SOLUTION: I am not understanding how to start this word problem. We are supposed to be using quadratic equations. The problem is..... The length of a rectangle is 1cm more than tw

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Question 686696: I am not understanding how to start this word problem. We are supposed to be using quadratic equations. The problem is.....

The length of a rectangle is 1cm more than twice it's width and 2cm less than the length of the diagonal of the rectangle. Find the dimensions of the rectangle.
I think the diagonal part is confusing me because we have never solved a problem with diagonal before.
Found 2 solutions by lwsshak3, ankor@dixie-net.com:
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
The length of a rectangle is 1cm more than twice it's width and 2cm less than the length of the diagonal of the rectangle. Find the dimensions of the rectangle.
**
let x=width
length=2x+1
diagonal=√(width^2+length^2) (by the pythagorean theorem)
=√[(x^2+(2x+1)^2]
=√[x^2+4x^2+4x+1]
=√(5x^2+4x+1)
..
(2x+1)+2=diagonal
2x+3=√(5x^2+4x+1)
square both sides
4x^2+12x+9=5x^2+4x+1
x^2-8x-8=0
solve for x by following quadratic formula:
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
a=1, b=-8, c=-8
ans:
x≈8.9
2x+1≈18.8
width≈8.9 cm
length≈18.8 cm

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
The length of a rectangle is 1cm more than twice it's width and 2cm less than the length of the diagonal of the rectangle.
Find the dimensions of the rectangle.
:
Let a = the width
Let b = the length
Let c = the diagonal
;
Write an equation for each statement, get each equation in terms of a:
"The length of a rectangle is 1cm more than twice it's width"
b = 2a + 1
:
"the length is 2cm less than the length of the diagonal of the rectangle."
b = c - 2
replace b with (2a+1)
2a + 1 = c - 2
2a + 1 + 2 = c
c = 2a + 3
:
The diagonal is the hypotenuse (c) of the rectangle where
a^2 + b^2 = c^2
Replace b with (2a+1) and replace c with (2a+3)
a^2 + (2a+1)^2 = (2a+3)^2
FOIL
a^2 + 4a^2 + 2a + 2a + 1 = 4a^2 + 6a + 6a + 9
Which is
5a^2 + 4a + 1 = 4a^2 + 12a + 9
Combine like terms on the left
5a^2 - 4a^2 + 4a - 12a + 1 - 9 = 0
a^2 - 8a - 8 = 0
Solve this quadratic equation using the quadratic formula
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
where x = a; b = -8; c = -8
a+=+%28-%28-8%29+%2B-+sqrt%28-8%5E2-4%2A1%2A-8+%29%29%2F%282%2A1%29+
a+=+%288+%2B-+sqrt%2864%2B32+%29%29%2F2+
a+=+%288+%2B-+sqrt%2896%29%29%2F2+
two solutions but we only need the positive solution which is approx
a = %288+%2B+9.8%29%2F2
a = 17.8%2F2
a = 8.9 cm is the width
then
b = 2(8.9) + 1
b = 18.8 cm is the length
:
:
Check this by finding the hypotenuse
c = 2(8.9) + 3
c = 20.8
:
Enter into calc sqrt%288.9%5E2+%2B+18.8%5E2%29 results: 20.800 confirms our solution
:
Find the dimensions of the rectangle. 18.8 by 8.9