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 .
Answers
Answer:
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 = 244X–4Y=24
4(X-Y) = 244(X−Y)=24
X – Y = 24/4X–Y=24/4
X – Y = 6X–Y=6
X=6+Y———(3)
Putting the value of X in equation (1)
(X)² + (Y)² = 468(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 = 18X=6+Y=6+12=18
Side of first square = X = 18 m
Side of second square = Y = 12 m.