# Thread: Height of an isosceles triangle, Easy question

1. ## [SOLVED]Height of an isosceles triangle, Easy question

Hi Forum

I've been kind of confused about how to get the height of an triangle only by the base and the angle of two intersecting medians.

Here is the question
Find the height of the isosceles triangle that has base of sqrt2; the medians to the legs form two right angles.

Calling P the point where the medians intersect we can get a right triangle on the base, with height sqrt2/2 (that goes to point P) and base sqrt2.
I can't remember exactly what to do to get the missing part of the height.

Thanks!

EDIT:

Ok, I got it, this one was easy!
In a isosceles triangle the median of the base is equal to the height.
And the centroid is exactly two thirds of the median!
So sqrt2/2 1/3
x 2/3
(Cross product)
x=sqrt2
h= sqrt2+sqrt2/2

2. Hi Zellator,
This problem has not been solved. It can be done using coordinate geometry.I suggest you use an even integer base and when you get that answer convert by ratio to terms of rad2

3. Hi again bjhopper!

But the answer is right, isn't it? (the original problem was actually after the area)
Can go into further details?
It is a good idea to use coordinate geometry, but I didn't see it as necessary.

Thanks

4. the answer is easy if you given an equilateral triangle but what you did is not correct.

bjh

5. Hi bjhopper.
I don't get it.
Why wouldn't the centroid theorem work here?
It works regardless of the triangle type.

If you can calculate the height of the point of the centroid, it's easy to calculate the rest of the height.
"The centroid is exactly two-thirds the way along each median. Put another way, the centroid divides each median into two segments whose lengths are in the ratio 2:1, with the longest one nearest the vertex."

Can you prove it otherwise?
Maybe by using coordinate geometry.
Thanks.

6. The altitude of an equilateral triangle side rad2 = rad6/2. The altitude of an isosceles triangle which you have described can be solved by coordinate geometry. What you are doing is mixing the two at the same time

bjh

7. Originally Posted by bjhopper
The altitude of an equilateral triangle side rad2 = rad6/2. The altitude of an isosceles triangle which you have described can be solved by coordinate geometry. What you are doing is mixing the two at the same time.
Hi bjhopper! Thanks for your reply again.

I agree that the height of a equilateral triangle is s*(sqrt3)/2.
If this is the case, we can assume that this triangle I am looking for is equilateral
This is what the question states:
The base of an isosceles triangle is $\sqrt2$; medians to the legs intersect at right angles. The area of the triangle is:

The centroid(yes, again) thing can't be false here.
If we trace a line in the height here, it will equal the median to the base. So we can consider it to be the centroid!

bjhopper, I searched over for your coordinate geometry idea, and I couldn't find anything on Google. Can you make yourself clearer, please?
How would you do this question?

Thanks

8. ## height of isosceles triangle

Hi Zellator,
The triangle you are looking for is isosceles and not equilateral. It has special properties. Can you graph ?Draw a line equal to 12 units on the x axis . This gives you two points (0,0) and (12,0)Draw two 45 degree lines from these points so that they cross above the axis. They intersect at 90 deg. Erect the perpendicular bisector of the 12 base.Where the lines meet go up higher and find a segment equal to 3 units parallel to base that fits between the 45 line and the perpendicular.Connect a line from base point (12,0) to the intersection of the segment and 45 deg line. Extend it to meet perpendicular.This intersection is the apex of the required isosceles triangle .Measure the height and then by factor find result in terms of rad2

bjh

### solved que. on isoceles

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