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 Trigonometry Trigonometry Math Forum

 July 10th, 2018, 10:27 AM #1 Newbie   Joined: Jul 2018 From: Cornwall Posts: 2 Thanks: 0 Sides of a triangle using ratios and area Hello, I have a question that I cannot find the solution to and would be grateful for any help. A right-angled triangle has an area of 2.8m^2. The sides are in the ratio of 5:12:13 and the triangle stands on the shortest side. Calculate the following: a) the perpendicular height b) the hypotenuse again thank you for any help. Last edited by skipjack; July 10th, 2018 at 10:53 AM. July 10th, 2018, 10:53 AM   #2
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 Originally Posted by Jamtrehope Hello, I have a question that I cannot find the solution to and would be grateful for any help. A right-angled triangle has an area of 2.8m^2. The sides are in the ratio of 5:12:13 and the triangle stands on the shortest side. Calculate the following: a) the perpendicular height b) the hypotenuse again thank you for any help.
You first need to know the formula for area of a triangle: $\displaystyle A = \frac{1}{2} b h$. Now you need an expression for the actual triangle sides. They are all in a ratio "x". Then b = 5x and h = 12x. Thus we know that
$\displaystyle 2.8 = \frac{1}{2} (5x)(12x)$

Solve this for x and then you can say what b = 5x and h = 12x are.

Can you finish?

-Dan

Last edited by skipjack; July 10th, 2018 at 10:59 AM. July 10th, 2018, 10:57 AM #3 Global Moderator   Joined: Dec 2006 Posts: 21,020 Thanks: 2255 If the "perpendicular height" is 12y, the other sides have lengths 5y and 13y. What is the triangle's area in terms of y? Oh, topsquark posted much the same while I was typing! Thanks from topsquark July 10th, 2018, 11:19 AM #4 Newbie   Joined: Jul 2018 From: Cornwall Posts: 2 Thanks: 0 Thank you for your quick response. I can sleep easy now. Last edited by skipjack; July 10th, 2018 at 11:48 AM. Tags area, ratios, sides, triangle hilbert cube is compact proof

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