2 }\) x 8 x 10 cm²
= 40 cm²
Area of the shaded portion = Area of ABCD – (Area of ΔAEF + Area of ΔCBE)
= 180 – (30 + 40)cm² = 180 cm² – 70 cm² = 110 cm²
(ii) Side of the square PQRS = 20 cm
Area of the square PQRS
= Side x Side
= 20 x 20 cm²
= 400 cm²
Area of the right ΔQPT
= \(\frac { 1 }{ 2 }\) x base x height
= \(\frac { 1 }{ 2 }\) x 20 x 10 cm²
= 100 cm²
Area of the rightΔTSU
= \(\frac { 1 }{ 2 }\) x 10 x 10 cm²
= 50 cm²
Area of the right AQRU
= \(\frac { 1 }{ 2 }\) x 10 x 20 cm²
= 100 cm²
Area of the shaded portion = Area of the square PQRS – (Area of ΔQPT + Area of ΔTSU + Area of ΔQRU)
= 400 cm² – (100 + 50 + 100) cm²
= (400 – 250) cm²
= 150 cm²

Question 11.
Find the area of the quadrilateral ABCD. Here, AC = 22 cm, BM = 3 cm, DN = 3 cm, and BM ⊥ AC, DN ⊥ AC
Answer:
Area of ΔABC

= \(\frac { 1 }{ 2 }\) x AC x BM
= \(\frac { 1 }{ 2 }\) x 22 x 3 cm²
= 33 cm²
Area of AACD = \(\frac { 1 }{ 2 }\) x AC x ND
= \(\frac { 1 }{ 2 }\) x 22 x 3 cm²
= 33 cm²
∴ Area of the quadrilateral ABCD = Area of ΔABC + Area of ΔACD
= 33 cm² + 33 cm²
= 66 cm²

These NCERT Solutions for Class 7 Maths Chapter 5 Lines and Angles Ex 5.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 5 Lines and Angles Exercise 5.1

Question 1.
Find the complement of each of the following angles.

Answer:
(i) Complement of
20° = 90° – 20° = 70°

(ii) Complement of
63° = 90° – 63° = 27°

(iii) Complement of 5 7° = 90° – 57°
= 33°

Quesrion 2.
Find the supplement of each of the following angles:


Answer:
(i) Supplement of
105°= 180°- 105° = 75°

(ii) Supplement of
87° = 180° – 87° = 93°

(iii) Supplement of
154° = 180° – 154° = 26°

Question 3.
Identify which of the following pairs of angles are complementary and which are supplementary.
(i) 65°, 115°
(ii) 63°, 27°
(iii) 112°, 68°
(iv) 130°, 50°
(v) 45°, 45°
(vi) 80°, 10°
Answer:
(i) 65°+ 115°= 180°
65 and 115° are supplementary angles.

(ii) 63°+ 27° = 90°
63 and 27° are complementary angles.

(iii) 112°+ 68°= 180°
112° and 68° are supplementary angles.

(iv) 130° + 50° = 180°
130° and 50° are supplementary angles.

(v) 45° + 45° = 90°
45° and 45° are complementary angles.

(vi) 80° + 10° = 90°
80° and 10° are complementary angles.

Question 4.
Find the angle which is equal to its complement.
Answer:
Let the required angle be x.
Given that the angle is equal to its complement
x = 90° – x
x + x = 90°
2x = 90°
x = \(\frac{90^{\circ}}{2}\) = 45°
Thus, 45° is equal to its complement.

Question 5.
Find the angle which is equal to its supplement.
Answer:
Let the required angle be ‘m’ and supplement of m = 180° – m
∴ m = 180° – m
(∵ m is equal to its supplement)
m + m = 180°
2m = 180°
m = \(\frac{180}{2}\) =90°
Thus, 90° is equal to its supplement.

Question 6.
In the given figure, ∠1 and ∠2 are supplementary angles.
If ∠1 is decreased, what changes should take place in ∠2 so that both the angles still remain supplementary.

Answer:
In case ∠1 is decreased, the same amount of degree measure is added to ∠2.
i.e. ∠2 be increased by same amount of degree measure.

