Home > CCG > Chapter 5 > Lesson 5.3.5 > Problem5-135

5-135.

Find the area and perimeter of $ΔABC$ at right. Give approximate (decimal) answers, not exact answers.

Separate $ΔABC$ into two separate triangles by dropping the height from vertex $B$.
Label the new point $D$, creating $ΔABD$ and $ΔDBC$.

Both of the triangles that have been created are special right triangles, because the height $BD$ creates two right triangles: a $45º\text{-}45º\text{-}90º$ triangle and a $30º\text{-}60º\text{-}90º$ triangle.

The side opposite the 45º in a $45º\text{-}45º\text{-}90º$ triangle is the the length of the leg times $\sqrt{2}$.
In a $30º\text{-}60º\text{-}90º$ triangle, the side opposite the $30^\circ$ is half the hypotenuse and the side opposite the $60^\circ$ is half the hypotenuse times $\sqrt{3}$.