Question

sum of the areas of two squares is 468 metres square if the difference of their perimeter is 24 metre find the side of the two square​

Answer 1

Answer:

Let the sides of first and second square be X and Y .

Area of first square = (X)²

And,

Area of 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

and,

Side of second square = Y = 12 m.

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Answer 2

$\huge\rm\underline\pink{GIVEN:-}$

• Sum of areas of two squares = 468m²
• Difference in their perimeter = 24 m

$\huge\rm\underline\pink{To\:FIND:-}$

• Side of two squares

$\huge\rm\underline\pink{SOLUTION:-}$

Let the side of one square be S₁

and Other side be S₂

$\large\rm\bold{\boxed{Area\:of\:square\:=\:(Side)^2}}$

$\large\rm{\implies{(S_1)^2+(S_2)^2\:=\:468}}$

$\large\rm{\implies{(S_1)\:=\:\sqrt{(468)-(S_2)^2}\rightarrow\:eq(1)}}$

$\large\rm\bold{\boxed{Perimeter\:of\:square\:=\:4(side)}}$

$\large\rm{\implies{4(S_1)+4(S_2)\:=\:24}}$

$\large\rm{\implies{S_1+S_2\:=\:6}}$

$\large\rm{\implies{S_1\:=\:6-S_2\rightarrow\:eq(2)}}$

Since LHS is equal RHS can be equated ,

$\large\rm{\implies{6-S_2\:=\:\sqrt{(468)-(S_2)^2}}}$

Squaring on both sides ,

$\large\rm{\implies{(6-S_2)^2\:=\:468-(S_2)^2}}$

$\large\rm\bold{\boxed{(a-b)^2\:=\:a^2+b^2-2ab}}$

$\large\rm{\implies{36+(S_2)^2-12S_2\:=\:468-(S_2)^2}}$

$\large\rm{\implies{36-12S_2\:=\:468+2(S_2)^2}}$

$\large\rm{\implies{-2(S_2)^2-12S_2-432\:=\:0}}$

$\large\rm{\implies{2(S_2)^2-12S-432}}$

$\large\rm{\implies{2(S_2)^2-36S+24S-432\:=\:0}}$

$\large\rm{\implies{2S_2(S_2-18)+24(S_2-18)}}$

$\large\rm{\implies{(S_2-18)(2S_2+24)\:=\:0}}$

$\large\rm{\implies{S_2\:=\:18\:(or)\:S_2\:=\frac{-24}{2}\:=\:-12}}$

Side of a square can't be negative so S₂ = 18 m

From eq(1) ,

$\large\rm{S_1\:=\:\sqrt{(468)-(18)^2}\:=\:\sqrt{468-324}\:=\:12\:m}$

∴ Sides of squares are 18 m , 12 m