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Q.

A room 5 m long and 4 m wide is surrounded by a verandah. If the verandah occupies an area of 22  m 2  , find the width of the verandah.


 R.D. Sharma Solutions Class 7 | Math Chapter 20 Mensuration I (Perimeter  and Area of Rectilinear Figures) Exercise 20.2

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a

1 m

b

2 m

c

4 m

d

5 m 

answer is A.

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Detailed Solution

Given a room 5 m long and 4 m wide is surrounded by a verandah.
R.D. Sharma Solutions Class 7 | Math Chapter 20 Mensuration I (Perimeter  and Area of Rectilinear Figures) Exercise 20.2Area of the room is,
A=length×breadth A=5 m×4 m   A=20  m 2  
The length PQ is,
PQ= 5+x+x   PQ= 5+2x m  
The breadth QR is,
QR= 4+x+x   QR= 4+2x m  
Area of veranda PQRS is,
A=(5+2x)×(4+2x)   A= 4 x 2 +18x+20 m 2  
Area of the veranda = Area of PQRS – Area of ABCD
22=4 x 2 +18x+2020   22=4 x 2 +18x  
11=2 x 2 +9x   2 x 2 +9x11=0   Now, mid-term splitting,
2 x 2 +11x2x11=0   x 2x+11 1 2x+11 =0   x1 2x+11 =0  
Therefore,
x1=0 x=1   Also,
2x+11=0 2x=11 x= 11 2   As, width can’t be negative, therefore neglecting x= 11 2  .
Therefore, the width is 1 m.
Hence, option 1 is correct.
 
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