right triangle JKL; <JKL right; KH perp. to JL; JL=20, JH=8  TOPIC_SOLVED

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right triangle JKL; <JKL right; KH perp. to JL; JL=20, JH=8

Postby Motherof8 on Sat Feb 12, 2011 1:31 am

JKL is a right triangle. Angle JKL is the right angle. Line KH is perpendicular to JL. If JH = 8, and JL = 20, we are told to find KH.

The book gives the answer 4 x the square root of 6. When I worked out the problem, I got a different answer. Could you tell me how they arrived at this answer and draw a diagram for me if you could? Thanks.
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Postby stapel_eliz on Sat Feb 12, 2011 12:07 pm

Motherof8 wrote: The book gives the answer 4 x the square root of 6. When I worked out the problem, I got a different answer. Could you tell me how they arrived at this answer and draw a diagram for me if you could?

Unless the solution manual shows the steps the solution author used, there is no way to know that for sure, so this question is not answerable. Sorry.

To draw the diagram, just draw the triangle with the listed vertices (corners), and label the listed angle and sides with the listed values.

Please show your solution method, so we can try to "see" where you may have gone astray. Thank you! :wink:
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Re: right triangle JKL; <JKL right; KH perp. to JL; JL=20, J

Postby Motherof8 on Sun Feb 13, 2011 8:08 pm

I drew a triangle where J was the right angle, K was the angle across from it, and L was the angle at the top. Line KH is drawn from angle K to the middle of line JL. JH is obviously a leg of another right angle,and JL is the side opposite the hypotenuse. JH would be the hypotenuse of the smaller right triangle. How would I find the hypotenuse of the larger triangle and and the smaller one using only these facts? If you know the measure of the hypotenuse and the hypotenuse segment, you can find the measure of the legs by cross multiplying. When I do that with 20 and 8, I get 4 times the square root of ten. I've tried using Pythagorean triples, 45 and 45 degree angles, and 30 60 degree angles and I still can't get the answer.
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  TOPIC_SOLVED

Postby stapel_eliz on Sun Feb 13, 2011 10:20 pm

Motherof8 wrote:I drew a triangle where J was the right angle, K was the angle across from it....

The only thing possible "across" from the right angle is the hypotenuse (which is also the longest side).

If the right angle is at vertex J, it then follows that KL is the hypotenuse. It would appear, since side KH is given (where H is not a vertex of the triangle), that H is on side KL and is the altitude of JKL.

Use the lengths of the two given sides (one leg and the hypotenuse) of the triangle JHL to find the length of the third side (the other leg) of JHL, which is the length of HL.

You are given the length of one leg of JHK. Label the other two sides with variables. Use these variables to create two equations, based on the Pythagorean Theorem, for triangles JHK and JKL.

This will give you two equations in two unknowns. Solve for the variables.

If you get stuck, please reply showing your work. Thank you. :wink:
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