Question 7.
Can two angles be supplementary if both of them are:
(i) acute?
(ii) obtuse?
(iii) right?
Answer:
(i) Sum of two acute angle is always less than 180°
∴ Two acute angles cannot be supplementary.

(ii) Sum of two obtuse angles is always more than 180°.
∴ Two obtuse angles cannot be supplementary.

(iii) Sum of two right angles =180°
∴ Two right angles can be supplementary.

Question 8.
An angle is greater than 45°. Is its complementary angle greater than 45° or equal to 45° or less than 45°?
Answer:
Complement of an angle (greater than 45°) is less than 45°.

Question 9.
In the figure given below:
(i) Is∠1 adjacent to ∠2?
(ii) Is ∠AOC adjacent to ∠AOE?
(iii) Do ∠COE and ∠EOD form a linear pair?
(iv) Are ∠BOD and ∠DOA supplementary?
(v) Is ∠1 vertically opposite to ∠4?
(vi) What is the vertically opposite angle of ∠5?

Answer:
(i) Yes, ∠1 and ∠2 are adjacent angles because both the angles have common arm OC and common vertex O.
(ii) No, ∠AOC is not adjacent to ∠AOE because ∠AOC is part of ∠AOE.
(iii) Yes, ∠COE and ∠EOD form a linear pair because COD is a straight line.
(iv) Yes, ∠BOD and ∠DOA are supplementary because ∠BOD + ∠DOA = 180°.
(v) Yes, because AB and CD intersect each other.
(vi) The vertically opposite angle of ∠5 is ∠BOC.

Q1uestion 10.
Indicate which pairs of angles are:
(i) Vertically opposite angles.
(ii) Linear pairs

Answer:
(i) Vertically opposite angles.
In the given figure, the following pairs are vertically opposite angles.
∠1 and ∠4
∠5 and (∠2 + ∠3)

(ii) Linear pairs:
∠4 and ∠5 form a linear pair.
∠1 and ∠5 form a linear pair.
∠1 and (∠3 + ∠2) form a linear pair.
∠4 and (∠3 + ∠2) form a linear pair.

Question 11.
In the following figure, is ∠1 adjacent to ∠2? Give reasons.

Answer:
∠1 and ∠2 are not adjacent angles because they do not have a common vertex.

Question 12.
Find the values of the angles x, y, and z in each of the following:

Answer:
(i) Since x and 55° are vertically opposite angles
x = 55°
Also, 55° + y = 180°
(Linear pair)
y = 180°- 55°
y = 125°
since y and z are vertically opposite angles
y = z = 125°
Thus x = 55°, y = 125° and z = 125°

(ii) since 40° and z are vertically
opposite angles
z = 40°
Again y and 40° form a Linear pair
y + 40° = 180°
y = 180°-40°
= 140°
y and (x + 25) are vertically opposite angles
(x + 25°) = y = 140°
∴ x + 25° = 140°
x = 140° -25°= 115°
Thus, x = 115°’ y = 140° and z = 40°

Question 13.
Fill in the blanks:
Answer:
(i) If two angles are complementary, then the sum of their measures is 90°.
(ii) If two angles are supplementary, then the sum of their measures is 180°.
(iii) Two angles forming a linear pair are supplementary.
(iv) If two adjacent angles are supplementary, they form a linear pair.
(v) If two lines intersect at a point, then the vertically opposite angles are always Equal.
(vi) If two lines intersect at a point, and if one pair of vertically opposite angles are acute angles, then the other pair of vertically opposite angles are obtuse angles.

