Question 238378: Hi,
Can you please help me with this problem?
Find the dimensions of a rectangle whose perimeter is 46m and whose area is 126m.
Also
The length of a rectangle is 8 cm greater then its width. Find the dimensions if its area is 105 cm
We need to solve it using quadratics and possibly using factoring!
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! Find the dimensions of a rectangle whose perimeter is 46m and whose area is 126m.
:
The perimeter equation:
2L + 2W = 126
Simplify divide by 2
L + W = 23
W = (23-L); use for substitution
:
The area
L * W = 126
Substitute (23-L) for W
L*(23-L) = 126
23L - L^2 = 126
Arrange as a quadratic equation
-L^2 + 23L - 126 = 0
Multiply by -1, easier to factor:
L^2 - 23L + 126 = 0
Factors to
(L-9)(L-14) = 0
Two solutions
L = 9, then W = 14
or
L = 14. then W = 9
:
Check: 2(14)+2(9) = 46
and area: 14 * 9 = 126
:
;
The length of a rectangle is 8 cm greater then its width. Find the dimensions if its area is 105 cm
:
Let x = the width
then
(x+8) = the length
:
Area
x(x+8) = 105
x^2 + 8x - 105 = 0; our old friend, the quadratic equation!
factors to
(x+15)(x-7) = 0
The positive solution is what we want here
x = 7 cm, the width
then
7 + 8 = 15 cm; the length
;
It this true?: 7 * 15 = 105 sq cm
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