Sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, find the sides of the two squares.
Answers
Step-by-step explanation:
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.
Answer:
Let the side of the first square be 'a'm and that of the second be
′
A
′
m.
Area of the first square =a
2
sq m.
Area of the second square =A
2
sq m.
Their perimeters would be 4a and 4A respectively.
Given 4A−4a=24
A−a=6 --(1)
A
2
+a
2
=468 --(2)
From (1), A=a+6
Substituting for A in (2), we get
(a+6)
2
+a
2
=468
a
2
+12a+36+a
2
=468
2a
2
+12a+36=468
a
2
+6a+18=234
a
2
+6a−216=0
a
2
+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. and the side of the second square is 18 m.