Question 14.
In the adjoining figure, name the following pairs of angles
(i) Obtuse vertically opposite angles
(ii) Adjacent complementary angles
(iii) Equal supplementary angles
(iv) Unequal supplementary angles
(v) Adjacent angles that do not form a linear pair

Answer:
(i) ∠BOC and ∠AOD are obtuse vertically opposite angles.
(ii) ∠AOB and ∠AOE are adjacent complementary angles.
(iii) ∠BOE and ∠EOD are equal supplementary angles.
(iv) ∠AOE and ∠EOC are unequal supplementary angles.
(v) (a) ∠BOA and ∠AOE
(b) ∠AOE and ∠EOD
(c) ∠EOD and ∠COD

These NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Ex 11.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area Exercise 11.3

Question 1.
Find the circumference of the circles with the following radius: (Take π = \(\frac{22}{7}\))
(a) 14 cm
(b) 28 mm
(c) 21 cm
Answer:
(a) Radius of the circle (r) = 14 cm
Circumference of the circle = 2πr units
= 2 × \(\frac{22}{7}\) x 14 cm
= 2 × 22 × 2 cm
= 88 cm

(b) Radius of the circle = 28 mm.
.’. Circumference of the circle = 2πr units
= 2 × \(\frac{22}{7}\) × 28 mm
= 2 × 22 × 4 mm
= 176 mm

(c) Radius of the circle = 21 cm
Circumference of the circle = 2πr 22
= 2 × \(\frac{22}{7}\) x 21
= 2 × 22 × 3 cm
= 132 cm

Question 2.
Find the area of the following circles, given that:
(a) Radius = 14 mm (π = \(\frac{22}{7}\))
(b) Diameter = 49 m
(c) Radius = 5 cm
Answer:
(a) Radius of the circle = 14 mm
Area of the circle = πr² sq.m
= \(\frac{22}{7}\) × 14²
= \(\frac{22}{7}\) × 196
= 22 × 28
= 616 mm²

(b) Diameter of the circle = 49 m.
Radius of the circle = \(\frac{49}{2}\) m
= 24.5 m
Area of the circle = πr²
= \(\frac{22}{7}\) × (24.5)²
= \(\frac{22}{7}\) × 600.25
= 22 × 85.75
= 1886.5 m²
Area of the circle = πr² sq units

(c) Radius of the circle = 5 cm
Area of the circle = πr² sq.unit
= \(\frac{22}{7}\) × 5 × 5 cm²
= \(\frac{22 \times 25}{7}\) cm²
= \(\frac{550}{7}\) cm² or 78.57 cm².

Question 3.
If the circumference of a circular sheet is 154 m, find its radius. Also find the area of the sheet. (Take π = \(\frac{22}{7}\))
Answer:
Circumference of a circle = 154 m
2πr = 154
2 × \(\frac{22}{7}\) × r = 154

Area of the circle = πr² sq.m
= \(\frac{22}{7}\) × (24.5)²
= \(\frac{22}{7}\) × 600.25
= 22 × 85.75 m²
= 1886.5 m²

Question 4.
A gardener wants to fence a circular garden of diameter 21 m. Find the length of the rope he needs to purchase, if he makes 2 rounds of fence. Also, find the cost of the rope, if it cost ₹ 4 per meter. (Take π = \(\frac{22}{7}\))

Answer:
Diameter of the circular garden = 21 m
radius of the circular garden = \(\frac{21}{2}\)
Circumference of the garden = 2πr units
= 2 × \(\frac{22}{7}\) × \(\frac{21}{2}\) m = 22 x 3m = 66m
Length of the rope required for one round fence = 66 m
Length of the rope required for two round fence = 66 m × 2 m = 132 m
Cost of rope per meter = ₹ 4
Total cost of the rope = ₹132 × 4 = ₹ 528

Question 5.
From a circular sheet of radius 4 cm, a circle of radius 3 cm is removed. Find the area of the remaining sheet. (Take π = 3.14)

Answer:
Radius of outer circle (R) = 4 cm
Radius of inner circle (r) = 3 cm
= Area of the outer circle – Area of the inner circle
Area of the remaining sheet = n (R² – r²)
= 3.14 (4² – 3²)
= 3.14 × (16-9)
= 3.14 x 7 = 21.98 cm²
So, the area of the remaining sheet is 21.98 cm².

