Sum of the areas of two squares is 468m^2. If the difference of their perimeters is 24m. Find the sides of two squares.
Answers
Step-by-step explanation:
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.
SOLUTION:-
Given:
•Sum of the areas of two square is 468m².
•If the difference of their perimeter is 24m.
To find:
The sides of two square.
Explanation:
Let R & M be the sides of two squares.
•The area of first square= R²
•The area of second square= M²
&
•Perimeter of the first square=4R m
•Perimeter of the second square=4M m
According to the question:
R² + M² = 468..............(1)
So,
=) 4R + 4M= 24
=) R + M = 6
=) R = M-6..................(2)
Now,
Putting the value of R in equation (1), we get;
=) (6-M)² + M² = 468
=) 6² + M² -2×6×M + M² = 468
=) 36 +2M² - 12M= 468
=) 2M² -12M =468 -36
=) 2M² -12M= 432
=) 2M² - 12M -432=0
=) M² - 6M - 216 =0
=) M² +12M - 18M -216=0
=) M(M+12) -18(M+12) =0
=) (M+12)(M -18) =0
=) M+12=0 or M -18=0
=) M= -12 or M= 18
Since, negative value isn't possible.
So,
M= 18
•Side of the first square= 18cm.
•Side of the second square=R= M-6
=) R= 18 -6