Question

Sum of the area of two squares is 640 ${\text{mete}}{{\text{r}}^2}$. If the difference of their perimeters is 64 meter, find the sides of the two squares.

Answer 1

Hint: In this question, two conditions first on area and other on perimeter of squares are given, so try to make two equations with the help of given condition and solve them by elimination or any other and find what is asked in the question.

Complete step-by-step answer:
Let the sides of the two squares be a meter and b meter.
Area of first square is ${{\text{a}}^2}$ ${\text{mete}}{{\text{r}}^2}$
Area of second square is ${{\text{b}}^2}$ ${\text{mete}}{{\text{r}}^2}$

Perimeter of first square is ${\text{4a}}$ ${\text{meter}}$
Perimeter of second square is ${\text{4b}}$ ${\text{meter}}$

 According to the given problem,
Sum of the area of two squares is 640 ${\text{mete}}{{\text{r}}^2}$
Hence, ${{\text{a}}^2}{\text{ + }}{{\text{b}}^2} = 640$………(1)
difference of their perimeters is 64 m
⇒${\text{4a - 4b = 64}}$
dividing by 4 on both side
⇒${\text{a - b = 16}}$…………. (2)
Using (2) in (1) we get,
⇒${{\text{a}}^2}{\text{ + (a - 16}}{{\text{)}}^2} = 640$ ( Using${\left( {{\text{a - b}}} \right)^2} = \left( {{{\text{a}}^2} + {{\text{b}}^2} - 2{\text{ab}}} \right)$
On squaring
⇒${{\text{a}}^2}{\text{ + }}{{\text{a}}^2} + 256 - 32{\text{a}} = 640$
⇒${\text{2}}{{\text{a}}^2} - 32{\text{a}} = 640 - 256$
⇒${\text{2}}{{\text{a}}^2} - 32{\text{a}} = 384$
⇒${\text{2}}{{\text{a}}^2} - 32{\text{a - }}384 = 0$
dividing by 2 on both side
⇒${{\text{a}}^2} - 16{\text{a - 192}} = 0$
We have, a quadratic equation ${{\text{a}}^2} - 16{\text{a - 192}} = 0$
Sridharacharya formula is actually the quadratic formula, used for finding the roots of a quadratic equation \[{\text{a}}{{\text{x}}^2}{\text{ + bx + c = 0}}\] , where a not equal to 0 , & a, b, c are real coefficients of the equation \[{\text{a}}{{\text{x}}^2}{\text{ + bx + c = 0}}\]
Being quadratic it has 2 roots.
⇒X = $\dfrac{{\left( { - {\text{b + }}\sqrt {{{\text{b}}^2} - 4{\text{ac}}} } \right)}}{{2{\text{a}}}}{\text{ & }}\dfrac{{\left( { - {\text{b - }}\sqrt {{{\text{b}}^2} - 4{\text{ac}}} } \right)}}{{2{\text{a}}}}$ (1)

On comparing the given equation ${{\text{a}}^2} - 16{\text{a - 192}} = 0$ with the general quadratic equation \[{\text{a}}{{\text{x}}^2}{\text{ + bx + c = 0}}\] we got values of coefficients a = 1, b = -16, c = 192
On putting the value of coefficients a, b, c in equation (1)
⇒${\text{a = }}\dfrac{{\left( { - ( - 16){\text{ + }}\sqrt {{{( - 16)}^2} - 4 \times (1) \times ( - 192)} } \right)}}{{2 \times 1}}{\text{ & }}\dfrac{{\left( { - ( - 16){\text{ - }}\sqrt {{{( - 16)}^2} - 4 \times (1) \times ( - 192)} } \right)}}{{2 \times 1}}$
⇒${\text{a = }}\dfrac{{\left( {{\text{16 + }}\sqrt {1024} } \right)}}{2}{\text{ & }}\dfrac{{\left( {{\text{16 - }}\sqrt {1024} } \right)}}{2}$
⇒${\text{a = }}\dfrac{{\left( {{\text{16 + 32}}} \right)}}{2}{\text{ & }}\dfrac{{\left( {{\text{16 - 32}}} \right)}}{2}$
⇒${\text{a = }}\dfrac{{\left( {48} \right)}}{2}{\text{ & }}\dfrac{{\left( {{\text{ - 16}}} \right)}}{2}$
⇒${\text{a = 24 & - 8}}$
Since a is the side of a square it can’t be negative so a=24
Now on putting a =24 in equation (1)
⇒${\text{24 - b = 16}}$
⇒b=8.
So the sides of the squares are 24 m and 8 m.

Note: In this particular type of question, the key concept is to remember the formulas of area and perimeter of a square, and arrange in such a way so that we can represent the following problem situation in the form of a quadratic equation. And roots of quadratic equations will give our desired results.