SOLUTION: The length of a rectangle is 1 cm longer than the width. If the DIAGONAL of the rectangle is 4 cm, what are the dimensions (of the length and width) of the rectangle?

Algebra ->  Rectangles -> SOLUTION: The length of a rectangle is 1 cm longer than the width. If the DIAGONAL of the rectangle is 4 cm, what are the dimensions (of the length and width) of the rectangle?      Log On


   

Question 101528: The length of a rectangle is
1 cm longer than the width.
If the DIAGONAL of the rectangle is 4 cm,
what are the dimensions
(of the length and width)
of the rectangle??
I can get my answer of
2.284 cm and 3.284 cm
but...
I need to show step by step how I got my answer,
this is where I keep writing it out wrong.
Please help...
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
Let say that:
Length=a
Width=b
DIAGONAL=d
Given:
a=b+1cm
d=4cm
you have to use Pythagorean theorem
a%5E2+%2B+b%5E2+=+d%5E2
%28b%2B1cm%29%5E2+%2B+b%5E2+=+4%5E2
b%5E2+%2B+2b%2B1+%2B+b%5E2=+16
2b%5E2+%2B2b-16%2B1+=+0
2b%5E2+%2B2b-15+=+0
2b%5E2+%2B2b-15+=0+ (now we can divide this by 2)
b%5E2+%2Bb-15%2F2+=0+
use quadratic formula
x=-b+%2B-+sqrt+%28b%5E2+-4%2Aa%2Ac+%29%29+%2F+%282%2Aa%29 where b=x1 and x1%3E0

or do it this way:
b%5E2+%2Bb-15%2F2+=0+
b%5E2+%2B+b+=+15%2F2
b%5E+2+%2B+b+%2B1%2F4+-1%2F4=15%2F2
b%5E+2+%2B+b+%2B1%2F4+=1%2F4+%2B15%2F2+
%28b%2B+1%2F2%29%5E2+=+1%2F4+%2B15%2F2++
%28b%2B+1%2F2%29%5E2+=++31%2F4
b%2B1%2F2= + - sqrt+%2831%2F4+%29

b=+-sqrt%2831%2F4%29 -+1%2F2
since the length is positive, we will use only +sqrt31/4 which is 2.784
then
b+=2.784%961%2F2
b+=2.284+ then a+=2.284+%2B1=3.284
check:
a%5E2+%2B+b%5E2+=+d%5E2
3.284%5E2+%2B2.284%5E2+=4%5E2
10.779+%2B+5.217++=16
15.996+is approximately 16