Geometry Question, Help Needed!?

Question

Hey guys, I need some help with a geometry problem. I know you people out there in Mathematics can do it, so I'm looking for some help. Here we go:

The medians to the legs of a right triangle are 3 x sq. rt. 17 and 6 x sq. rt 2. What is the perimeter of the triangle.


sq. rt: Square root of

Also, if youd help with this one, Ill be thankful:

http://answers.yahoo.com/question/index?qid=20070226065124AAFYUvM&r=w

Answer 1

Let right triangle ABC have angle B = 90, AC is the hypotenuse
Let base = BC = 2y, Perpendicular = AB = 2x
Let D be the midpoint of BC and E the midpt. of AB

AD is hypotenuse for right triangle ABD (angle B = 90, remember?)
Similarly, CE is hypotenuse for triangle EBC

By Pythagorean theorem,
(2x)^2 + y^2 = [3 sqrt(17)]^2..........(1)
(2y)^2 + x^2 = [6 sqrt(2)]^2..........(2)

Work with equation (1),
(2x)^2 + y^2 = [3 sqrt(17)]^2
4x^2 + y^2 = 9*17
4x^2 + y^2 = 153
4x^2 = 153 - y^2
x^2 = (153 - y^2)/4..........(3)

Substitute (3) in (2)
(2y)^2 + x^2 = [6 sqrt(2)]^2
4y^2 + (153 - y^2)/4 = 36*2
4y^2 + (153 - y^2)/4 = 72
(16y^2 - y^2 + 153)/4 = 72
16y^2 - y^2 + 153 = 72*4
16y^2 - y^2 + 153 = 288
16y^2 - y^2 = 288 - 153
16y^2 - y^2 = 135
15y^2 = 135
y^2 = 135/15 = 9..........(4)
y = 3
2y = 3*2 = 6
BC = 6 cm

Substitute (4) in (1),
(2x)^2 + y^2 = [3 sqrt(17)]^2
4x^2 + 9 = 153
4x^2 = 153 - 9
4x^2 = 144
x^2 = 144/4 = 36
x = 6
2x = 12
AB = 12 cm

AB^2 + BC^2 = AC^2
6^2 + 12^2 = AC^2
36 + 144 = AC^2
AC^2 = 180
AC = sqrt 180
= 6 * sqrt(5)

Perimeter of triangle = AB + BC + AC
= 12 + 6 + 6 sqrt 5
= 18 + 6 sqrt(5) cm

Answer 2

Let the legs length to be x & y
So 1/4 x^2 + y^2 = 9×17 = 153
and 1/4 y^2 + x^2 = 36×2 = 72
Solving for x & y you get
x= 6 , y= 12
So its perimeter = 6+12+6 sqrt(5) = 18 + 6 sqrt(5)

Answer 3

Its just pythagoras....

X^2 = (3sqrt17)^2 + (6sqrt2)^2

X^2 = 225

X = 15

Then just add up the 3 values....

15 + 3sqrt17 + 6sqrt2 = 35.855 units