SOLUTION: the width of a rectangle is x centimetres the length of a diagonal of the rectangel is 12 cm. the perimeter of the rectangle is 28cm. fine the value of x . give your values correc

Algebra ->  Surface-area -> SOLUTION: the width of a rectangle is x centimetres the length of a diagonal of the rectangel is 12 cm. the perimeter of the rectangle is 28cm. fine the value of x . give your values correc      Log On


   

Question 1075802: the width of a rectangle is x centimetres the length of a diagonal of the rectangel is 12 cm. the perimeter of the rectangle is 28cm. fine the value of x . give your values correct to 3 significant figures.
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Found 2 solutions by josgarithmetic, jorel1380:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Length is sqrt%2812%5E2-x%5E2%29 from Pythagorean Theorem formula.


If L is length,
2L%2B2x=28
L%2Bx=14
L=14-x
Find L^2.
L%5E2=14%5E2-28x%2Bx%5E2
.
L%5E2=144-x%5E2

Equate the two expressions of L^2.
x%5E2-28x%2B196=-x%5E2%2B144
2x%5E2-28x%2B196-144=0
x%5E2-14x%2B26=0
you can finish from here.

Answer by jorel1380(3719) About Me  (Show Source):
You can put this solution on YOUR website!
If the width is x, then 2x+2*length=28
2*length=28-2x
length=14-x
So:
x²+(14-x)²=12²=144
x²+x²-28x+196-144=0
2x²-28x+52=0
x²-14x+26=0
x²-14x+49=49-26
(x-7)²=23
x-7=+/-√23
x=7+/-√23=11.796 or 2.204 ☺☺☺☺