extreme help needed (medians of a right triangle)

crazygirlpsf

New member
Joined
Apr 25, 2006
Messages
2
Here is the math problem:

The medians drawn from the vertices of the actue angles of a right triangle are 5 and sq root of 40. What is the length of the hypotenuse.



SO far
I know that you have to split the big triangle into to seperate triangles using the two given medians as hypotenuses for new trianlges. but once i have that i dont know how to solve the problem.I know you have to use pythagorean theorm but how can you when both a squared and b squared are unknown and you only know c squared?

help would be greatly appreciateddd
 

tkhunny

Moderator
Staff member
Joined
Apr 12, 2005
Messages
10,028
What's a Median?

Draw the '5' median to side 'a'.
Label side 'a' twice. The whole thing is 'a'. Each half is 'a/2'.
Draw the sqrt(40) median to side 'b'
Label side 'b' twice. The whole thing is 'b'. Each half is 'b/2'.

Then the Pythagorean Party!!

a^2 + b^2 = c^2
(a/2)^2 + b^2 = 5^2
a^2 + (b/2)^2 = 40

That should be enough.
 

crazygirlpsf

New member
Joined
Apr 25, 2006
Messages
2
so i dont understand how i solve for A or B when there are two variables
can anyone solve it out more for me?
 

stapel

Super Moderator
Staff member
Joined
Feb 4, 2004
Messages
15,946
crazygirlpsf said:
i dont understand how i solve...when there are two variables
Have you not learned how to solve systems of equations?

Thank you.

Eliz.
 

tkhunny

Moderator
Staff member
Joined
Apr 12, 2005
Messages
10,028
crazygirlpsf said:
so i dont understand how i solve for A or B when there are two variables
can anyone solve it out more for me?
Did you overlook 'c'?

I'll get you started along the road of "Substitution".

a^2 + b^2 = c^2
(a/2)^2 + b^2 = 5^2
a^2 + (b/2)^2 = 40

Rewrite just a bit.

a^2 + b^2 = c^2
(a^2)/4 + b^2 = 25
a^2 + (b^2)/4 = 40

Solve one equation for something

a^2 = c^2 - b^2

Substitute into all other equations.

(c^2 - b^2)/4 + b^2 = 25
(c^2 - b^2) + (b^2)/4 = 40

Simplify

(c^2)/4 - (b^2)/4 + b^2 = 25
c^2 - b^2 + (b^2)/4 = 40

(c^2)/4 + 3*(b^2)/4 = 25
c^2 - 3*(b^2)/4 = 40

Solve one equation for something

c^2 = 40 + 3*(b^2)/4

Substitute into all other equations.

(40 + 3*(b^2)/4)/4 + 3*(b^2)/4 = 25

Simplify

10 + 3*(b^2)/16 + 12*(b^2)/16 = 25
15*(b^2)/16 = 15
(b^2)/16 = 1
(b^2) = 16

Substitute into previous equations to find the other parts

c^2 = 40 + 3*(16)/4 = 40 + 12 = 52
a^2 = 52 - 16 = 36

It's a little tedious, but a very good exercise in paying attention.
 
Top