Question

Sum of the areas of two squares is 468 m square if the difference of their perimeter is 24 m find the sides of the two square .​

Answer 1

$\huge{\color{t}{\textsf{\textbf {\underline{\underline{ Anѕwєr : }}}}}}$

Let the sides of the first and second square be X and Y . Area of the first square = (X)²

Area of the second square = (Y)²

• According to question, (X)² + (Y)² = 468 m² ——(1).

Perimeter of first square = 4 × X and Perimeter of second square = 4 × Y

According to question,

• 4X – 4Y = 24 ——–(2)

From equation (2) we get,

4X – 4Y = 24, 4(X-Y) = 24

X – Y = 24/4 , X – Y = 6

• X = 6+Y ———(3)

Putting the value of X in equation (1)

(X)² + (Y)² = 468, (6+Y)² + (Y)² = 468

(6)² + (Y)² + 2 × 6 × Y + (Y)² = 468

36 + Y² + 12Y + Y² = 468

2Y² + 12Y – 468 +36 = 0

2Y² + 12Y -432 = 0

2( Y² + 6Y – 216) = 0

Y² + 6Y – 216 = 0

Y² + 18Y – 12Y -216 = 0

Y(Y+18) – 12(Y+18) = 0 (Y+18) (Y-12) = 0

(Y+18) = 0 Or (Y-12) = 0 Y = -18 OR Y = 12

Putting Y = 12 in EQUATION (3)

X = 6+Y = 6+12 = 18

• Side of first square = X = 18 m

• Side of second square = Y = 12 m.

Answer 2

$\fbox\pink{★ Solution :}$

Let the side of the first square be $\bold{'a'm}$ and that of the second be $\bold{`A` m}$

• Area of the first square = a² sqm

• Area of the second square = A² sqm

Their perimeters would be $\bold{4a}$ and $\bold{4A}$ respectively.

Given ,

4A - 4a = 24

A - a = 6 -- (1)

A² + a² = 468 --(2)

$\bold{From\: (1) \:a\: = a\: +\: 6}$

$\bold{Substituting \:for \:A \:in\: (2), \:we\: get}$

→ (a +6)² + a² = 468

→ a² + 12a + 36 + a² = 468

→ 2a² + 12a + 36 = 468

→ a² + 6a + 18 = 234

→ a² + 6a - 216 = 0

→ a² + 18a - 12a - 216 = 0

→ a(a+18) − 12(a+18) = 0

→ (a−12) (a+18) = 0

→ a = 12, -18

• So, the side of the first square is $\bold{12 m.}$

• The side of the second square is $\bold{18 m.}$