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