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
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.
Area of the first square = a²sqm
Area of the second square = A² sqm
Their perimeters would be 4a and 4A respectively.
Given ,
4A - 4a = 24
A - a = 6 -- (1)
A² + a² = 468 --(2)
From (1) a = a + 6
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 12 m.
The side of the second square is 18 m.