The DAV Maths Class 7 Solutions and **DAV Class 7 Maths Chapter 11 Worksheet 6** Solutions of Perimeter and Area offer comprehensive answers to textbook questions.

## DAV Class 7 Maths Ch 11 WS 6 Solutions

Question 1.

Find the area of a circle of diameter 8.4 m.

Answer:

Diameter = 8.4 m

radius = \(\frac{8.4}{2}\) = 4.2 m

Area of circle = πr^{2}

= \(\frac{22}{7}\) × 4.2 × 4.2

= 22 × 0.6 × 4.2 = 55.44 m^{2}

Hence the required area = 55.44 m^{2}.

Question 2.

The area of a circular lawn is 154 m^{2}. Find its circumference.

Answer:

Area = 154 m^{2}

πr^{2} = 154

\(\frac{22}{7}\) × r^{2} = 154

r^{2} = \(\frac{154 \times 7}{22}\)

r^{2} = 7 × 7

r = 7m

Circumference = 2πr = 2 × \(\frac{22}{7}\) × 7 = 44 m

Hence the required circumference = 44 m.

Question 3.

A circular plot is of radius 77 m. Find its value at ₹ 57 per square meter.

Answer:

Area of the circular plot = πr^{2}

= \(\frac{22}{7}\) × 77 × 77

= 22 × 11 × 77

= 18634 m^{2}

Value of the plot = ₹ 57 × 18634 = ₹ 10,62,138

Hence the value of the plot = ₹ 10,62,138

Question 4.

The diameters of two silver discs are in the ratio 2:3. What will be the ratio of their areas?

Answer:

Let the diameters of the two silver discs be 2x cm and 3x cm respectively.

Their radii are x cm and \(\frac{3x}{2}\) cm

If A_{1} and A_{2} be their respective areas

∴ \(\frac{A_1}{A_2}=\frac{\pi(x)^2}{\pi(3 x / 2)^2}\)

⇒ \(\frac{A_1}{A_2}=\frac{x^2}{\frac{9}{4} x^2}=\frac{4}{9}\)

Hence the required ratio is 4 : 9.

Question 5.

If the perimeter of a semi-circular park is 72 m,find its area and radius.

Answer:

AC + ABC = 72 m

2r + πr = 72

r (2 + π) = 72

r (2 + \(\frac{22}{7}\)) = 72

r × \(\frac{36}{7}\) = 72

r = 72 × \(\frac{7}{36}\) = 14 m

Area of the park = \(\frac{1}{2}\)πr^{2}

= \(\frac{1}{2} \times \frac{22}{7}\) × 14 × 14

= \(\frac{1}{2}\) × 616 m^{2}

= 308 m^{2}

Hence the required area is 308 m^{2} and the radius = 14 m.

Question 6.

If a wire of 440 m length is moulded in the form of a circle and a square turn-by-turn, find the ratio of the area of the circle to that of square.

Answer:If it is moulded into a circle,

then the circumference C = 2πr

440 = 2 × \(\frac{22}{7}\) × r

r = \(\frac{440 \times 7}{2 \times 22}\) = 70 cm

Area of the circle = πr^{2}

= \(\frac{22}{7}\) × 70 × 70 = 15400 cm^{2}

If the wire is moulded into a square,

then its perimeter = 440 m

4 × Side = 440 m

Side = \(\frac{440}{4}\) = 110 m

Area of the square = (side)^{2}

= 110 × 110 = 12100 m^{2} .

∴ \(\frac{\text { Area of the circle }}{\text { Area of the square }}=\frac{15400}{12100}=\frac{14}{11}\)

Hence the required ratio is 14 :11.

Question 7.

Find the area of a square that can be inscribed in a circle of radius 8 cm.

Answer:

Let ABCD be a square inscribed in a circle of radius ‘r’.

∴ AC = 2 r

In right ΔABC,

AB^{2} + BC^{2} = (2r)^{2} (By Pythagoras theorem)

AB^{2} + AB^{2} = 4r^{2}

2AB^{2} = 4r^{2}

AB^{2} = 2r^{2}

∴ AB = √2r = √2 × 8 = 8√2 cm

Area of the square = (side)^{2}

= (8√2)^{2} = 64 × 2 = 128 cm^{2}

Hence the required area = 128 cm^{2}.

Question 8.

The diameter of a circular park is 24 m. A 3 m wide path runs around it from inside. Find the cost of paving the path at ₹ 25 per square meter.

Answer:

Here, diameter of circular park i.e., AB = 24 m

Radius of circular park, r_{1} = \(\frac{24}{2}\) = 12 m

So, outer radius, r_{1} = 12 m

and inner radius, r_{2} = 12m – 3m = 9m

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

= πr_{1}^{2} – πr_{2}^{2}

= π[r_{1}^{2} – r_{2}^{2}]

= \(\frac{22}{7}\)[12^{2} – 9^{2}]

= \(\frac{22}{7}\)(12 + 9)(12 – 9)

= \(\frac{22}{7}\) × 21 × 3

= \(\frac{1386}{7}\)

= 198 m^{2}

Cost of paving the path = Area of path × Rate

= 198 × 25 = ₹ 4950

Hence the cost of paving the path is ₹ 4950.

Question 9.

A paper is in the form of a rectangle PQRS in which PQ = 10 cm and QR Two semi-circular portions with QR and PS as diameters are cut-off (As shown in figure), area of the, remaining part.

Answer:

Area of the paper = 10 × 7 = 70 cm^{2}

Radius of each semi-circular ends = \(\frac{7}{2}\) cm.

