Math Help - Question on RightTriangles(Median)

1. Question on RightTriangles(Median)

The legs of a right triangle are 6 and 8.
(a) what is the length of the median?
(b)What is the length of the Altitude?

Can anyone show me how to find the answer?

2. By median, I an assuming that you mean hypotenuse.

If two sides of a right triangle are 6 and 8, the hypotenuse is (using Pythagoras Theorem):

sqrt(6^2 + 8^2) = sqrt(36 + 64) = sqrt(100) = 10

Now, by using sine or cos formulas for a right triangle, you can easily find the other angles of the triangle.

Once, you have done this, using either of the same formulas will allow you to find the altitude of the triangle - which I assume is a perpendicular line from the hypotenuse to the point of the right angle of the triangle. You will need to use one of the sides of the original triangle as the hypotenuse of the smaller triangle obtained when drawing the altitude line.

I hope that is clear. You need to draw a diagram to see what I mean. I am new to this site, so I don't know how to add diagrams yet.

3. Originally Posted by ewojmath
The legs of a right triangle are 6 and 8.
(a) what is the length of the median?
(b)What is the length of the Altitude?

Can anyone show me how to find the answer?

2. You posted this question in three different sub-fori which is against the rules.

3. Please don't double-triple-multiple post: It's only a waste of time for us.

4. Hello, ewojmath1!

I have a different interpretation of your wording.

The legs of a right triangle are 6 and 8.

(a) What is the length of the median to the hypotenuse ?
(b) What is the length of the altitude to the hypotenuse ?

Using Pythagorus: . $hyp^2 \:=\:6^2+8^2\quad\Rightarrow\quad hyp \:=\:10$

Code:
            *
* \ *
6 *   \   *   8
*     \     *
*       \       *
*         \         *
* - - - - - * - - - - - *
5           5

The right triangle can be inscribed in a semicircle of radius 5.

Since the median to hypotenuse is also a radius, its length is 5.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Let $CD$ be the altitude to the hypotenuse.

Code:
            C
*
*|  *
6 * |     *   8
*  |h       *
*   |           *
*    |D             *
A * - - * - - - - - - - - * B
10

Since $\Delta CDA \sim \Delta BCA$, we have: . $\frac{h}{6} \:=\:\frac{8}{10} \quad\Rightarrow\quad h \:=\:4.8$

5. Ohhh,

OK, I didn't realise that the "median" meant median of the hypotenuse. I sort of assumed that perhaps that was another way to say hypotenuse in your country.