MHB Area of a Triangle from 3 sides

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The discussion revolves around calculating the area of a triangle using the Pythagorean theorem, despite initial thoughts of using trigonometric methods. The user successfully divides the triangle into two right triangles by drawing an altitude, leading to the identification of Pythagorean triples. Through calculations, they derive the values of the base segments and height, ultimately finding the area to be 84. The exchange highlights the importance of understanding different methods for solving geometric problems. The solution effectively demonstrates the application of the Pythagorean theorem in this context.
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Can I have an opinion on this question, please? Personally I would use the cosine & sine rules to work out the angles then use trig to calculate the height. However, the question asks for Pythag to be used. Can someone please explain what method I should be using to answer this? Thanks

Thanks
Carla
 

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Draw an altitude from the top vertex to the base who length is 21. The altitude divdes the triangle into two right triangles satisfying special Pythagorean triples. Which Pythagorean triples do these triangles satisfy?
 
Euge said:
Draw an altitude from the top vertex to the base who length is 21. The altitude divdes the triangle into two right triangles satisfying special Pythagorean triples. Which Pythagorean triples do these triangles satisfy?

Ok, I think I've got it.

So I have 2 right angle triangles with triples $(17,h,x)$ and $(10, 21-x,h)$. From pythag I get:

$$h^2+x^2=289$$
$$h^2+x^2-42x=-341$$

Subbing in $h^2$ into the second equation gives me $x=15$ and $h=8$. This gives me an area of $84$.

Thanks for the help!
 
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