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Question

If the area of the rectangle is 9x212x32, then find the perimeter of the given rectangle. [4]

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Solution

Given,
The area of the rectangle =9x212x32

Now,
9x212x32 can be written as
=(3x)22×3x×2+436
=(3x2)262
[Using the identity, (a+b)×(ab)=a2b2]
=(3x2+6)×(3x26)
=(3x+4)×(3x8) [2 marks]

Now,
Length =3x+4; Breadth =3x8
Perimeter =2×(length+breadth)
=2×(3x+4+3x8)
=2×(6x4) [2 marks]

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