Apiece of wire of length 60m is cut into two parts. each part is then bent to form a square. it is found that the combined area of the two squares is 120m2. find the sides of the two squares.

s1= 9.44, s2 = 5.56m

Step-by-step explanation:

P1 + P2 = 60m

A1 + A2 = 120 sq m

Let s be the length of 1 side of the square

P = 4s

A = s^2

4s1 + 4s2 = 60

Divide the whole equation by 4

s1 + s2 = 15 -eqn1

(s1)^2 + (s2)^2 = 120 -equation2

From equation 1

s1 = 15-s2

Substitute to equation 2

(15-s2)^2 +(s2)^2 = 120

225 -30s2 +(s2)^2 +(s2)^2 = 120

2(s2)^2 -30s2 +225-120 =0

2(s2)^2 -30s2 + 105 = 0

s2 = 9.44m and 5.56m

Therefore the sides of the 2 squares are 9.44m and 5.56m

10 m and 3 m

Step-by-step explanation:

If we make a square with the total length of the wire of length 52 m,

the perimeter, 4X = 52 m ,

so one side, X = 52/4 = 13 m

When we make two squares of sides y and z, the sum of perimeter will be 52 m.

ie, 4y + 4z = 52

4(y+z) = 52

y+z = 52/4 =13

Sum of their area = y² + z² = 109

So the sides of the two squares are such that their sum is 13 and sum og their squares is 109.

When we do trial and error we can find that 10 + 3 = 13

and 10² + 3² = 100 + 9  = 109

yes it simple we solve them