CBSE students can refer to NCERT Solutions for Class 7 Maths Chapter 11 Perimeter and Area InText Questions and Answers are provided by experts in order to help students secure good marks in exams.

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

Try These (Page 205)

Question 1.

Ayush and Deeksha made pictures. Ayush made his picture on a rectangular sheet of length 60 cm and breadth 20 cm while Deeksha made hers on a rectangular sheet of length 40 cm and breadth 35 cm. Both of these pictures have to be separately framed and laminated. Who has to pay more for framing, if the cost of framing is ₹ 3.00 per cm?

If the rate of lamination is ₹ 2.00 per cm^{2}, who has to pay more for lamination?

Solution:

For finding the cost of framing, we need to find the perimeter and then multiple it by the rate for framing.

In case of Ayush :

Perimeter of the rectangular sheet = 2 (Length + Breadth)

= 2 × (60 + 20) cm = 2 × 80 cm = 160 cm

∴ Cost of framing at the rate of ₹ 3 per cm

= ₹ (3 × 160) = ₹ 480

In case of Deeksha :

Perimeter of the rectangular sheet

= 2 × (40 + 35) cm

= 2 × 75 cm = 150 cm

∴ Cost of framing at the rate of ₹ 3 per cm

= ₹ (3 × 150) = ₹ 450

Thus, Ayush paid more for framing (as 480 > 450).

For finding the cost of lamination, we need to find the area and then multiply it by the rate for lamination.

In the case of Ayush :

Area of the rectangular sheet

= (60 × 20) cm^{2}

=1200 cm^{2}

∴ Cost of lamination at the rate of ₹ 2 per cm^{2}

= ₹ (2 × 1200)

= ₹ 2400

In the case of Deeksha :

Area of the rectangular sheet

= (40 × 35) cm^{2} = 1400 cm^{2}

∴ Cost of lamination at the rate of ₹ 2 per cm^{2}

= ₹ (2 × 1400)

= ₹ 2800

Thus, Deeksha paid more for lamination (as 2800 > 2400).

Try These (Page 205)

Question 1.

What would you need to find, area, or perimeter to answer the following?

(i) How much space does a blackboard occupy?

Solution:

For finding the space occupied by a blackboard, we should find the area of the blackboard.

(ii) What is the length of a wire required to fence a rectangular flower bed?

Solution:

For fencing a rectangular flower bed, we need to find the perimeter of the flower bed.

(iii) What distance would you cover by taking two rounds of a triangular park?

Solution:

For finding the distance covered by taking two rounds of a triangular park, we need to find the perimeter of the park and then multiply it by 2.

(iv) How much plastic sheet do you need to cover a rectangular swimming pool?

Solution:

For finding the plastic sheet required to cover a rectangular swimming pool, we need to find the area of the pool.

Try These (Page 206)

Question 1.

Tanya needed a square of side 4 cm for completing college. She had a rectangular sheet of length 28 cm and breadth 21 cm [See fig. (i)] She cuts off a square of side 4 cm from the rectangular sheet. Her friend saw the remaining sheet

[See fig.(ii)] and asked Tanya, “Has the perimeter of the sheet increased or decreased now?”

(i) Has the total length of side AD increased after cutting off the square?

Solution:

There has been a change in the length of side AD after cutting the square.

(ii) Has the area increased or decreased?

Solution:

The area has decreased.

(iii) Tanya cuts off one more square from the opposite side. (See fig.) Will (the perimeter of the remaining sheet increase further?

Solution:

Yes, the perimeter increases.

(iv) Will the area increase or decrease further?

Solution:

The area will decrease.

(v) So, what can we infer from this?

Solution:

Clearly, if the perimeter increases, it is not necessary that the area also increases.

Try These (Page 206)

Question 1.

Experiment with several such shapes and cut-outs. 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.

Solution:

Do yourself (Based on the experiment).

Question 2.

Give two. examples where area increases as the perimeter increases.

Solution:

(i)

Consider a rectangle ABCD.

Area (ABCD) = A_{1} = 8 × 3 cm^{2} = 24 cm

Perimeter (ABCD) = P_{1} = 2(8 + 3) cm = 22 cm

Let us consider another rectangle ABEF.

Area (ABEF) = A_{2} = 10 × 3 cm^{2} = 30 cm^{2}

Perimeter (ABEF) = P_{2} = 2(10 + 3) cm = 26 cm

Thus, A_{2} > A_{1} => P_{2} > P_{1}

i.e., as area increases, perimeter increases.

(ii)

Consider a square ABCD.

Area (ABCD) = 3 × 3 cm^{2} = 9 cm^{2}

Perimeter (ABCD) = 3 × 3 cm = 9 cm

Let us consider a rectangle ABEF.

Area (ABEF) = 3 × 5 cm^{2} = 15 cm^{2}

Perimeter (ABEF) = 2(3 + 5) cm = 16 cm

Thus, A_{2} > A_{1} => P_{2} > P_{1}

i.e., as area increases, perimeter increases.

Question 3.

