#Those who'll solve this I will agree on your intelligence.
The Perimeter of a square S1 is 12 more than perimeter of square S2. If the area of Sl is 3 times the area of S2 minus 11,
then, find the perimeters of Sl&S2?
Answers
Answer:
The perimeter of the square s1 is = 4x = 32 units
The perimeter of the square s2 is = 4y = 20 units
Step-by-step explanation:
let the side of square s1 is = x
then perimeter = 4x
and the side of square s2 is = y
then perimeter = 4y
According to given condition
4x = 4y + 12
divide whole equation by 4 we have
x = y +3 ______ ( 1 )
we have given that the, the area of s1 is 3 times the area of s2 minus 11
Therefore
x² = 3 ( y² ) -11
x² = 3y²- 11
( y + 3 ) ² = 3y² -11 { from =n(1) }
y² + 9 + 6y = 3y² -11
[ (a+b)²= a²+b²+2ab ]
2y² = 20+6y
y² = 10 +3y
y²-3y-10 =0
y²- 5y + 2y - 10 = 0
y ( y - 5 ) + 2 ( y - 5 ) = 0
( y + 2 ) ( y - 5 ) = 0
y = -2 or y = 5
because y can't be negative y = 5
therefore from =n(1)
x = 5 + 3
x = 8
hence the perimeters of squares s1 and s2 is 32 units and 20 units