Question

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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 a house can be constructed.​

Answer 1

Step-by-step explanation:

LettheplotberectangleABCDwherelengthAD=BC=40m&widthAB=CD=15m.

Effective lengthofplotafterleaving3meachinthefrontandback=EH=FG=40−(3+3)=34m

Effectivewidthofplotafterleaving2meachonthesides=EF=HG=15−(2+2)=11m

∴Thelargestareawherethehousecanbeconstructed

=AreaofrectangleEFGH

=EH×HG

=34×11

=374

Answer 2

Step-by-step explanation:

${ \pmb{ \underline{ \green{Question:-}}}}$

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 a house can be constructed.

${ \pmb{ \underline{ \green{Answer:-}}}}$

• The largest area where a house can be constructed is 374m²

${ \pmb{ \underline{ \green{Solution:-}}}}$

Formula Used:

• Area of Rectangle = l×b

ATQ,

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.

$OD=CL=AP=BK= 3m$

$DN=CM=BJ=AI=2m$

So, actual length where house can be constructed,

$\pink{EH = AD-(DO+AP)}$

$\pink{EH = 40 - (3 + 3)}$

$\qquad \qquad \red{EH = 34 \: m}$

The actual width where house can be constructed,

$\pink{EF = AB-(AI+BJ)}$

$\pink{EF =15 - (2 + 2)}$

$\qquad \qquad \red{EF =11 \: m}$

Area of the house constructed = 34*11 = 374m²