Area of the two semi-circular portion

= 2 × \(\frac{1}{2}\)πr^{2} = πr^{2}

= \(\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}=\frac{77}{2}\)cm

Area of the remaining portion = 70 – \(\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}=\frac{77}{2}\)

= \(\frac{140-77}{2}=\frac{63}{2}\)cm^{2} = 31.5 cm^{2}

Hence the required area = 31.5 cm^{2}.

Question 10.

A horse is tied to a pole with 21m long string. Find the area where the horse can graze.

Answer:

Radius of the circular field = 21 m

∴ Its area = πr^{2}

= \(\frac{22}{7}\) × 21 × 21

= 1386 m^{2}

Hence the required area = 1386 m^{2}.

Question 11.

An iron washer is made by cutting out a concentric circular plate of radius 2 cm from a circular plate of radius 5 cm. Find the area of face of the washer.

Answer:

Radius of the circular plate (r_{1}) = 5 cm

Radius of the circular plate (r_{2}) = 2 cm

∴ Area of the face of the washer = Area of the outer plate – Area of the inner plate

= πr_{1}^{2} – πr_{2}^{2}

= π[r_{1}^{2} – r_{2}^{2}]

= \(\frac{22}{7}\)[(5)^{2} – (2)^{2}]

= \(\frac{22}{7}\)[25 – 4]

= \(\frac{22}{7}\) × 21

= 66 cm^{2}

Hence the required area = 66 cm^{2}.

Question 12.

Two concentric circles have radii of 14 cm and 7 cm respectively. Find the area of the space between them.

Answer:

Here r_{1} = 14 cm and r_{2} = 7 cm

Area of the space between them = πr_{1}^{2} – πr_{1}^{2}

= π[r_{1}^{2} – r_{2}^{2}]

= \(\frac{22}{7}\)[(14)^{2} – (7)^{2}]

= \(\frac{22}{7}\)(196 – 49) = \(\frac{22}{7}\) × 147

= 22 × 21 = 462 cm^{2}

Hence the required area = 462 cm^{2}.

Question 13.

The sum of the circumferences of four small circles is equal to the circumference of a bigger circle. Find the ratio of the area of the bigger circle to that of the smaller circle.

Answer:

Let the radius of each smaller circle be r_{1} cm

∴ Circumferences of the 4 smaller circles = 4 × 2πr_{1} = 8πr_{1}

Circumferences of 4 smaller circles = circumference of the bigger circle = 8πr_{1}

Let r_{2} be the radius of the bigger circle

∴ 2πr_{2} = 8πr_{1}

r_{2} = 4r_{1}

\(\frac{\text { Area of the bigger circle }}{\text { Area of the smaller circle }}=\frac{\pi \mathrm{r}_2^2}{\pi \mathrm{r}_1^2}=\frac{r_2^2}{r_1^2}=\frac{(4 n)^2}{r_1^2}=\frac{16}{1}\)

Hence the required ratio = 16:1.

Question 14.

The area of a circle is equal to the area of a rectangle with sides 112 m and 88 m respectively. Find the circumference of the circle.

Answer:

Area of the circle = Area of the rectangle

πr^{2} = 112 × 88

\(\frac{22}{7}\) × r^{2} = 112 × 88

r^{2} = 112 × 88 × \(\frac{7}{22}\)

r^{2} = 112 × 4 × 7

r^{2} = 4 × 4 × 7 × 4 × 7

∴ r = 2 × 2 × 2 × 7 = 56m

∴ Circumference of the circle = 2nr

= 2 × \(\frac{22}{7}\) × 56

= 44 × 8

= 352 m

Hence the required circumference = 352 m.

### DAV Class 7 Maths Chapter 11 Value Based Questions

Question 1.

Jia’s mother prepared three pizzas of same size for her daughter and her two friends. Just then two more friends joined them. They decided to share the three pizzas among themselves.

(i) If radius of each pizza is 7 cm, find the area of the share of pizza that each friend gets.

(ii) What value of the children is depicted here?

Answer:

(i) Radius of each pizza (r) = 7 cm

∴ Area of each pizza = πr^{2} = \(\frac{22}{7}\)(7)^{2}

= \(\frac{22}{7}\) × 7 × 7

= 154 cm^{2}

So, area of three pizzas of same size = 3 × 154 cm^{2} = 462 cm^{2}

Total number of persons = Jia + Her two friends + Two more friends

= 1 + 2 + 2 = 5

Now, the area of the share of pizza that each person gets = \(\frac{1}{5}\) (462) cm^{2} = 92.4 cm^{2}

Hence, the area of the share of pizza that each friend gets is 92.4 cm^{2}

(ii) Value: Sharing the things equally among friends.

Question 2.

There was a deserted land near a colony where people used to throw garbage. Colony people united to develop a pond in triangular shape as shown in figure.

The land is in the shape of a parallelogram ABCD. In the rest of the portion medicinal plants are grown. Area of the parallelogram is 200 m2.

(i) Calculate the area where medicinal plants are grown.

(ii) Which value is depicted here?

Answer:

Area of the parallelogram ABCD = 200 m^{2}

If we cut the triangle DAP and join this triangle as shown in the figure given below.

The figure obtained now is a rectangle with length DC and breadth DP.

So, Area of parallelogram ABCD = Area of rectangle DCPaP

⇒ 200m2 = Area of rectangle DCP^{1}P

⇒ Area of rectangle DCP^{1}P

= 200m^{2}

From the figure, rectangle DCP^{1}P is divided into two traingles ΔDCP and ΔCP^{1}P of equal areas.

∴ Area of A CP^{1}P = \(\frac{1}{2}\) (Area of rectangle DCP^{1}P)

⇒ Area of A CP^{1}P = \(\frac{1}{2}\) (200) m^{2} = 100 m^{2}

Therefore, the area where medicinal plants are grown is 100 m^{2}.

(ii) Values: Colony people are aware about the cleanliness of the environment. So, they are caring towards the environment.