What is the perimeter of the rectangle whose width is two third the length and the area is 96

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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

Let the length of the rectangle be L and the width be W.

Given that the width is 2/3 of the length.

Then, we will write:

W = (2/3)*L................(1)

Also, given that the area of the rectangle is 96.

Then,

A = w*L = 96..............(2)

Now we will substitute (1) into (2).

==> W*L = 96

==> (2/3)*L * L = 96

==> (2/3)*L^2 = 96

Now we will multiply by 3/2.

==> L^2 = 96*3/2

==> L^2 = 144

Now we will take the root of both sides;

==> L = 12 .

==> W = (2/3)L = (2/3)*12 = 8

==> W = 8.

Now we will calculate the perimeter;

==> P = 2L + 2W = 2*12 + 2*8 = 24 + 16 = 40.

Then, the perimeter of the rectangle is 40 units.

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neela | High School Teacher | (Level 3) Valedictorian

Posted on

What is the perimeter of the rectangle whose width is two third the length and the area is 96.

Let the length of the rectangle be l. Then the width is (2/3) of length.

=> width w = (2/3)l.

Therefore perimeter p = 2(l+w) = 2{l+(2/3)l}....(1).

The area A of the rectangle is given by:

A = l*w = l*(2/3)l = (2/3)l^2.

A = 96 sq units.

=> A = (2/3)l^2 = 96

=> l^2 = 96*3/2 = 144 sq units.

=> l = sqrt(144) = 12 units.

=> w = (2/3)l = (2/3)*12 = 8 units.

Therefore p = 2(l+w) = 2(12+8) = 40 units.

So the perimeter of the rectangle p = 40 units.

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