sum of area of two circle is 468 metre square if the difference of their perimeter is 24 find the sides of two square
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Answers
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.
Answer:
18 m and 12 m
Step-by-step explanation:
Here, there are areas of two circles.
Let us consider the larger circle side as x.
Let us consider the smaller circle side as y.
Sum of areas of two circle is 468 m^2.
=> x^2 + y^2 = 468 ------ (1)
Difference of their perimeters = 24.
=> 4x - 4y = 24
=> x - y = 6
=> x = 6 + y ----- (2)
Place x in (1)
=> (6 + y)^2 + y^2 = 468
=> 36 + y^2 + 12y + y^2 = 468
=> 2y^2 + 12y + 36 = 468
=> 2y^2 + 12y = 432
=> 2y^2 + 12y - 432 = 0
=> y^2 + 6y - 216 = 0
=> y^2 + 18y - 12y -216 = 0
=> y(y + 18) - 12(y + 18) = 0
=> y = 12,-18[It wont be negative]
=> y = 12
Place y = 12 in (2).
=> x = 6 + y
=> x = 6 + 12
=> x = 18 m
Hence, sides are 18 m and 12 m.
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