Question 6.
Saima wants to put a lace on the edge of a circular table cover of diameter 1.5 m. Find the length of the lace required and also find its cost if one meter of the lace costs ₹ 15. (Take π = 3.14)
Answer:
Diameter of the table cover = 1.5 m
Radius of the table cover = \(\frac { 1.5 }{ 2 }\) m
Circumference of the table cover = 2πr units
= 2 x 3.14 x \(\frac { 1.5 }{ 2 }\) = 3.14 x 1.5 m = 4.71 m
Cost of lace for one metre = ₹15
Cost of lace for table cover = ₹ 4.71 x 15
= ₹ 70.71

Question 7.
Find the perimeter of the given figure, which is a semicircle including its diameter.

Answer:
Diameter of the semicircle = 10 cm
Radius of the semicircle = \(\frac { 10 }{ 2 }\) = 5 cm
Circumference of the semicircle = \(\frac { 1 }{ 2 }\) x 2πr unit
= πr = \(\frac { 22 }{ 7 }\) x 5
= 15.71 cm
∴ Perimeter of the semicircle
= 15.71 + 10 cm
= 25.71 cm

Question 8.
Find the cost of polishing a circular table-top of diameter 1.6 m, if the rate of polishing is ₹ 15/m². (Take π = 3.14)
Answer:
Diameter of the table-top = 1.6 m
Radius of the table-top = \(\frac { 1.6 }{ 2 }\) = 0.8m
Area of the table-top = πr² sq. m
= 3.14 × 0.8 × 0.8 m² = 2.0096 m²
Rate of Polishing = ₹ 15 per m²
Cost of polishing the table-top = ₹ 2.0096 × 15 =₹ 30.14 (approx.)

Question 9.
Shazli took a wire of length 44 cm and bent it into the shape of a circle. Find the radius of that circle. Also, find its area. If the same wire is bent into the shape of a square, what will be the length of each of its sides? Which figure encloses more area, the circle or the square?
(Take π = \(\frac { 22 }{ 7 }\))
Answer:
Length of the wire = 44 cm
Let the radius of the circle be r.
Circumference of the circle = 44 cm
2πr = 44
2 × \(\frac { 22 }{ 7 }\) × r = 44
r = \(\frac{44 \times 7}{2 \times 22}\) cm =7 cm
Area of the circle = πr² sq.m
= \(\frac { 22 }{ 7 }\) × 7 × 7 cm²
= 154 cm

Since, the wire is rebent to form a square. Perimeter of the square = Length of the wire
4a = 44
[Perimeter of a square = 4a] 44
a = \(\frac { 44 }{ 6 }\) = 11 cm 6
Side of a square =11 cm
Area of the square = side × side
= 11 × 11 cm² = 121 cm²
154 cm² > 121 cm²
Area of the circle > Area of the square
∴ The circle encloses greater area.

Question 10.
From a circular card sheet of radius 14cm, two circles of radius 3.5 cm and a rectangle of length 3 cm and breadth 1 cm are removed, (as shown in the given figure). Find the area of the remaining 22
sheet. (Take π = \(\frac { 22 }{ 7 }\))

Answer:
Radius of the circular card sheet =14 cm
Area of the sheet = πr² sq m
= \(\frac { 22 }{ 7 }\) × 14 × 14 cm²
= 22 × 2 × 14 cm2 = 616 cm²
Radius of a small circle = 3.5 cm
Area of 2 small circles = 2 × πr² sq m
= 2 × \(\frac { 22 }{ 7 }\) × 3.5 × 3.5 cm²
= 2 × 22 × 0.5 × 3.5 cm = 77 cm²
Length of a small rectangle = 3 cm
Breadth of a small rectangle = 1 cm.
Area of small rectangle = l × b sq.m
= 3 × 1 cm² = 3 cm²
Area of the remaining sheet = Area of the circular sheet – (Area of two small circles + Area of rectangle)
= 616 cm²– (77 + 3)cm²
= 616 cm² – 80 cm² = 536 cm²
∴ Required area of the remaining sheet = 536 cm²

