Question 60124This question is from textbook Elementry and Intermediate Algebra
: If the width of a rectangle is 1 inch less than its length and the diagonal is 1 inch longer than its length, find the length and the width.
This question is from textbook Elementry and Intermediate Algebra
Answer by ptaylor(2198)   (Show Source):
You can put this solution on YOUR website! Let L=length
then width (W)=L-1
and diagonal (D)=L+1
but the diagonal is actually the hypotenuse of a right triangle that has a length of L and a width of L-1
and Pythagoras tells us that D=sqrt{L^2+(L-1)^2}
Thus, the equation that we need to solve is:
sqrt{L^2+(L-1)^2}=L+1
Lets go ahead and solve it.
First, we will square both sides to get rid of the radical which gives us:
{L^2+(L-1)^2}=L^2+2L+1: Simplifying we get:
2L^2-2L+1=L^2+2L+1: and further simplifyiing we get:
L^2-4L=0 and since it's a quadratic, we have two solutions:
L=0 and L=4; obviously L has to be 4, therefore
Length (L)=4
Width (W)=3
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