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A rectangular plot is given for constructing a house, having a measurement of 40 m long and 15 m in the front. According to the laws, a minimum of 3 m, wide space should be left in the front and back each and 2 m wide space on each of other sides. Find the largest area where house can be constructed.

Answers (1)

[374\; m^{2}]

Let ABCD be the rectangular plot,

AB = 40 cm, AD = 15 cm

                                        

Given that minimum of 3 m wide space should be left in the front and back

length \; of \; PQ = [AB - (3 + 3)] \; m

PQ = [40 - 6] \; m

PQ = 34 \; m

Similarly, RS = 34 m

Given that 2 m wide space on each of other sides is to be left

Length\; of \; PS = [AD - (2 + 2)] \; m

PS = [15 - 4] \; m

PS = 11 \; m \; and

QR = 11 \; m

So here PQRS is another rectangle formed in the rectangle ABCD

So, Area of rectangle PQRS = length \times breadth

= PQ\times PS= \left ( 34\times 11 \right )\; m^{2}= 374\; m^{2}

Hence the area of house can be constructed in 374 m^{2}

 

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