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 .
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Its the Last one !!​

Answer 1

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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)

$(X)² + (Y)² = 468$

$= >  (6+Y)² + (Y)² = 468$

$= > (6)² + (Y)² + 2 × 6 × Y + (Y)² = 468$

$= > 36 + Y² + 12Y + Y² = 468$

$= > 2( Y² + 6Y – 216) = 0$

$= > Y² + 6Y – 216 = 0$

$= > Y² + 18Y – 12Y -216 = 0$

$= > Y(Y+18) – 12(Y+18) = 0$

$= >(Y+18) (Y-12) = 0$

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

Answer:

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