Question 11.
A circle of radius 2 cm is cut out from a square piece of an aluminium sheet of side 6 cm. What is the area of the left over aluminium sheet? (Take π = 3.14)
A circle of radius 2 cm is cut out from a square piece of an aluminium sheet of side 6 cm. What is the area of the left over aluminium sheet? (Take π = 3.14)
Answer:
Side of the square = 6 cm
Area of the square = side × side
= 6 cm × 6 cm = 36 cm²

Radius of the circle cut out from the sheet = 2 cm
Area of the circle = πr² sq.m
= 3.14 × 2 × 2 cm² = 12.56 cm²
Area of the remaining sheet
= 36 cm² – 12.56 cm² = 23.44 cm²

Question 12.
The circumference of a circle is 31.4 cm. Find the radius and the area of the circle? (Take n = 3.14)
Answer:
Let the radius of the circle be ‘r’
Circumference of a circle = 31.4 cm
2πr² = 31.4
2 × 3.14 × r = 31.4
r = \(\frac{31.4}{2 \times 3.14}\)
= \(\frac{314 \times 10}{2 \times 314}\)
= \(\frac{10}{2}\)
= 5 cm
Area of the circle = 2πr² sq. units = 3.14 × 5 × 5 cm² = 78.5 cm²

Question 13.
A circular flower bed is surrounded by a path 4 m wide. The diameter of the flower
bed is 66 m. What is the area of this path? (π = 3.14)
Answer:
Diameter of the flower bed = 66 m
Radius of the flower bed = \(\frac { 66 }{ 2 }\) m
(r) = 33 m

width of the surrounding path = 4 m
Radius of the outer circle (R)
= 33 m + 4 m
(R) = 37 m
Area of the pathway = Area of the outer circle – Area of the inner circle
= πR² – πr²
= π(R² – r²) sq. units
= 3.14 (37² – 33² ) m²
= 3.14 (37 + 33) (37 – 33) m²
= 3.14 × 70 × 4 m²
= 3.14 × 280 m²
= 879.2 m²
Thus, the area of the path = 879.2 m²

Question 14.
A circular flower garden has an area of 314 m². A sprinkler at the centre of the garden can cover an area that has a radius of 12 m. Will the sprinkler water
the entire garden? (Take π = 3.14)
Answer:
Let the radius of the garden be ‘r’
Area of the circular garden = 314 m²
πr² =314
3.14 × r² = 314
r² = \(\frac{314}{3.14}=\frac{314 \times 100}{314}\)
r² = 100
r²= 10²
r = 10 m
Radius of the area covered by the sprinkler = 12 m
Since 12 m > 10 m
The sprinkler covers an area beyond the garden.
Yes, the entire garden is covered by the sprinkler.

Question 15.
Find the circumference of the inner and the outer circles, shown in the adjoining figure? (Take π = 3.14)

Answer:
Radius of the outer circle (R) = 19 m.
Circumference of the outer circle
= 2πr unit
= 2 × 3.14 × 19 m
= 119.32 m
Radius of the inner circle (r)
= 19- 10 = 9m
Circumference of the inner circle
= 2π = 2 × 3.14 × 9 m = 56.52 m

Question 16.
How many times a wheel of radius 28 cm must rotate to go 352 m? (Take π = \(\frac { 22 }{ 7 }\))
Answer:
Radius of a wheel = 28 cm
∴ Circumference of the wheel = 2πr units
= 2 × \(\frac { 22 }{ 7 }\) × 28
= 2 × 22 × 4 cm = 176 cm
Total distance covered
= 352 m
= 352 × 100 cm
= 35200 cm
Number of rotations
= \(\begin{aligned}
&\frac{\text { Total distance }}{\text { umference of the wheel }}\\&\text { Circum }
\end{aligned}\)
= \(\frac{35200}{176}\) = 200
Thus, the distance of 352 m will be covered in 200 rotations.

