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Home » Sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, find the sides of the two squares.
Sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, find the sides of the two squares.
CBSE | Class 10 | Exercise 4.3 | Quadratic Equations
Sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, find the sides of the two squares.

Solution:

Let the sides of the two squares be x m and y m.

Therefore, their perimeter will be 4x and 4y respectively

And area of the squares will be {{x}^{2}}~and~{{y}^{2}}~ respectively.

Given,

4x~\text{ }4y~=\text{ }24

x~~y~=\text{ }6

x~=~y~+\text{ }6

Also, {{x}^{2~}}{{+}^{~}}{{y}^{2}}~=\text{ }468

\Rightarrow (6~+~{{y}^{2}})~+~{{y}^{2}}~=\text{ }468

\Rightarrow 36~+~{{y}^{2}}~+\text{ }12y~+~{{y}^{2}}~=\text{ }468

\Rightarrow 2{{y}^{2}}~+\text{ }12y~+\text{ }432\text{ }=\text{ }0

\Rightarrow ~{{y}^{2}}~+\text{ }6y\text{ }\text{ }216\text{ }=\text{ }0

\Rightarrow ~{{y}^{2}}~+\text{ }18y~\text{ }12y~\text{ }216\text{ }=\text{ }0

\Rightarrow ~y\left( y~+18 \right)\text{ }-12\left( y~+\text{ }18 \right)\text{ }=\text{ }0

\Rightarrow \left( y~+\text{ }18 \right)\left( y~\text{ }12 \right)\text{ }=\text{ }0

\Rightarrow ~y~=\text{ }-18,\text{ }12

As we know, the side of a square cannot be negative.

Hence, the sides of the squares are 12 m and (12 + 6) m = 18 m.

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