Question

Sum of the areas of two squares is 468 m. If the difference of their
parameters is 24 m, find the sides of two squares.​

Answer 1

Answer:

The sides are 12 m and 18 m.

Step-by-step explanation:

Given :

Sum of the areas = 468 m²

Difference between their perimeters = 24 m

To find :

The sides of each square

Solution :

Let the sides be -

• One as x
• Second as y

★ Sum of their areas = 468 m²

$\boxed{\sf{Area=Side \times Side}}$

$\sf{\longrightarrow} \: (x \times x) + (y \times y) = 468 \\ \\ \sf{\longrightarrow} \:{x}^{2} + {y}^{2} = 468 ----\rm{\gray{(1)}}$

$\rule{300}{1.5}$

★ Difference between the perimeter = 24 m

$\boxed{\sf{Perimeter = Side \times 4}}$

$\sf{\longrightarrow} \:(4 \times x) - (4 - y) = 24\\ \\ \sf{\longrightarrow} \:4x - 4y = 24 \\ \\ \sf{\longrightarrow} \:x - y = 6 \\ \\ \sf{\longrightarrow} \:x = 6 + y----\rm{\gray{(2)}}$

$\rule{300}{1.5}$

★ Substitute the value of 'x' from equation (2) in equation (1) -

$\sf{\longrightarrow} \: {x}^{2} + {y}^{2} = 468 \\ \\ \sf{\longrightarrow} \:(y + 6)^{2} + {y}^{2} = 468 \\ \\ \sf{\longrightarrow} \: {y}^{2} + {6}^{2} + 2 \times y \times 6 + {y}^{2} = 468 \\ \\ \sf{\longrightarrow} \:2 {y}^{2} + 36 + 12y = 468 \\ \\ \sf{\longrightarrow} \: {y}^{2} + 36 + 12y - 468 = 0 \\ \\ \sf{\longrightarrow} \: {y}^{2} + 12y - 432 = 0 \\ \\ \sf{\longrightarrow} \: {y}^{2} + 6y - 216 = 0 \\ \\ \sf{\longrightarrow} \: {y}^{2} + 18y - 12y - 216 = 0 \\ \\ \sf{\longrightarrow} \:y(y + 18) - 12(y + 18) = 0 \\ \\ \sf{\longrightarrow} \:(y + 18)(y - 12) = 0 \\ \\ \sf{\longrightarrow} \:y + 18 = 0 \: \: or \: \: y - 12 = 0 \\ \\ \sf{\longrightarrow} \:y = - 18 \: \: or \: \: y = 12$

Side can't be negative, hence the side is 12 m

$\rule{300}{1.5}$

★ The side of the second Square -

$\sf{\longrightarrow} \:x = y + 6 \\ \\ \sf{\longrightarrow} \:x = 12 + 6 \\ \\ \sf{\longrightarrow} \:x = 18$

Side of the second Square = 18 m

$\therefore$ The sides are 12 m and 18 m.

Answer 2

$\bf{ \underline{ \: \: Given :}}$

• Sum of the area of two squares = 468 m²
• Difference of their perimeter = 24 m

$\bf{ \underline{ \: \: To Find :}}$

• Side of both squares

Let the side of one square is x

and that of second square is y

★ Sum of the area of two squares = 468 m²

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \sf{ \underline{Area = Side \times Side}}}$

$\rightarrow \: \tt{(x \times x) + (y \times y) = 468} \\ \\ \rightarrow \: \tt{ {x}^{2} + {y}^{2} = 468 - - - - - (1) }$

★ Difference of their perimeter = 24 m

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \sf{ \underline{Perimeter = 4 \times Side}}}$

$\rightarrow \: \tt{(4 \times x) - (4 \times y) = 24} \\ \\ \rightarrow \: \tt{4x - 4y = 24} \\ \\ \rightarrow \: \tt{ \cancel{ \:4}(x - y) = \cancel{24}} \\ \\ \rightarrow \: \tt{x - y = 6} \\ \\ \rightarrow \: \tt{x = 6 + y - - - - - (2)}$

★ Putting the value of x in equation (1)

$\rightarrow \: \tt{ {x}^{2} + {y}^{2} = 468 } \\ \\ \rightarrow \: \tt{ {(y + 6)}^{2} + {y}^{2} = 468 } \\ \\ \rightarrow \: \tt{ {y}^{2} + 36 + 12y + {y}^{2} = 468 }\\ \\ \rightarrow \: \tt{2 {y}^{2} + 12y + 36 - 468 = 0 }\\ \\ \rightarrow \: \tt{ 2{y}^{2} + 12y - 432 = 0}\\ \\ \rightarrow \: \tt{ {y}^{2} + 6y - 216 = 0}\\ \\ \rightarrow \: \tt{ {y}^{2} + (18 - 12)y - 216 = 0 }\\ \\ \rightarrow \: \tt{ {y}^{2} + 18y - 12y - 216 = 0}\\ \\ \rightarrow \: \tt{ y(y + 18) - 12(y + 18) = 0}\\ \\ \rightarrow \: \tt{(y + 18)(y - 12) = 0}\\ \\ \rightarrow \: \tt{y + 18 = 0 \: \: or \: \: y - 12 = 0}\\ \\ \rightarrow \: \tt{y = - 18 \: \: or \: \: y = 12}$

Side of any square can not be negative, so the side is 12 m...

★ Putting ( y = 12 ) in equation (2)

$\rightarrow \: \tt{x = 6 + y}\\ \\ \rightarrow \: \tt{x = 12 + 6}\\ \\ \rightarrow \: \tt{x = 18}$

Side of the second square is 18 m...

$\blue{\boxed{\text{\small{\red{So, the sides of both the squares are 12 m and 18 m respectively}}}}}$