Sum of the areas of two squares is 157 m2. If the sum of their perimeters is 68 m, find the sides of the two squares.

Sum of the areas of two squares is 157 m². If the sum of their perimeters is 68 m, find the sides of the two squares.

Answers (1)

Given Area of square 1+ Area of square 2 = 157 m²
Perimeter of square 1 + Perimeter of square 2 = 68 m
Let's assume square 1 has side S₁
square 2 has side S₂
$S_{1}^{2}+S_{2}^{2}= 157---(1)$ (area of square = side²)
$4S_{1}+4S_{2}= 68---(2)$ (perimeter of square = 4 $\times$ side)
$\Rightarrow 4\left ( S_{1}+S_{2} \right )= 68$
$\Rightarrow S_{1}+S_{2} = \frac{68}{4}= 17$
$\Rightarrow S_{2} = 17-S_{1}---(3)$
Put value of S₂ in equation (1)
$S_{1}^{2}+\left ( 17-S_{2} \right )^{2}= 157$
$S_{1}^{2}+17^{2}+S_{1}^{2}-2\times 17\times S_{1}= 157$
$2S_{1}^{2}-34S_{1}+289-157= 0$
$2S_{1}^{2}-34S_{1}+132= 0$
$S_{1}^{2}-17S_{1}+66= 0$ (divide by 2)
$S_{1}= \frac{-\left ( -17 \right )\pm \sqrt{\left ( 17 \right )^{2}-4\times 1\times 66}}{2\times 1}$ (shri dharacharya formula)
$S_{1}= \frac{17\pm \sqrt{289-264}}{2}$
$S_{1}= \frac{17\pm \sqrt{25}}{2}$
$S_{1}= \frac{17\pm 5}{2}= \frac{17+5}{2};\frac{17-5}{2}$
$S_{1}= 11 ; 6$
$S_{1}$ could be 11 or 6
Putting $S_{1}= 11$ in equation (3)
$S_{2}= 6\; \; S_{1}= 11$
Sides of the square are 11 cm and 6 cm.