Perimeter and Area Maths class 7 NCERT Solutions || GetStudySolution.com

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NCERT Solutions for class 7 Maths chapter 11 – Perimeter and Area

Back Exercise

Exercise 11.1

Question 1.
The length and the breadth of a rectangular piece of land is 500 m and 300 m respectively. Find
(i) its area
(ii) the cost of the land, if 1 m² of the land cost ₹ 10,000.
Solution:

Here, length = 500 m, breadth = 300 m
(i) Area = length × breadth = (500 × 300) m² = 1,50,000 m²(ii) Cost of land at the rate of ₹ 10,000 per 1 m²= ₹ (10,000 × 1,50,000) = ₹ 1,50,00,00,000.

Question 2.
Find the area of a square park whose perimeter is 320 m.
Solution:

Question 3.
Find the breadth of a rectangular plot of land, if its area is 440 m² and the length is 22 m. Also, find its perimeter.
Solution:

Question 4.
The perimeter of a rectangular sheet is 100 cm. If the length is 35 cm, find its breadth. Also find the area.

Solution:

Question 5.
The area of a square park is the same as a rectangular park. If the side of the square park is 60 m and the length of the rectangular park is 90 m, find the breadth of the rectangular park.

Solution:

Hence, the breadth of the rectangular park is 40 m.

Question 6.
A wire is in the shape of a rectangle. Its length is 40 cm and breadth is 22 cm. If the same wire is rebent in the shape of a square, what will be the measure of each side? Also, find which shape encloses more area?
Solution:

Question 7.
The perimeter of a rectangle is 130 cm. If the breadth of the rectangle is 30 cm, find its length. Also, find the area of the rectangle.

Solution:

Question 8.
A door of length 2 m and breadth 1 m is fitted in a wall. The length of the wall is 4.5 m and the breadth is 3.6 m (figure). Find the cost of whitewashing the wall, if the rate of whitewashing the wall is ₹ 20 per m².
Solution:

Exercise 11.2

Question 1.
Find the area of the following parallelograms:

Solution:

Question 2.
Find the area of each of the following triangles:

Solution:

Question 3.
Find the missing values:

Solution:

Question 4.
Find the missing values:

Solution:

Question 5.
PQRS is a parallelogram (in Figure). QM is the height from Q to SR and QN is the height from Q to PS. If SR = 12 cm and QM = 7.6 cm. Find:
(a) the area of the parallelogram PQRS
(b) QN, if PS = 8 cm.

Solution:
(a) Area of the parallelogram PQRS = base × height = SR × QM = 12 × 7.6 cm² = 91.2 cm²
(b) Area of the parallelogram PQRS = base × height = PS × QN
⇒ 91.2 = 8 × QN
⇒ QN = $\frac { 912 }{ 8 }$ cm
⇒ QN = 11.4 cm.

Question 6.
DL and BM are the heights on sides AB and AD respectively of parallelogram ABCD (in Figure). If the area of a parallelogram is 1470 cm2, AB = 35 cm, and AD = 49 cm, find the length of BM and DL.

Solution:

Question 7.
∆ ABC is right-angled at A (in Figure). AD is perpendicular to BC. If AB = 5 cm, BC = 13 cm and AC = 12 cm, find the area of ∆ ABC. Also, find the length of AD.

Solution:

Question 8.
∆ ABC is isosceles with AB = AC = 7.5 cm, and BC = 9 cm (in Figure). The height of AD from A to BC is 6 cm. Find the area of ∆ ABC. What will be the height from C to AB i.e., CE?

Solution:

Exercise 11.3

Find the circumference of the circles with the following radus (Take π = $\frac { 22 }{ 7 }$ )
(a) 14 cm
(b) 28 mm
(c) 21 cm
Solution:

Find the area of the following circles, given that: ( Take π = $\frac { 22 }{ 7 }$ )
(a) radius = 14 mm
(b) diameter = 49 m
(c) radius = 5 cm.
Solution:

If the circumference of a circular sheet is 154 m, find its radius. Also find the area of the sheet ( Take π = $\frac { 22 }{ 7 }$ )
Solution:
Circumference of the circular sheet = 154 m
Let the radius of the circular sheet be r cm
Then, its circumference = 2nr m According to the question,
Circumference = 2πr = 154

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 offense. Also find the cost of the rope, if it costs ₹ 4 per meter. ( Take π = $\frac { 22 }{ 7 }$ )
Solution:
The diameter of the circular garden (r) = 21 m
Radius of the circular garden (r) = $\frac { 21 }{ 2 }$ m
∴ Circumference of the circular garden = 2πr
= 2 × $\frac { 22 }{ 7 }$ × $\frac { 21 }{ 2 }$ m = 66m
⇒ Length of the rope needed to make 1 round of fence = 66 m
⇒ Length of the rope needed to make 2 rounds of fence
= 66 × 2 m = 132 m
Cost of rope per meter = ₹ 4
∴ 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)
Solution:

