plzzz solve in copy , Sum of the areas of 2 squares is 468 sqm,if the difference between their perimeters is 24m.find the sides of the square ♥️
Answers
let sides of squares be a and b
Sum of areas = 468
a^2 + b^2 = 468
Difference between perimeters = 24
4( a-b)= 24 , a>b
a- b= 24/4 = 6
a= 6 + b
( 6+ b)^2 + b^2 = 468
36 + b^2 + 12b + b^2 -468= 0
2b^2 + 12b - 432 = 0
b^2 + 6b - 216= 0
b^2 +18 b- 12b -216= 0
b( b+18) -12( b+18)= 0
b-12)(b+18)= 0
b= 12,-18
neglect -18
So b= 12
a= 6+b= 6+12= 18
Step-by-step explanation:
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 .