Question 17.
The minute hand of a circular clock is 15 cm long. How far does the tip of the minute hand move in 1 hour. (Take π = 3.14)
Answer:
Length of the minute hand =15 cm
radius of the circle (r) = 15 cm
(made by the tip of the minute hand)
Perimeter of the circle = 2πr units
= 2 × 3.14 × 15 cm = 94.2 cm
Distance covered by the minute hand = 94.2 cm

These NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations InText Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations InText Questions

NCERT In-text Question Page No. 78
Question 1.
The value of the expression (10y – 20) depends on the value of y. Verify this by giving five different values to y and finding for each y the value of (10y – 20). From the different values of (10y – 20) you obtain, do you see a solution to 10y – 20 = 50? If there is no solution, try giving more values to y and find whether the condition 10y – 20 = 50 is met.
Answer:

Thus the condition 10y – 20 is true for y = 7

NCERT In-text Question Page No. 80
Question 1.
Write atleast one other form for each equation (ii), (iii) and (iv)

Statement of other form of the equation:
(a) Taking away 4 from five times p gives 16.
(b) Add 9 to four times n to get 45.
(c) Add 8 to one-third of m to get 17.

NCERT In-text Question Page No. 88
Question 1.
Start with the same step x = 5 and make two different equations. Ask two of your classmates to solve the equations. Check whether they get the solution x = 5.
Answer:
(i) x = 5
Multiplying both sides by 2, we have
2x = 10
Adding 6 to both sides, we have
or 2x + 6 = 10 + 6
or 2x + 6 = 16 is an equation.

(ii) x = 5
Dividing both sides by 3, we have
\(\frac{x}{3}=\frac{5}{3}\)
Subtracting 2 from both sides, we have
\(\frac{\mathrm{x}}{3}\) – 2 = \(\frac{5}{3}\) – 2
or
\(\frac{\mathrm{x}}{3}\) – 2 = \(\frac{5-6}{3}=\frac{-1}{3}\)
Thus,
\(\frac{\mathrm{x}}{3}\) – 2 = \(\frac{-1}{3}\) is an equation.

Solution to I
2x + 6 = 16
Subtracting 6 from both sides, we have
2x + 6- 6 = 16 – 6
or 2x = 10
Dividing both sides by 2, we have
\(\frac{2 \mathrm{x}}{2}=\frac{10}{2}\) or x = 5

Solution to II
\(\frac{\mathrm{x}}{3}\) – 2 = \(\frac{-1}{3}\)
Adding 2 to both sides, we have
\(\frac{\mathrm{x}}{3}\) – 2 + 2 = \(\frac{-1}{3}\) + 2 or \(\frac{x}{3}=\frac{5}{3}\)
Multiplying both sides by 3, we have
\(\frac{\mathrm{x}}{3}\) × 3 = \(\frac{5}{3}\) × 3 or x = 5

NCERT In-text Question Page No. 88
Question 1.
Try to make two number puzzles, one with the solution 11 and another with 100.
Answer:
I. A puzzle having the solution as 11:
Think of a number. Multiply bit by 3 and add 7. Tell me the sum.
If the sum is 40, then the number is 11.

II. A puzzle having the solution as 100:
Think of a number. Divide it by 4 and add 5. Tell me what you get.
If you get 30, then the number is 100.

Note:
Instead of making the same operation on both sides, we can move a number from one side to another by changing its sign from (+) to (-) and (-) to (+). This is called ‘transposing a number’. Thus, transposing a number is the same as adding or subtracting the number from both sides. In doing so, the sign of the number has to be changed.

NCERT In-text Question Page No. 90
Question 1.
(i) When you multiply a number by 6 and subtract 5 from the product, you get 7. Can you tell what the number is?

(ii) What is that number one-third of which added to 5 gives 8?
Answer:
(i) Let the number be x.
∴ Multiply the number by 6, we have 6x.
Now, according to the condition, we have
6x = 7 + 5 = 12
Dividing both sides by 6, we have
\(\frac{6 x}{6}=\frac{12}{6}\)
or x = 2
∴ The required number = 2

(ii) Let the required number be x.
∵ One-third of the number = \(\frac{1}{3}\)x
∴ According to the condition, we have
5 + \(\frac{1}{3}\)x = 8
Transposing 5 from L.H.S. to R.H.S., we have
\(\frac{1}{3}\)x = 8 – 5 = 3
Multiplying both sides by 3, we have
3 × \(\frac{1}{3}\)x = 3 × 3
or x = 9
Thus, the required number = 9.

