Question 31040
Hello!
There appears to be a typo in your question. You wrote "Find the length and width of the triangle", but it looks like you meant "rectangle" instead of triangle. I'll asume you meant that.

As in all word problems, the main difficulty is in "translating" the given data into algebraic terms.

Let's call X to the rectangle length, and Y to its width. We're told that:

- "The length of a rectangle is 2 cm less than twice the width". Twice the width is 2Y. Two less than twice the width is 2Y - 2. So we have: {{{X=2Y-2=2(Y-1)}}}

- "The area of the rectangle is 180 cm^2". The formula for the area of a rectangle is length times width (XY). So we get the equation {{{X*Y=180}}}

What we have here is a system of equations:
{{{system(X=2(Y-1),X*Y=180)}}}

Replacing the 1st equation into the second one, we get:
{{{2*(Y-1)*Y = 180}}}
{{{(Y-1)*Y = 180/2=90}}}
{{{Y^2-Y = 90}}}
{{{Y^2-Y-90=0}}}

So we're left with a quadratic equation, which can be solved with the standard procedure:
*[invoke solve_quadratic_equation 1, -1, -90]

The solutions are 10 and -9; but -9 clearly doesn't make sense (width can't be negative). So we conclude that the width of this rectangle is 10. Finally, we knew that {{{X*Y=180}}}. Plugging Y = 10, we get:
{{{10X=180}}}
{{{X = 18}}}

So the length is 18 cm and the width is 10 cm


I hope this helps!

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