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Rd sharma solutions

A room 5 m long and 4 m wide is surrounded by a verandah. If the verandah occupies an area of 22 m2, find the width of the verandah.

Let the width of the verandah be x m.

Given Length of the room AB = 5 m and breadth of the room, BC = 4 m

We know that area of rectangle = length x breadth

Area of the room = 5 m x 4 m

= 20 m2

From the figure, it is clear that

Length of the veranda PQ = (5 + x + x) = (5 + 2x) m

Breadth of the veranda QR = (4 + x + x) = (4 + 2x) m

Area of veranda PQRS = (5 + 2x) x (4 + 2x)

= (4×2 + 18x + 20) m^2

Area of veranda = Area of PQRS – Area of ABCD

22 = 4x2 + 18x + 20 – 20

22 = 4x2 + 18x

On dividing above equation by 2 we get,

11 = 2x2 + 9x

2x2 + 9x – 11 = 0

2x2 + 11x – 2x – 11 = 0

x (2x+11) - 1 (2x+11) = 0

(x- 1) (2x+11)= 0

When x – 1 = 0, x = 1

When 2x + 11 = 0, x = (-11/2)

= The width cannot be a negative value. So, width of the veranda = x = 1 m

Let the width of the verandah be x m.

Given Length of the room AB = 5 m and breadth of the room, BC = 4 m

We know that area of rectangle = length x breadth

Area of the room = 5 m x 4 m

= 20 m2

From the figure, it is clear that

Length of the veranda PQ = (5 + x + x) = (5 + 2x) m

Breadth of the veranda QR = (4 + x + x) = (4 + 2x) m

Area of veranda PQRS = (5 + 2x) x (4 + 2x)

= (4×2 + 18x + 20) m^2

Area of veranda = Area of PQRS – Area of ABCD

22 = 4x2 + 18x + 20 – 20

22 = 4x2 + 18x

On dividing above equation by 2 we get,

11 = 2x2 + 9x

2x2 + 9x – 11 = 0

2x2 + 11x – 2x – 11 = 0

x (2x+11) - 1 (2x+11) = 0

(x- 1) (2x+11)= 0

When x – 1 = 0, x = 1

When 2x + 11 = 0, x = (-11/2)

= The width cannot be a negative value. So, width of the veranda = x = 1 m

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