Sum of the area of two square is 468 meter square. If there differences of their parameters is 24m .find its side.
Answers
Let two side be s and s' m.
A/Q
s^2+s'^2=468m^2. -----(i)
4s-4s'=24m
4(s-s')=24
s-s'=6
s=6+s'. ------(ii)
Putting value of s in I from ii.
(6+s')^2+s'^2=468
36+s'^2+12s'+s'^2=468
2s'^2+12s'-432=0
2(s'^2+6s'-216)=0
s'^2+18s'-12s'-216=0
s'(s'+18)-12(s'+18)=06+12
(s'+18)(s'-12)=0
,', s'=12 m
And, s = 6+12 =18m
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 .