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The sum of areas of two squares is $2650 \text{ cm}^2$. If the sum of their perimeters is $280 \text{ cm}$, find the sides of the two given squares.
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Solution: (a) Let the sides of squares be $x$ and $y$
A.T.Q.
$x^2 + y^2 = 2650$ --------(i) [1 mark]
$4x + 4y = 280 \Rightarrow x + y = 70$ ---(ii) [1 mark]
getting $2x^2 - 140x + 2250 = 0$ or $x^2 - 70x + 1125 = 0$ [1 mark]
$\Rightarrow (x - 25)(x - 45) = 0$
$\Rightarrow x = 25$ and $x = 45$ [1 mark]
$\therefore y = 45$ and $y = 25$
sides of square are $25 \text{ cm}$ and $45 \text{ cm}$. [1 mark]
A.T.Q.
$x^2 + y^2 = 2650$ --------(i) [1 mark]
$4x + 4y = 280 \Rightarrow x + y = 70$ ---(ii) [1 mark]
getting $2x^2 - 140x + 2250 = 0$ or $x^2 - 70x + 1125 = 0$ [1 mark]
$\Rightarrow (x - 25)(x - 45) = 0$
$\Rightarrow x = 25$ and $x = 45$ [1 mark]
$\therefore y = 45$ and $y = 25$
sides of square are $25 \text{ cm}$ and $45 \text{ cm}$. [1 mark]