# Find Area of Right Triangle

• Feb 12th 2009, 06:09 PM
magentarita
Find Area of Right Triangle
Right triangle, where one of the legs equals 10, the bisector of the opposite angle cuts the leg into the ratio of 2:3. Find the area of the triangle.
• Feb 12th 2009, 09:03 PM
Soroban
Hello, magentarita!

Quote:

In a right triangle, where one of the legs equals 10,
the bisector of the opposite angle cuts the leg into the ratio of 2:3.
Find the area of the triangle.

This problem requires a seldom-used theorem about angle bisectors.

In a triangle, the angle bisector of an angle divides the opposite side
. . into segments proportional to the other two sides.

Code:

B
*
*  *  2a
*    *
2 *        *  D
*          *
*        *    *  3a
*    *          *
*  *                *
A *  *  *  *  *  *  *  C
3

Suppose we have $\displaystyle \Delta ABC$ with $\displaystyle AB = 2,\:AC = 3$

If $\displaystyle AD$ bisects $\displaystyle \angle A$, then $\displaystyle BD:DC \,=\,2:3$

We have the following right triangle . . .
Code:

B
o  -
* *  |
*  *  |
*  3 *  |
3a  *      *  10
*        o D |
*      *  *  |
*    *      *  |
*  *        2 *  |
* *              *  |
A o  *  *  *  *  o  -
2a        C

We are given: .$\displaystyle BC = 10$
$\displaystyle AD$ bisects $\displaystyle \angle A$ and divides $\displaystyle BC$ in the ratio 2:3.

Hence, $\displaystyle AC:AB \,=\,2:3$

Let: $\displaystyle AC = 2a,\;AB = 3a$

Using Pythagorus, we have: .$\displaystyle (2a)^2 + 10^2 \:=\:(3a)^2$

Then: .$\displaystyle 4a^2 + 100 \:=\:9a^2\quad\Rightarrow\quad 5a^2 \:=\:100 \quad\Rightarrow\quad a^2 \:=\:20$

. . Hence: .$\displaystyle a \:=\:\sqrt{20} \:=\:2\sqrt{5}$

So we have: .$\displaystyle AC \:=\:2a \:=\:4\sqrt{5}$

The area of the triangle is: .$\displaystyle \tfrac{1}{2}\text{(base)(height)} \:=\: \tfrac{1}{2}(4\sqrt{5})(10) \;=\; 20\sqrt{5}$

• Feb 13th 2009, 07:58 AM
Just out of curiousity , is there a name for this theorem and is there a proof for this theorem . THanks .
• Feb 15th 2009, 12:35 PM
magentarita
Soroban...
Quote:

Originally Posted by Soroban
Hello, magentarita!

This problem requires a seldom-used theorem about angle bisectors.

In a triangle, the angle bisector of an angle divides the opposite side
. . into segments proportional to the other two sides.
Code:

B
*
*  *  2a
*    *
2 *        *  D
*          *
*        *    *  3a
*    *          *
*  *                *
A *  *  *  *  *  *  *  C
3

Suppose we have $\displaystyle \Delta ABC$ with $\displaystyle AB = 2,\:AC = 3$

If $\displaystyle AD$ bisects $\displaystyle \angle A$, then $\displaystyle BD:DC \,=\,2:3$

We have the following right triangle . . .
Code:

B
o  -
* *  |
*  *  |
*  3 *  |
3a  *      *  10
*        o D |
*      *  *  |
*    *      *  |
*  *        2 *  |
* *              *  |
A o  *  *  *  *  o  -
2a        C

We are given: .$\displaystyle BC = 10$
$\displaystyle AD$ bisects $\displaystyle \angle A$ and divides $\displaystyle BC$ in the ratio 2:3.

Hence, $\displaystyle AC:AB \,=\,2:3$

Let: $\displaystyle AC = 2a,\;AB = 3a$

Using Pythagorus, we have: .$\displaystyle (2a)^2 + 10^2 \:=\:(3a)^2$

Then: .$\displaystyle 4a^2 + 100 \:=\:9a^2\quad\Rightarrow\quad 5a^2 \:=\:100 \quad\Rightarrow\quad a^2 \:=\:20$

. . Hence: .$\displaystyle a \:=\:\sqrt{20} \:=\:2\sqrt{5}$

So we have: .$\displaystyle AC \:=\:2a \:=\:4\sqrt{5}$

The area of the triangle is: .$\displaystyle \tfrac{1}{2}\text{(base)(height)} \:=\: \tfrac{1}{2}(4\sqrt{5})(10) \;=\; 20\sqrt{5}$

Wonderful job as usual. Thanks.