Area of a Triangle from 3 sides

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SUMMARY

The discussion centers on calculating the area of a triangle using the Pythagorean theorem and special Pythagorean triples. Carla suggests drawing an altitude from the top vertex to the base, which measures 21 units, effectively dividing the triangle into two right triangles. The calculations reveal that the right triangles satisfy the triples (17, h, x) and (10, 21-x, h), leading to the conclusion that the area of the triangle is 84 square units.

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Carla1985
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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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