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 cm^{2} = 24 cm^{2}

(i) Here, number of congruent polygons = 6

∴ Area of each polygon = \(\frac{24}{6}\) cm^{2} = 4 cm^{2}

(ii) Here, number of congruent polygons = 4

Area of each polygon = \(\frac{24}{4}\)cm^{2} = 6 cm^{2}

(iii) Here, number of congruent polygons = 2

∴ Area of each polygon = \(\frac{24}{2}\)cm^{2} = 12 cm^{2}

(iv) Number of congruent polygons = 2

∴ Area of each polygon = \(\frac{24}{2}\)cm^{2} = 12 cm^{2}

(v) Number of congruent polygons = 8

∴ Area of each polygon = \(\frac{24}{8}\)cm^{2} = 3 cm^{2}

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 cm^{2}

(ii) Base = 8 cm, Height = 2.5 cm

Area of the parallelogram = Base x Height = 8 cm x 2.5 cm = 20 cm^{2}

(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

= \(\frac{3240}{100}\) cm^{2} = 32.40 cm^{2}

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 cm^{2} in mm^{2}

(ii) 2 ha in m^{2}

(iii) 10 m2 in cm^{2}

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

Answer:

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

∵ 1 cm^{2} = 100 mm^{2}

∴ 50 cm^{2} = 50 x 100 mm^{2} = 5000 mm^{2}

(ii) 2 ha in m^{2}

1 ha = 1000 m^{2}

∴ 2 ha = 2 x 1000 m2 = 20000 m^{2}

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

∵ 1 cm^{2} = 10000 cm^{2}

∴ 10 m^{2}= 10 x 10000 cm^{2} = 100000 cm^{2}

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

∵ 10000 cm^{2} = 1 m^{2}

∴ 1 cm^{2} = \(\frac{1}{10000}\) m^{2}

So, 1000 cm^{2} = \(\frac{1}{10000}\) x 1000 m^{2}

= \(\frac{1}{10}\) m^{2} = 0.1m^{2}