the sum of the area of two squares is 468 m^2 and the perimeters are 24 m. tell the sides of the squares?
Answers
Given,
Sum of the area of two squares is 468 m^2
Difference between the perimeter of these sqaures is 24 m.
To find ,
length of the sides of these sqaures.
Main solution :-
Let the side of 1st sqaures be equal to x metre.
Let the side of the second sqaure be equal to y metre.
So,
Area of sqaure 1 = x^2
Area of sqaure 2 = y^2
Perimeter of sqaure 1 = 4x
Perimeter of sqaure 2 = 4y
According to question,
x^2 + y^2 = 468 m^2 (i)
4x - 4y = 24m
To find, x and y
4x - 4y = 24
4(x - y) = 24
x -y = 24/4
x-y = 6 (ii)
x = 6 +y (iii)
Putting this in (i)
x^2 + y^2 = 468
(6+y)^2 +y^2 = 468
36 + y^2 + 12 y +y^2 = 468
2y^2 + 12y -432 = 0
2(y^2 +6y -216 ) = 0
y^2 +6y -216 = 0
y^2+18y-12y-216=0
y(y+18)-12(y+18)=0
(y+18)(y-12)=0
Now, y = -18 or y = +12
Since, length can't be negative. We will take only the positive value of y .
Now,
From equation (iii)
x = 6 + y
= 6 + 12
= 18
Answer :- Length of side of one sqaure is 18 m and length of side of other sqaure is 12 m
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 .