NCERT In-text Question Page No. 90
Question 1.
There are two types of boxes containing mangoes. Each box of the larger type contains 4 more mangoes than the number of mangoes contained in 8 boxes of he smaller type. Each larger box contains 100 mangoes. Find the number of mangoes contained in the smaller box?
Answer:
Let the number of mangoes contained in the smaller box be x.
∴ Number of mangoes in 8 smaller boxes = 8x
Now, according to the condition, we have
8x + 4 = 100
Transposing 4 to R.H.S., we have
8x = 100 – 4 or 8x = 96
Dividing both sides by 8, we have
\(\frac{8 x}{8}=\frac{96}{8}\)
or x = 12
Thus, the number of mangoes in the smaller box = 12.

These NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Exercise 4.4

Question 1.
Set up equations and solve them to find the unknown numbers in the following cases:
(a) Add 4 to eight times a number; you get 60.
(b) One-fifth of a number minus 4 gives 3.
(c) If I take three-fourths of a number and add 3 to it, I get 21.
(d) When I subtracted 11 from twice a number, the result was 15.
(e) Munna subtracts thrice the number of notebooks he has from 50, he finds the result to be 8.
(f) Ibenhal thinks of a number. If she adds 19 to it and divides the sum by 5, she will get 8.
(g) Anwar thinks of a number. If he takes away 7 from \(\frac{5}{2}\) of the number, the result is 23.
Answer:
(a) Let the required number be x
∴ Eight times of the number = 8x
According to the given question, we get
8x + 4 =60
Transposing 4 to R.H.S
8x = 60 – 4
8x = 56
Dividing both sides by 8, we get
\(\frac{8 x}{8}=\frac{56}{8}\)
x = 7
∴ The required number = 7

(b) Let the number be x
one-fifth of the number = \(\frac{1}{5}\) x
According to the given question, we get
\(\frac{1}{5}\) x – 4 = 3
Transposing -4 to R.H.S
\(\frac{1}{5}\)x = 3 + 4
\(\frac{1}{5}\)x = 7
Multiplying both sides by 5, we get 1
\(\frac{1}{5}\)x × 5 = 7 × 5
x = 35
∴ The required number = 35

(c) Let the number be x
Three-fourths of the number = \(\frac{3}{4}\) x
According to the given question, we get
\(\frac{3}{4}\) x + 3 = 21
Transposing 3 to R.H.S
\(\frac{3}{4}\) x = 21 – 3
\(\frac{3}{4}\) x = 18
Multiplying both sides by 4, we get
\(\frac{3}{4}\) x × 4 = 18 × 4
3x = 72
Dividing both sides by 3, we get
\(\frac{3 x}{3}=\frac{72}{3}\)
x = 24
∴ Thus, the required number = 24

(d) Let the number be x
Twice a number = 2x
According to the given question
2x – 11 = 15
Transposing -11 to R.H.S
2x = 15 + 11
2x = 26
Dividing both sides by 2, we get
\(\frac{2 \mathrm{x}}{2}=\frac{26}{2}\)
x = 13
∴ The required number 13.

(e) Let the number of notebooks with Munna be ‘x’
Thrice the number of notebooks = 3x
According to the given question, we get
50 – 3x = 8
Transposing 50 to R.H.S
-3x = 8 – 50
-3x = -42
Dividing both sides by (-3), we get
\(\frac{-3 x}{-3}=\frac{-42}{-3}\)
x = 14
∴ The required number of notebooks is 14.

(f) Let the number be x
According to the given question, we get
\(\frac{x+19}{5}\) = 8
Multiplying both sides by 5, we get
\(\frac{x+19}{5}\) × 5 = 8 × 5
x + 19 = 40
Transposing 19 to R.H.S
x = 40 – 19
x = 21
∴ The required number = 21

(g) Let the number be x
\(\frac{5}{2}\) of the number = \(\frac{5}{2}\) x
According to the given question, we get
\(\frac{5}{2}\) x – 7 = 23
Transposing -7 to R.H.S
\(\frac{5}{2}\)x = 23 + 7
\(\frac{5}{2}\)x = 30
Multiplying both sides by 2, we get
\(\frac{5}{2}\) x × 2 = 30 × 2
5x = 60
Dividing both sides by 5, we get
\(\frac{5 x}{5}=\frac{60}{5}\)
x = 12
The required number = 12.

