• Dec 11th 2009, 04:59 AM
reiward
The altitude of a triangle is 30 cm and its base is 80 cm. Find the area of the trapezoid formed by a line parallel to the base of the triangle and is 12 cm from the vertex.

Help please, I really have no idea. Can you show it with illustrations, thanks! :)
• Dec 11th 2009, 06:40 AM
Soroban
Hello, reiward!

Quote:

The altitude of a triangle is 30 cm and its base is 80 cm.
Find the area of the trapezoid formed by a line
parallel to the base of the triangle and is 12 cm from the vertex.

Code:

                B   -            *   :            *| *   :          * |  *   :          *  |12  *   :        *  |      *  30      D *----+---------* E   :      *    |          *   :      *      |            *   :    *      |18            *   :    *        |                *   - A *---------*-------------------* C       : - - - - - - 80- - - - - - - :
The area of a trapezoid is: .$\displaystyle A \;=\;\tfrac{h}{2}(b_1+b_2)$

. . where $\displaystyle h$ is the altitude and $\displaystyle b_1,b_2$ are the lengths of the parallel sides.

We already know: .$\displaystyle h = 18,\;b_1 = 80$
. . We need: .$\displaystyle b_2 \,=\,DE$

Since $\displaystyle \Delta DBE \sim \Delta ABC\!:\;\;\frac{DE}{12} \,=\,\frac{80}{30} \quad\Rightarrow\quad DE \,=\,32$

Therefore: .$\displaystyle A \;=\;\frac{18}{2}(80 + 32) \;=\;1008\text{ cm}^2$