Question

the side of a square exceed the side of another square by 4 cm and the sum of areas of two squares is 400 square centimetre find the side of two squares​

Answer 1

☯ Let $\sf S_1$ and $\sf S_2$ be two squares.

Let the side of the square $\sf S_2$ be x cm in length. Then,The side of square $\sf S_1$ is (x + 4) cm.

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

We know that,

$\star\;{\boxed{\sf{\purple{Area_{\;(square)} = (side)^2}}}}$

Therefore,

Area of square, $\sf S_1$ = (x + 4)²

Area of square, $\sf S_2$ = x²

$\underline{\bigstar\:\boldsymbol{According\:to\: Question\::}}$

Area of Square $\sf S_1$ + Area of Square $\sf S_2$ = 400 cm²

$:\implies\sf (x + 4)^2 + x^2 = 400$

$:\implies\sf (x^2 + 8x + 16 + x^2) = 400$

$:\implies\sf 2x^2 + 8x + 16 = 400$

$:\implies\sf 2x^2 + 8x + 16 - 400 = 0$

$:\implies\sf 2x^2 + 8x - 384 = 0$

$:\implies\sf 2(x^2 + 4x - 192) = 0$

$:\implies\sf x^2 + 4x - 192 = 0$

$:\implies\sf x^2 + 16x - 12x^2 - 192 = 0$

$:\implies\sf x(x + 16) - 12(x + 16) = 0$

$:\implies\sf (x + 16)(x - 12) = 0$

$:\implies\sf x = 12\;or,\; x = -16$

Length of the side of a square cannot be negative.

So, x = 12

Therefore,

• Side of square $\sf S_1 = x + 5 = 12 + 4 = \bf{16\;cm}$

• Side of square $\sf S_2 = \bf{12\;cm}$