Give two examples where the area does not increase when the perimeter increases.

Solution:

(i)

Consider a rectangle ABCD.

Area (ABCD) = A_{1} =10 × 8 cm^{2} = 80 cm^{2}

Perimeter (ABCD) = P_{2} = 2(8 +10) cm = 36 cm

Cut-off a triangle from the rectangle ABCD as shown.

Its area = A_{2}

Its perimeter = P_{2} = (8 + 10 + 8 + 3 + 3 + 5 + 3) cm = 40 cm

Clearly, A_{1} < A_{2}, i.e., area does not increases, but P_{2} > P_{2}, i.e., perimeter increases.

(ii)

Let us have another cut as shown.

Its area = A_{3}

= (A_{2} – 4 × 3) cm^{2}

= (74 – 12) cm^{2}

= 62 cm^{2}

Its perimeter = P_{3}

= (2 + 3 + 4 + 3 + 2 + 10 + 8 + 3 + 3 + 5 + 3) cm = 46 cm

Clearly, A_{3} < A_{2}, but P_{3} > P_{2}, i.e., area decreases and perimeter increases.

Try These (Page 209)

Question 1.

Take a rectangle of sides 8 cm and 5 cm. Cut the rectangle along its diagonal to get two triangles. Superpose one triangle on the other.

ing

(i) Are they exactly the same in size?

Solution:

These triangles are exactly the same in size.

(ii) Can you say that both the triangles are equal in area?

Solution:

Yes, both the triangles are equal in area.

(iii) Are the triangles congruent also?

Solution:

Yes, by the SSS criterion of congruence, these triangles are congruent.

(iv) What is the area of each of these triangles.

Solution:

The area of each triangle

Where l = length and b = breadth

∴ Area of each triangle = \(\frac{1}{2}\) (8 × 5) cm^{2} = 20 cm^{2}

Question 2.

Take a square of side 5 cm and divide it into four triangles as shown.

(i) Are the four triangles equal in area?

Solution:

Yes, the four triangles are equal in area.

(ii) Are they congruent to each other?

Solution:

Yes, by superimposing, we find that the triangles are congruent.

(iii) What is the area of each triangle?

Solution:

The area of each triangle

= \(\frac{1}{4}\) (Area of the square)

= \(\frac{1}{4}\) (Side)^{2} = \(\frac{1}{4}\) (5)^{2} cm^{2} = 6.25 cm^{2}

Try These (Page 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.

Solution:

Given, length of rectangle = 6 cm

and breadth of rectangle = 4 cm

∴ Area of a rectangle = Length × Breadth = 6 × 4 = 24 cm^{2}

(a)

From the figure, it is clear that the rectangle is divided into 6 congruent polygons or parts.

∴ Area of each polygon = \(\frac{1}{6}\) × (Area of a rectangle)

= \(\frac{1}{6}\) × 24 = 4 cm^{2}

(b)

From the figure, it is clear that the rectangle is divided into 4 congruent polygons.

∴ Area of each polygon = \(\frac{1}{4}\) × (Area of a rectangle)

= \(\frac{1}{4}\) × 24 = 6 cm^{2}

(c)

From the figure, it is clear that the rectangle is divided into 2 congruent polygons.

∴ Area of each polygon = \(\frac{1}{2}\) × (Area of a rectangle)

= \(\frac{1}{2}\) × 24 =12 cm^{2}

(d)

From the figure, it is clear that the rectangle is divided into 2 congruent polygons.

∴ Area of each polygon = \(\frac{1}{2}\) × (Area of a rectangle)

= \(\frac{1}{2}\) × 24 = 12 cm^{2}

From the figure, it is clear that the rectangle is divided into 6 congruent polygons.

∴ Area of each polygon = \(\frac{1}{8}\) × (Area of a rectangle)

= \(\frac{1}{8}\) × 24 = 3 cm^{2}

Try These (Page 211)

Question 1.

Consider the following parallelograms

Find the areas of the parallelograms by counting the squares enclosed within the figures and also find the perimeters by measuring the sides.

Complete the following table :

Solution:

On counting the squares enclosed within the figures of the parallelogram, we find in each case these are 15 in numbers. So, the area of each parallelogram = 15 sq. units. Let us draw a perpendicular DM to the base AB, produced if necessary, as shown :

Let us find the other side AD of the parallelogram ABCD in each case in order to find the perimeter in each case.

From right-angled ΔADM, using Pythagoras theorem,

On filling the above data in the given table, we have

We observe that all these parallelograms have equal areas but different perimeters.

Try These (Page 212)

Question 1.

Consider the following parallelograms with the sides 7 cm and 5 cm.

(a)

(b)

(C)

(d)

Find the perimeter and the area of each of these parallelograms. Analyze your results. What do you infer from this?

Solution:

The perimeter and the area of each parallelogram are as under :

We observe that all these parallelograms have different areas but equal perimeters.

Thus, to find the area of a parallelogram, we need to know only the base and the corresponding height of the parallelogram.

Try These (Page 212)

Question 1.