Question 2.
Solve the following:
(a) The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7. The highest score is 87. What is the lowest score?

(b) In an isosceles triangle, the base angles are equal. The vertex angle is 40°. What are the base angles of the triangle? (Remember, the sum of three angles of a triangle is 180°).

(c) Sachin scored twice as many runs as Rahul. Together, their runs fell two short of a double century. How many runs did each one score?
Answer:
(a) Let the lowest marks scored be x
Twice the lowest marks = 2x
According to the given question, we get
2x + 7 = 87
Transposing 7 to R.H.S
2x = 87 – 7
2x = 80
Dividing both sides by 2, we get
\(\frac{2 x}{2}=\frac{80}{2}\)
x = 40
∴ The lowest marks = 40

(b) Let the base angle be x°
∴ The other base angle = x°
The vertex angle = 40°
∴ Sum of the angles = x + x + 40
= 2x + 40
According to the given question, we get
2x + 40 = 180°
(Sum of the three angles of a triangle)
2x = 180° – 40°
2x = 140°
Dividing both sides by 2, we get
\(\frac{2 x}{2}=\frac{140}{2}\)
x = 70°
∴ The base angle of the triangle
= 70°

(c) Let the runs scored by Rahul be x
∴ Sachins runs = 2x
According to the given question, we get
x + 2x =2(100) -2
3x = 200 – 2
3x = 198
Dividing both sides by 3, we get
\(\frac{3 x}{x}=\frac{198}{3}\)
x = 66
∴ Rahul score = 66 runs and Sachins score = 2 × 66 = 132 runs

Question 3.
Solve the following:
(i) Ifran says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles. How many marbles does Parmit have?

(ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. What is Laxmi’s age?

(iii) People of Sundargram planted trees in the village garden. Some of the trees were fruit trees. The number of non-fruit trees were two more than three times the number of fruit trees. What was the number of fruit trees planted if the number of non-fruit trees planted was 77?
Answer:
(i) Let the number of marbles with Parmit be x
Number of marbles with Irfan = 37
According to the given question, we get
5x + 7 = 37
Transposing 7 to R.H.S
5x = 37 – 7
5x = 30
Dividing both sides by 5, we get
\(\frac{5 x}{5}=\frac{30}{5}\)
\(\frac{5 x}{5}=\frac{30}{5}\) ⇒ x = 6
∴ Parmit has 6 marbles.

(ii) Let Laxmi’s age be x years
Three times Laxmi s age = 3x
According to the given question, we get
3x + 4 = 49
Transposing 4 to R.H.S
3x = 49 – 4
3x = 45
Dividing both sides by 3, we get
\(\frac{3 x}{3}=\frac{45}{3}\)
x = 15
∴ Laxmi’s age= 15 years

(iii) Let the number of fruit trees be x
Three times fruit trees = 3x
Number of non-fruit trees = 2 + 3x
According to the given question, we get
2 + 3 x = 77
Transposing 2 to R.H.S
3x = 77 – 2
3x = 75
Dividing both sides by 3, we get
\(\frac{3 x}{3}=\frac{75}{3}\)
⇒ x = 25
∴ The number of fruit trees = 25

Question 4.
Solve the following riddle :
I am a number,
Tell my identity!
Take me seven times over
And add a fifty!
To reach a triple century
You still need forty!
Answer:
Let the required number be x
7 times the number = 7x
each triple century = 3 × 100 = 300
According to the given riddle, we get
7x + 50 = 300 – 40
7x + 50 = 260
Transposing 50 to R.H.S
7x = 260 – 50
7x = 210
Dividing both sides by 7, we get
\(\frac{7 \mathrm{x}}{7}=\frac{210}{7}\)
x = 30
∴ The required number is 30.