Here, Outer radius, r = 4 cm
Inner radius, r = 3 cm
Area of the remaining sheet = Outer area – Inner area
= π (R² – r²) = 3.14 (4² – 3²) cm²
= 3.14 (16 – 9) cm²
= 3.14 × 7 cm² = 21.98 cm²

Question 6.
Saima wants to put lace on the edge of a circular table cover of a diameter of 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)
Solution:
Diameter of the table cover = 1.5 m
⇒ Radius of the table cover (r) = $\frac { 1.5 }{ 2 }$ m
⇒ Circumference of the table cover = 2πr
= 2 × 3.14 × $\frac { 1.5 }{ 2 }$ m = 4.71 m
⇒ Length of the lace required = 4.71 m
∵ Cost of lace per meter = ₹ 15
∴ Cost of the lace = ₹ 4.71 × 15 = ₹ 70.65

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

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)
Solution:
Diameter of the table-top = 1.6 m
⇒ Radius of the table-top (r) = $\frac { 1.6 }{ 2 }$ m = 0.8 m
∴ Area of the table-top = πr²
= 3.14 × (0.8)2 m²
= 3.14 × 0.64 m²
= 2.0096 m²
∵ Rate of polishing = ₹ 15 per m²
∴ Cost of polishing the table-top = ₹ 2.0096 × 15
= ₹ 30.144
= ₹ 30.14 (approx.).

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 side? Which figure encloses more area, the circle or the square? ( Take π = $\frac { 22 }{ 7 }$ )
Solution:

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

Solution:

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 leftover aluminium sheet? (Take π = 3.14)
Solution:

Question 12.
The circumference of a circle is 31.4 cm. Find the radius and the area of the circle? (Take π = 3.14)
Solution:

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)

Solution:

Question 14.
A circular flower garden has an area of about 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? (Taken π = 3.14)
Solution:
The circular area of the sprinkler = πr²
= 3.14 × 12 × 12
= 3.14 × 144 = 452.16 m²
Area of the circular flower garden = 314 m²
Since the area of the circular flower garden is smaller than by sprinkler
Therefore, the sprinkler will water the entire garden.

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

Solution:
Radius of inner circle = 19 – 10 = 9 m
∴ Circumference of the inner circle = 2 πr = 2 × 3.14 × 9 m = 56.52 cm
The radius of the outer circle = 19 m
∴ Circumference of the outer circle = 2πr = 2 × 3.14 × 19 m = 119.32 m.

How many times a wheel of radius 28 cm must rotate to go 352 m? ( Take π = $\frac { 22 }{ 7 }$ )
Solution:

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)
Solution:
We know that the minute hand describes one complete revolution in one hour.
∴ Distance covered by its tip = Circumference of the circle of radius 15 cm
= (2 × 3.14 × 15) cm
= 94.2 cm

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 a hectare.
Solution:

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.
Solution:
PQ = 125 m + 3m + 3m = 131 m
QR = 65m + 3m + 3m = 71 m

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.
Solution:

Question 4.
A verandah of width 2.25 m is constructed all along the outside of 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.
Solution:

Question 5.
A path 1 rn 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 f 40 per m².
Solution:

Question 6.
Two crossroads, each of width 10 m, cut at right angles through the centre of a rectangular park of the length 700 m and the breadth 300 m parallel to its sides. Find the area of the roads. Also, find the area of the park excluding crossroads. Give the answer in hectares.
Solution:

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.
Solution:
(i) Area covered by the roads = Area of the rectangle PQRS + Area of the rectangle EFGH – Area of the square KLMN
= (90 × 3 + 60 × 3 – 3 × 3) m²
= (270 + 180 – 9) m²
= 441 m²
(ii) Cost of constructing the roads at the rate of ₹ 110 per m² ₹ 441 × 110 = ₹ 48510.

Question 8.
Pragya wrapped a cord around the circular pipe of the radius 4 cm (adjoining figure) and cut off the length required of the cord. Then she wrapped it around the square box of the side 4 cm (also shown. Did she hate any cord left? = ( π = 3.14)

Solution:
Radius of the circular pipe (r) = 4 cm
∴ Circumference of the circular pipe = 2πr = 2 × 3.14 × 4 cm = 25.12 cm
Perimeter of the square box of the side 4 cm = 4 × side = 4 × 4 cm = 16 cm
∵ 25.12 cm > 16 cm
∴ She had extra cord left and the length of the cord left = (25.12 – 16) cm = 9.12 cm.

Question 9.
The following figure represents a rectangular lawn with a circular flower bed in the middle. Find:
(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.

Solution:

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

Solution:

Question 11.
Find the area of the quadrilateral ABCD.
Here, AC = 22 cm, BM = 3 cm, DN = 3 cm, and BM ⊥ AC, DN ⊥ AC.
Solution:
Area of the quadrilateral ABCD = Area of the triangle ABC + Area of the triangle ADC