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# Sum of the areas of two squares is 468 ${m^2}$. If the difference of their perimeters is 24 m, find the sides of the two squares.

Last updated date: 17th Jul 2024
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Hint: Assume the sides of the squares as some variables. Form two equations using the conditions given in the question and solve them.

Let the side of the first square be $a$ and that of the second square be $A$.
Then the area of the first square $= {a^2}$
And the area of the second square $= {A^2}$
Their perimeters would be $4a$ and $4A$ respectively.

The difference of the perimeters of the squares is given as 24 m. So, we have:
$\Rightarrow 4A - 4a = 24 \\ \Rightarrow 4\left( {A - a} \right) = 24 \\ \Rightarrow A - a = 6 .....(i) \\$
And sum of their areas is given as 468 ${m^2}$:
$\Rightarrow {A^2} + {a^2} = 468 .....(ii)$
Putting $A = a + 6$ from equation $(i)$ in equation $(ii)$, we’ll get:
$\Rightarrow {\left( {a + 6} \right)^2} + {a^2} = 468 \\ \Rightarrow {a^2} + 36 + 12a + {a^2} = 468 \\ \Rightarrow 2{a^2} + 12a + 36 = 468 \\ \Rightarrow {a^2} + 6a + 18 = 234 \\ \Rightarrow {a^2} + 6a - 216 = 0 \\ \Rightarrow {a^2} + 18a - 12a - 216 = 0 \\ \Rightarrow a\left( {a + 18} \right) - 12\left( {a + 18} \right) = 0 \\ \Rightarrow \left( {a - 12} \right)\left( {a + 18} \right) = 0 \\ \Rightarrow a = 12{\text{ or }}a = - 18 \\$
But the side of the square cannot be negative, $a = 12$ is the valid solution.
Putting the value of $a$ in equation $(i)$ we’ll get:
$\Rightarrow A - 12 = 6 \\ \Rightarrow A = 18 \\$
Thus, the side of the first square is 12 m and the side of the second square is 18 m.

Note: If we face any difficulty finding the roots of the quadratic equation $a{x^2} + bx + c = 0$ by simple factorization, we can apply the formula for finding roots:
$\Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.