I'm not sure if this is the easiest way of doing this, but it surely work.
Let the base of the triangle be AB, and let CH be the height. Just for reference, we have
Moreover, let CH=y and BC=z
Now, AHC, CHB and ABC are all right triangles. If we write the pythagorean theorem for each of them, we have the following system
If we solve the first two equations for y squared, we have
And we can deduce
So that the third equation becomes
(we can't accept the negative root because negative lengths make no sense)
Alright, so we start off my subtracting 4a from both sides, separating the x and getting 10-4a=2x. Next, we divide both sides by 2, getting that 5-2a=X
Answer:
This is a simple question for simple answer
9/3=3
3 seeds
Step-by-step explanation:
Step-by-step explanation:
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Answer:
No
Step-by-step explanation:
We can check with the Pythagorean Theorem.
a^2 + b^2 = c^2
a and b are always the smaller sides, the legs
c is the longer side, the hypotenuse
3^2 + 3^2 > 6^2
9 + 9 > 36
18 < 36
18 is not greater than 36, therefore this triangle can not have the side lengths 3, 3, and 6.
