geometry question...help? urgent?

Question

the medians of a right triangle that are drawn from the verticies of the acute angles have lengths of 2squareroot13 and squareroot73. what is the length of the hypotenuse?

anyone get it? im sort of lost...please help asap

Answer 1

Take the legs of the triangle to be x and y. so, clearly, x, y/2 and 2 sqrt(13) form a right triangle and x/2, y, sqrt(73) form a right triangle, so we get 2 equations

x^2 + y^2/4 = 52
x^2/4 + y^2 = 73

Add them

5/4(hypoteneuse^2) = 125
hypoteneuse = 10

Answer 2

The length of the hypotenuse is 10

You can make like this:
The first median divides the side BC to two segments a + a, so its full length is 2a (conventionally)
The second median divides the side CA into two segments b + b, so its full length will make 2b
It this case the hypotenuse will make 2c length, and the line which comes across the middle points of sides BC and CA makes a segment with these points of length c (as the triangles ABC and the less one which is formed with the point C and the two middle side points are similar)

Then we can draw two other triangles (as in you task), and to express the length of their hypotenuses (using the Pythagoras theorem) in terms of a and b:

(2a)^2 + b^2 = (2sq.root{13})^2 4a^2 + b^2 = 52
a^2 + (2b)^2 = (sq.root{73})^2 a^2 + 4b^2 = 73

By adding the equations we get 5a^2 + 5b^2 = 125,
and a^2 + b^2 = 25
But we know that c^2 = a^2 + b^2 = 25, so
c = 5,
and the side AB = 2c = 10

Answer 3

Base = a
Height = b
Hypotenuse = c = sqrt(a^2 + b^2)

Given:
a^2 + b^2 / 4 = (2sqrt13)^2 = 52 .... (1)
a^2 / 4 + b^2 = (sqrt73)^2 = 73 ..... (2)

Solving equations (1) and (2)
a^2 = 36 and b^2 = 64

c^2 = a^2 + b^2 = 36 + 64 = 100

c = 10 (take the positive value, since it is length only)

Answer 4

I can try.. I took geomentry last year. Since its a triangle it all equal 180 and the equation is a square + b square = c square

Since two of the angles are 2 square root 13 and square root 73.

so.. 2 square root 13 + square root 73 = c square

square 2 square root 13 + square root 73 = c square

2 x 2 = 4, and square 13 is 13.
square 73 is 73. since the square root cancel out.

4(13) + 73 = 125

put it in square root term its 5 square root of 25.
I am not sure though, but I tried.
hope it give you some thought!

Answer 5

Let x = one leg and y the other
Then (y/2)^2 +x^2 = (2sqrt(13))^2, and y^2 + (x/2)^2= (sqrt (73))^2.
y^2/4 +x^2 = 52 <--- Eq 1
y^2 + x^2/4 = 73 <---- Eq 2
Multiply Eq 1 by -4 getting:
-y^2 -4x^2 = -208 <---- Eq 3
Add Eq 2 and Eq 3 getting:
x^2/4 - 4x^2 = -135
-15x^2/4 = -135
-15x^2= -540
x^2 = 540/15=36
plug this value of x^2 into Eq 3 getting:
-y^2 -4(36) = -208
-y^2= -64
y^2=64
Finally hypotenuse = sqrt(x^2 + y^2)= sqrt(64+36)=10