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

Last updated date: 18th Jun 2024
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Hint: Here we will discuss the area of the path. It is observed that in square or rectangular gardens, parks, etc., some space in the form of path is left inside or outside or in between as cross paths. We will apply this concept for the areas of rectangle and square to determine the areas of different paths.

Complete step by step answer:
Length of the room, $l=5m$
Breadth of the room, $b=4m$
The area of the rectangular room is $A=l\times b$
  & \therefore A=5\times 4 \\
 & \therefore A=20\,{{m}^{2}} \\

Now, assume that the unknown width be $=x$
As the verandah is surrounding the room, it increases both the length and the width two times by the value of its width.
The length of the verandah,
  & l=5+x+x \\
 & l=5+2x \\

The breadth of the verandah,
  & b=4+x+x \\
 & b=4+2x \\

Also, given that Area of the verandah is $=22\,{{m}^{2}}$
We get,
Area of room is = length and breadth of the verandah - Area of the verandah
Simplify the above equation and find the value “x”
$(5+2x)(4+2x)-22-20=0$ [Take ‘20’ to the RHS, hence it becomes ‘-20’]

  & 5(4+2x)+2x(4+2x)-42=0 \\
 & 20+10x+8x+4{{x}^{2}}-42=0 \\
Rearrange the above equation in the form of quadratic equations –
$2{{x}^{2}}+9x-11=0$ [Take '2' common from all the terms]

Find factors of the above equations –
  & 2{{x}^{2}}+11x-2x-11=0 \\
 & \underline{2{{x}^{2}}-2x}+\underline{11x-11}=0 \\
 & 2x(x-1)+11(x-11)=0 \\
 & (2x+11)(x-1)=0 \\
So, $x=\dfrac{-11}{2}\ or\ x=1$
Since, measure of the width can never be negative, $x=\dfrac{-11}{2}$ is not possible.

Therefore, $x=1$ is the width of the verandah and the required solution.

Note: One golden rule to find the unknowns is to set the given word problem in the mathematical form and find correlations between the known and unknown terms and solve it accordingly. Remember the basic standard formulas for areas i.e the formula to find the area of a rectangle.