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Sum of the areas of two squares is 400 cm$^{2}$. If the difference of their perimeter is 16 cm, Find the sides of the two squares.
Given: Sum of the areas of two squares $=400\ cm^{2}$ and the difference of their perimeter $=16\ cm$.
To do: To find the sides of both the squares.
Solution:
Let the sides of the two squares be a cm and b cm.
Then, Their areas are $a^{2}$ and $b^{2}$ and their perimeters are 4a and 4b.
By the given condition
$a^{2} +b^{2} \ =\ 400\ ........\ ( 1)$
and
$4a-4b=16$
$\Rightarrow a-b=\frac{16}{4} =4\ \ .............( 2)$
$\Rightarrow a=b+4$
Substituting the value of a in $( 1)$ ,
We get,
$( b+4)^{2} +b^{2} =400$
$\Rightarrow b^{2} +16+8b+b^{2} =400$
$\Rightarrow 2b^{2} +8b=400-16$
$\Rightarrow 2\left( b^{2} +4b\right) =384$
$\Rightarrow b^{2} +4b-192=0$
$\Rightarrow b^{2} +16b-12b-192=0$
$\Rightarrow b( b+16) -12( b+16) =0$
$\Rightarrow ( b-12)( b+16) =0$
$\Rightarrow b=12,\ -16$
$\because$ Side of the square can't be negative,
$\therefore$ We accept only $b=12\ \ \ \ \ \ \ ( we\ reject\ b=-160)$
If $b=12$, On subtituting this value in $( 2)$
We get $a=b+4$
$=12+4$
$=16$
Thus, The sides of the two squares are 16 cm and 12 cm.
Updated on: 10-Oct-2022
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