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
Answer:
→ 18m and 12 m .
Step-by-step explanation:
Let the sides of two squares be x m and y m respectively .
Case 1 .
→ Sum of the areas of two squares is 468 m² .
A/Q,
∵ x² + y² = 468 . ...........(1) .
[ area of square = side² . ]
Case 2 .
→ The difference of their perimeters is 24 m .
A/Q,
∵ 4x - 4y = 24 .
[ ∵ Perimeter of square = 4 × side . ]
⇒ 4( x - y ) = 24 .
⇒ x - y = 24/4.
⇒ x - y = 6 .
∴ y = x - 6 ..........(2) .
From equation (1) and (2) , we get
∵ x² + ( x - 6 )² = 468 .
⇒ x² + x² - 12x + 36 = 468 .
⇒ 2x² - 12x + 36 - 468 = 0 .
⇒ 2x² - 12x - 432 = 0 .
⇒ 2( x² - 6x - 216 ) = 0 .
⇒ x² - 6x - 216 = 0 .
⇒ x² - 18x + 12x - 216 = 0 .
⇒ x( x - 18 ) + 12( x - 18 ) = 0 .
⇒ ( x + 12 ) ( x - 18 ) = 0 .
⇒ x + 12 = 0 and x - 18 = 0 .
⇒ x = - 12m [ rejected ] . and x = 18m .
∴ x = 18 m .
Put the value of 'x' in equation (2), we get
∵ y = x - 6 .
⇒ y = 18 - 6 .
∴ y = 12 m .
Hence, sides of two squares are 18m and 12m respectively .
Answer:
Step-by-step explanation:
let a^2 and b^2 be the area of the 2 squares and 4a and 4b be the perimeters
a^2+b^2=468-------------(1)
4a-4b=24
a-b=6
a=6+b ---------(2) (bringing 4 to rhs side)
substituting 2 in 1 we get
(6+b)^2+b^2=468
36+b^2+12b+b^2=468 [ (a+b)^2 formula]
2b^2+12b-432=0
b^2+6b-216=0
solving the equation we get 18 and -12
so one side is 18 and the other one is 12 this happens when we substitute the value of 18 in eqn 2
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