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Let's say I have a right triangle (ABC) with two medians drawn A to the side of BC, which cuts it in half, and the median from BC that cuts side AC in half. Then, how would I find the length of the hypotenuse? Thank you! smiley

 Jun 26, 2019

Wait, I can label the bottom parts as 1/2a and 1/2a, and I can label the top parts as 1/2b and 1/2b.

There are also two triangles at the top, so I can use the Pythagorean Theorem? 

 Jun 26, 2019

Are those numbers 6 and 8?

Where do they belong?

 Jun 27, 2019

Ant, I assume we know the length of the medians....if so....


Call one median AD   and the other BE


Then,,,,we have two right triangles  ADC and BEC   


So.....this sets up a system of equations using the Pythagorean Theorem


[ (1/2)AC]^2 + [BC]^2  =  [BE]^2     ⇒   (1/4)AC^2  + BC^2  = BE^2    (1)

[ (AC]^2  +  [ [(1/2)BC]^2  = [ AD]^2  ⇒   AC^2  + (1/4)BC^2  = AD^2  (2)   


Multiply (1) by  -4       and we have


-AC^2   - 4BC^2  = - 4BE^2

AC^2  + (1/4)BC^2  = AD^2          add these


(-15/4)BC^2  =  -4BE^2  + AD^2


Since we assume that we know BE and AD, we can solve this for BC   and then use either (1) or (2) to find AC


Then.....just use thePythagorean Theorem again to find  AB




cool cool cool

 Jun 27, 2019

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