Find the area of the following parallelograms :

(i)

Solution:

Area = base x height

= (8 × 3.5) cm^{2} = 28 cm^{2}

(ii)

Solution:

Area = base x height

= (8 × 2.5) cm^{2} = 20 cm^{2}

(iii) In a parallelogram ABCD, AB = 7.2 cm and the perpendicular from C on AB is 4.5 cm.

Solution:

Area of parallelogram ABCD

= AB × length of perpendicular from C on AB

= (7.2 × 4.5) cm^{2}

= 32.4 cm^{2}

Try These (Page 213)

Question 1.

Try the above activity with different types of triangles.

Solution:

Here, AABC is the given triangle. We have fixed another AACP in such a way that we get a parallelogram ABCP in each case as shown in the figure below:

Question 2.

Take different parallelograms. Divide each of the parallelograms into two triangles by cutting along any of its diagonals. Is the triangle congruent?

Solution:

Here, ABCD is the given parallelogram and each of the parallelograms is divided into two triangles by cutting along its diagonal \(\frac{A C}{B D}\). The triangles thus formed are congruent.

Try These (Page 213)

Question 1.

In the adjacent figures, all the triangles are on the base AB = 6 cm. What can you say about the height of each of the triangles corresponding to the base AB?

Can we say all the triangles are equal in area?

Solution:

Clearly, from the given figures, the heights of each of the triangles corresponding to base AB are equal.

Since the area of a triangle = \(\frac{1}{2}\) × base × height

So, we can say that all the triangles are equal in the area as they have the same base and equal height.

Try These (Page 214)

Question 1.

Consider the obtuse-angled triangle ABC of base 6 cm. It’s height AD which is perpendicular from the vertex A is outside the triangle. Can you find the area of a triangle?

Solution:

Yes, its area can be found out.

Area = \(\frac{1}{2}\) × base × height

Try These (Page 219)

Question 1.

In the given figure,

(a) Which square has the larger perimeter?

Solution:

The outer square has a larger area.

(b) Which is larger, the perimeter of a smaller square or the circumference of the circle.

Solution:

The circumference of the circle is larger than the perimeter of the smaller square.

Try These (Page 219)

Question 1.

Take one each of the quarter plate and half plate. Roll once each of these on a table-top. Which plate covers more distance in one complete revolution? Which plate will take fewer revolutions to cover the length of the table-top?

Solution:

Doing as directed, we observe that the half plate covers more distance in one complete revolution than the distance covered by the quarter plate. Also, the half plate takes fewer revolutions to cover the length of the table-top.

Try These (Page 221)

Question 1.

Consider the following:

(i) A farmer dug a flower bed of a radius of 7 m at the centre of a field. He needs to purchase fertilizer. If 1 kg of fertilizer is required for 1 square meter area, how much fertilizer should he purchase?

Solution:

Area of the circular flower bed = π r^{2}, where r = 7 m

Fertilizer required for the flower bed at the rate of 1 kg per 1 m^{2} = (154 × 1) kg = 154 kg

(ii) What will be the cost of polishing a circular table-top of radius 2 m at the rate of ₹ 10 per square meter?

Solution:

Area of the top of circular table = π r^{2}, where r = 2 m

Cost of polishing at the rate of ₹ 2 per sq. meter

=₹ 25.14 (approx)

Try These (Page 222)

Question 1.

Draw circles of different radii on graph paper. Find the area by counting the number of squares. Also, find the area by using the formula. Compare the two answers.

Solution:

Let us draw two circles of radius 1 cm and 2 cm on the graph paper of 1 cm^{2} square.

By using graph paper, we find that their respective areas are 4 cm^{2} and 12 cm^{2} respectively.

By using the formula, we. find that their respective areas are

= 12.57 cm^{2} (approx.)

Thus, we see that the two values differ.

Try These (Page 224)

Question 1.

Can you convert 1 km^{2} into m^{2}?

Solution:

We know that 1 km = 1000 m

1 km^{2} =1 km × 1 km

1 km^{2} = 1000 m × 1000 m

= 1000000 m^{2}

Try These (Page 225)

Question 1.

Convert the following:

(i) 50 cm^{2} in mm^{2}

Solution:

1 cm^{2} = 1 cm x 1 cm

= 10 mm × 10 mm (∵ 1 cm = 10 mm)

= 100 mm^{2}

50 cm^{2} = (50 × 100) mm^{2} = 5000 mm^{2}

(ii) 2 ha in m^{2}

Solution:

We know that a square of side 100 m has an area of 1 hectare (ha).

1 hectare -100 m × 100 m

=10000 m^{2}

(iii) 10 m^{2} in cm^{2}

Solution:

2 hectares =2 × 10000 m^{2} = 20000 m^{2}

1 m^{2} =1 m × 1 m

= 100 cm × 100 cm (∵ 1 m = 100 cm)

= 10000 cm^{2}

10 m^{2} = 10 × 10000 cm^{2} = 100000 cm^{2}

(iv) 1000 cm^{2} in m^{2}

Solution:

∴ 1000 cm^{2} =