# Perimeter And Area

## Square and Rectangle

Perimeter of square = 4 × Side

Area of square = Side × Side = (Side)^{2}

Perimeter of Rectangle = 2 × (Length + Breadth)

Area of Rectangle = Length × Breadth

### Exercise 11.1

**Question 1:** The length and the breadth of a rectangular piece of land are 500m and 300 m respectively. Find (i) Its areas (ii) the cost of the land, if 1m^{2} of the land costs Rs 10,000.

**Answer:** Given: Length = 500 m and breadth = 300 m, cost of 1m2 of land = Rs. 10,000

Area = Length `xx` Breadth

`= 500 xx 300 = 150000` sq m

Price of Land = Area `xx` cost per sq m

`= 150000 xx 10000` = Rs. 1,500,000,000

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

**Answer:** Given, Perimeter of a square park = 320 m

Side = Perimeter ÷ 4

`= 320 ÷ 4 = 80 m`

Area of Square = Side^{2}

`= 80^2= 6400` sq m

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

**Answer:** Given: Length = 22m and area of the field = 440 m^{2}

Area = Length `xx` Breadth

So, Breadth = Area ÷ Length

`= 440 ÷ 22 = 20 m`

Now, perimeter = 2(length + breadth)

`= 2(22 + 20) = 2 xx 42 = 84 m`

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

**Answer:** Given perimeter = 100 cm and length = 35 cm

Perimeter = 2(Length + Breadth)

Or, `100 = 2(35 + text(breadth))`

Or, `35 + text(breadth) = 50`

Or, breadth `= 50 – 35 = 15 cm`

Area = length `xx` breadth

`= 35 xx 15 = 525` sq cm

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

**Answer:** Given; side of square park = 60 m and length of rectangular park = 90 m

Area of square park = Side^{2}

`= 60^2 = 3600` sq m

Area of rectangular park = Area of rectangular park

Hence, `3600 = 90 xx \text(breadth)`

Or, breadth `= 3600 ÷ 90 = 40 m`

**Question 6:** A wire is in the shape of a rectangle. Its length is 40cm and breadth is 22cm. 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?

**Answer:** Given; length = 40 cm and breadth = 22 m

Perimeter = 2(length + breadth)

`= 2(40 + 22) = 2 xx 62 = 124` cm

Perimeter of rectangle = Perimeter of square

Or, `124 = 4 xx \text(Side)`

Or, Side `= 124 ÷ 4 = 31 cm`

Now, area of square = Side^{2}

`= 31^2 = 961` sq cm

Area of rectangle = Length `xx` Breadth

`= 40 xx 22 = 880` sq cm

Hence, area of square > area of rectangle

**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.

**Answer:** Given; perimeter = 130 cm and breadth = 30 cm

Perimeter = 2(length + breadth)

Or, `130 = 2(text(length) + 30)`

Or, `text(length) + 30 = 65`

Or, length `= 65 – 30 = 35 cm`

Area of rectangle = Length `xx` Breadth

`= 35 xx 30 = 1050` sq cm

**Question 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 (See Figure). Find the cost of white washing the wall, if the rate of white washing the wall is Rs 20 per m^{2}.

**Answer:** Given; length of wall = 4.5 m, breadth of wall = 3.6 m, length of door = 2 m and breadth of door = 1 m

Rate of whitewashing = Rs. 20 per sq m

Area of wall = Length `xx` Breadth

`= 4.5 xx 3.6 = 16.2` sq m

Area of door = Length `xx` Breadth

`= 2 xx 1 = 2` sq m

Area of wall – Area of door

`= 16.2 – 2 = 14.2` sq m

Cost of whitewashing = Area of wall (leaving the door) `xx` rate

`= 14.2 xx 20 = Rs. 284`