NCERT Solutions for Class 7 Maths (Chapter 11 - Perimeter and Area) are available here. With this article, you can access solutions for all the answers of Class Maths NCERT Textbook, Chapter 11. This chapter is one of the most important chapters of Class 7 Maths NCERT textbook and students must prepare this chapter well to secure good marks in the exam.

**NCERT Book for Class 7 Mathematics (PDF): Hindi & English (New Edition)**

**NCERT Solutions for Class 7 Maths****:**** Chapter 11**** - **** Perimeter and Area**

__EXERCISE 11.1__

**1.** The length and the breadth of a rectangular piece of land are 500 m and 300 m respectively. Find

(i) its area

(ii) the cost of the land, if 1 m^{2} of the land costs Rs. 10,000.

**Solutions:**

(i) Area = (Length × Breadth) = 500 × 300 = 1,50,000 m^{2}

(ii) As, Cost of 1 m^{2} land = Rs. 10000

So, cost of 1,50,000 m^{2} land = 10,000 × 1,50,000 = Rs. 1,500,000,000

**2. **Find the area of a square park whose perimeter is 320 m.

**Solutions:**

Given, Perimeter = 320 m

Now, Perimeter = 4 × Length of the side of park = 320

Also, length of the side of the park = 320/4 = 80 m

Therefore, area = (Length of the side of the park)^{2} = (80)^{2} = 6400 m^{2}

**3. **Find the breadth of a rectangular plot of land, if its area is 440 m^{2} and the length is 22 m. Also find its perimeter.

**Solutions:**

Given, Area = 440 m^{2}

Also, Area = Length × Breadth

So, 22 × Breadth = 440

Or Breadth = 440/22 = 20 m

Now, Perimeter = 2 (Length + Breadth) = 2 (22 + 20) = 2 (42) = 84 m.

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

**Solutions:**

Given, Perimeter = 100 cm

Also, Perimeter = 2 (Length + Breadth)

So, 2 (35 + Breadth) = 100

⇒ 35 + Breadth = 50

⇒ Breadth = 50 − 35 = 15 cm

Now, Area = Length × Breadth = 35 × 15 = 525 cm^{2}

**5. **The area of a square park is the same as of 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.

**Solutions:**

Given** **side of the square park is 60 m

So, area of square park = (Side)^{2} = (60)^{2} = 3600 m^{2}

Now, area of rectangular park = Length × Breadth = 3600

⇒ 90 × Breadth = 3600

⇒ Breadth = 40 m

**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?

**Solutions:**

According to the question, Perimeter of rectangle = Perimeter of square

Perimeter of rectangle = 2 (Length + Breadth) & Perimeter of square = 4 × Side

So, according to the question, we have, 2 (Length + Breadth) = 4 × Side

⇒ 2 (40 + 22) = 4 × Side

⇒ 2 × 62 = 4 × Side

⇒ Side = 124/4 = 31 cm

⇒ Area of rectangle = 40 × 22 = 880 cm^{2}

⇒ Area of square = (Side)^{2} = 31 × 31 = 961 cm^{2}

Clearly, the square-shaped wire encloses greater area.

**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.

**Solutions:**

Given, Perimeter = 2 (Length + Breadth) = 130

⇒ 2 (Length + 30) = 130

⇒ Length + 30 = 65

⇒ Length = 65 − 30 = 35 cm

So, Area = Length × Breadth = 35 × 30 = 1050 cm^{2}

**8. **A door of length 2 m and breadth 1m is fitted in a wall. The length of the wall is 4.5 m and the breadth is 3.6 m (Fig11.6). Find the cost of white Fig 11.6 washing the wall, if the rate of white washing the wall is Rs. 20 per m^{2}.

**Solutions:**

Given, the length of the wall is 4.5 m and the breadth is 3.6 m

So, area of wall = 4.5 × 3.6 = 16.2 m^{2}

Also, area of door = 2 × 1 = 2 m^{2}

Now, Area to be white-washed = 16.2 − 2 = 14.2 m^{2}

Given, cost of white-washing of 1 m^{2} area = Rs. 20

∴ Cost of white-washing 14.2 m^{2} area = 14.2 × 20 = Rs. 284.

**Solutions of exercises 11.1, 11.2, 11.3, 11.4 will be available shortly.**