**Concise Mathematics – Class 8 Selina Exercise 20B Solution by experts | 20B Selina Concise Solution Class 8 | Class 8 Selina Concise Ex 20B Page 227 Solution | RK Bansal Class 8 ICSE Book Solution Online.**

**Exercise 20B Solution**

**(1) Find the length and perimeter of a rectangle, whose area = 120 cm ^{2} and breadth = 8 cm.**

**(2) The perimeter of a rectangle is 46 m and its length is 15 m. Find its:**

**(i) breadth**

**(ii) area**

**(iii) diagonal**

**Solution: **

**(3) The diagonal of a rectangle is 34 cm. If its breadth is 16 cm; find its:**

**(i) length**

**(ii) area**

**(4) The area of a small rectangular plot is 84 m ^{2}. If the difference between its length and the breadth is 5 m; find its perimeter.**

**Solution: **

or, x – 7 = 0

x = 7

Perimeter = 2 (l + b) m

= 2 (12 + 7) m

= 2 x 19 m

= 38 m

**(5) The perimeter of a square is 36 cm; find its area.**

**(6) Find the perimeter of a square whose area is 1.69 m ^{2}.**

**(7) The diagonal of a square is 12 cm long; find its area and length of one side.**

**Solution: **

**(8) The diagonal of a square is 15 m, find the length of its one side and perimeter.**

**(9) The area of a square is 169 cm square. Find its :**

**(i) one side**

**(ii) perimeter**

**(10) The length of a rectangle is 16 cm and its perimeter is equal to the perimeter of a square with side 12.5 cm. Find the area of the rectangle.**

**Solution: **

According to statement,

32 + 2x = 50

Or, 2x = 50 – 32

Or, x = 18/2

x = 9

∴ Breadth of rectangle = 9 cm

∴ Area of the rectangle = (16 x 9) cm square

= 144 cm square.

**(11) The perimeter of a square is numerically equal to its area. Find its area.**

**(12) Each side of a rectangle is doubled is doubled. Find the ration between:**

**(i) perimeters of the original rectangle and the resulting rectangle.**

**(ii) areas of the original rectangle and the resulting rectangle.**

**(13) In each of the following cases, ABCD is a square and PQRS is a rectangle. Find in each case, the area of the shaded portion.**

**Solution: **

Ans. Area of shaded portion is 18.72 m square.

**(14) A path of uniform width 3 m, runs around outside of a square field of side 21 m. Find the area of the path.**

**(15) A path of uniform width 2.5 m, runs around the inside of a rectangular field 30 m by 27 m. Find the area of the path.**

**Solution: **

Area of the rectangular field excluding the path = (25 x 22) m square

= 550 m square

∴ Area of the path = (810 – 550) m square

= 260 m square.

**(16) The length of a hall is 18 m and its width is 13.5 m. Find the least number of square tiles. Each of side 25 cm, required to cover the floor of the hall.**

**(i) without leaving any margin.**

**(ii) leaving a margin of width 1.5 m all around.**

**In each case, find the cost of the tiles required at the rate of Rs. 6 per tile.**

**Solution: **

**(17) A rectangular field is 30 m in length and 22 m in width. Two manually perpendicular roads, each 2.5 m wide, are drawn inside the field so that one road is parallel to the length of the field and the other road is parallel to its width. Calculate the area of the crossroads.**

**(18) The length and the breadth of a rectangular field are in the ratio 5:4 and its 3380 m ^{2}. Find the cost of fencing it at the rate of Rs. 75 per. m.**

**(19) The length and the breadth of a conference hall are in the ratio 7:4 and its perimeter 110 m. Find:**

**(i) area of the floor of the hall.**

**(ii) number of tiles, each a rectangle of size 25 cm x 20 cm, required for flooring of the hall.**

**(iii) the cost of the tiles at the rate of Rs. 1400 per hundred tiles.**

**Solution: **

**__ x __**