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area of a triangle...

 
 
Reply Mon 21 Sep, 2009 06:20 pm
the area of a triangle is the 5th root of 243x^5 y^10 square units. if the length of a side of the triangle is the 5th root of x^5, express the length of the altitude to that side in simplest radical form.
i know that the 5th root of x^5 is x...but where do i go from there?
 
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Reply Mon 21 Sep, 2009 06:28 pm
Let's look at the area term.

243x^5 y^10

As you said, the fifth root of x^5 is x. Did you know that the fifth root of y^10 is y^2? Here's the fun part: 243 = 3^5, so the fifth root of 3^5 is 3. That means the area is

3xy^2

You already figured out the length is x. The area of a triangle is 1/2 base x height, so plugging in,

A = 1/2 B x H
3xy^2 = 1/2 x H
6y^2 = H
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