the measure of the base of an isosceles triangle is radical 2. the medians to the congruent sidesof the triangle meet at the right angles. find the area of the triangle

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- Jan 2nd 2007, 02:50 PManime_maniatriangles
the measure of the base of an isosceles triangle is radical 2. the medians to the congruent sidesof the triangle meet at the right angles. find the area of the triangle

- Jan 2nd 2007, 05:02 PMThePerfectHacker
Look at diagram below.

The two medians upon the equal sides of an isoceles triangle are equal.

Since we have $\displaystyle \sqrt{1}$ by Pythagorean theorem those red segments are equal to 1.

**Theorem**: The length of the median upon the side $\displaystyle a$ is:

$\displaystyle l=\frac{1}{2}\sqrt{2(b^2+c^2)-a^2}$

In this case,

$\displaystyle l=\frac{1}{2} \sqrt{2(2+a^2)-a^2}=\frac{1}{2}\sqrt{4-a^2}$

That means the length of blue is:

$\displaystyle \mbox{BLUE }=\frac{1}{2}\sqrt{4-a^2}-1$

By Pythagorean theorem:

$\displaystyle \mbox{BLUE}^2+\mbox{RED}^2=\mbox{GREEN}^2$

Thus,

$\displaystyle \left(\frac{1}{2}\sqrt{4-a^2} \right)^2+1=(a/2)^2$

The equation solving is for thee. - Jan 2nd 2007, 05:23 PManime_mania
thanks so much but is that the area of only that part of the triangle or the whole triangle, also in the theorem you stated what is Asquare, Bsquare, and Csquare

- Jan 2nd 2007, 05:34 PMThePerfectHacker
- Jan 2nd 2007, 07:59 PManime_mania
sorry but it still doesnt show me how to find the area for the whole thing but still thanks for all your hard work

- Jan 2nd 2007, 09:55 PMticbol
Here is one way, the way where we will use the properties of, or what about, the medians of a triangle.

The 3 medians of any triangle intersect at the centroid of the triangle. These medians divide each other in the ratio of 2:1, meaning, the centroid lies at 1/3 of the length of any median---measured from the side of the triangle, not from the corner.

Imagine, or draw the figure on paper. So the two given medians are from the base corners of the isosceles triangle. Draw those, and draw also the 3rd median---the one from the apex to the midpoint of the base. If the base were horizontal, then this 3rd median is vertical. This 3rd median, call it "h", is the altitude of the isosceles triangle from the sqrt(2) long base.

Area of the whole triangle, A = (1/2)(base)(altitude)

A = (1/2) * sqrt(2) * h

So we find h and we are done.

In the figure, in any of the two right triangles below the centroid, we have:

---base = (1/2)[sqrt(2)]

---angle oppsite base = (1/2)(90deg) = 45deg.

Hence, angle adjacent to base is also 45deg.

So any of those two right triangles is 45-45-90 right triangle or an isoceles right triangle.

Then, if one of the legs--the base--is (1/2)[sqrt(2)], then the other leg should also be (1/2)[sqrt(2)].

This other leg is 1/3 of the length of h----centroid thing, remember.

So (1/3)h = (1/2)[sqrt(2).

And so, h = (3/2)[sqrt(2)]

Therefore,

A = (1/2)[sqrt(2)][(3/2)sqrt(2)] = 3/2 = 1.5 sq.units -----------answer. - Jan 2nd 2007, 10:06 PMearboth
Hello,

to calculate the area of a triangle you need the length of the base (which you already kow) and the length of the height. To solve your problem it is necessary to get the length of the height.

I've attached a sketch of the triangle:

The base is AB, the top of the triangle is T.

The medians of a triangle intercept in the centroid C. The centroid divides each median with a ratio of 2:1 measured from the vertex of the triangle where the median starts.

The triangle ABC is an isosceles**right**triangle. The height in this triangle is half as long as the base**and**it must be $\displaystyle \frac{1}{3}$ of the height of triangle ABT.

$\displaystyle \frac{1}{3} \cdot h = \frac{1}{2} \cdot \sqrt{2} \ \Longleftrightarrow \ h=\frac{3}{2} \cdot \sqrt{2}$

Now you can calculate the area. The general formula is:

$\displaystyle A_{triangle}=\frac{1}{2} \cdot {base} \cdot {height}$. Plug in all known values:

$\displaystyle A_{triangle}=\frac{1}{2} \cdot \underbrace{\sqrt{2}}_{base} \cdot \underbrace{\frac{3}{2} \cdot \sqrt{2}}_{height} = \frac{3}{2} $

EB