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December 8, 2016
Posted by **Nabil ** on Monday, October 10, 2011 at 12:06am.

Each piece bent into a square.

Find the length of the two pieces so the sum of the areas of the two squares in 10 square inches.

- algebra -
**Anonymous**, Monday, October 10, 2011 at 2:01amL = Length of a wire

x = Length of the first pice

y = Length of the cecond pice

x / 4 = Length of the first square

y / 4 = Length of the second square

L = 16 in

16 = x + y

y = 16 - x

Area of the first pice = ( x / 4 ) ^ 2

Area of the second pice = ( y / 4 ) ^ 2

( x / 4 ) ^ 2 + ( y / 4 ) ^ 2 = 10

x ^ 2 / 16 + y ^ 2 / 16 = 10

( x ^ 2 + y ^ 2 ) / 16 = 10 Multiply both sides with 16

x ^ 2 + y ^ 2 = 160

x ^ 2 + ( 16 - x ) ^ 2 = 160

( Remark: ( 16 - x ) ^ 2 = 16 ^ 2 - 32 x + x ^ 2

x ^ 2 + 16 ^ 2 - 32 x + x ^ 2 = 160

2 x ^ 2 + 256 - 32 x = 160

2 x ^ 2 + 256 - 160 - 32 x = 0

2 x ^ 2 + 96 - 32 x = 0

2 x ^ 2 - 32 x + 96 = 0 Divide both sides with 2

x ^ 2 - 16 x + 48 = 0

The exact solutions are:

x = 12

and

x = 4

Length of a wire = 16 in

y = 16 - x

When: x = 12 ; y = 16 - 12 = 4

When x = 4 ; y = 16 - 4 = 12

Length of the two pieces :

12 in

and

4 in

Proof:

( 12 / 4 ) ^ 2 + ( 4 / 4 ) ^ 2 = 10

3 ^ 2 + 1 ^ 2 = 10

9 + 1 = 10 in ^ 2

P.S.

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x ^ 2 - 16 x + 48 = 0

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x ^ 2 - 16 x + 48 = 0

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You wil see solution step-by-step - algebra -
**Anonymous**, Sunday, December 14, 2014 at 7:28pmeff