Hello, 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. 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 $