Discussion
The area of the Triangle ABC shown in the figure is 30. The area of Triangle ADC is
*This question is included in Nova Math - Problem Set J: Geometry, question #65
| (A) | 5 |
| (B) | ... |
| (C) | ... |
| (D) | ... |
| (E) | ... |
| (F) | ... |
The solution is
Posted: 09/29/2012 20:32
How is it that we can say that the height if ABC is equal to AF? Are we just plugging in a variable? I don't see how adding triangle ABF gives us extra information.
Posted: 10/01/2012 14:01
Mike, we are given the area as 30. Recall that the area of a triangle is 1/2 * base * height, with 1 condition: the height has to be perpendicular to the base. Here, the base is BC, which is =3, but the height is unknown. So we create a segment AF which is perpendicular to BC.
1/2 * BC * AF = 30. 1/2 * 3 * AF = 30. AF = 20.
Now, since DC is perpendicular to AF, AF is also the height for triangle ADC, so we can calculate the area ADC.
1/2 * BC * AF = 30. 1/2 * 3 * AF = 30. AF = 20.
Now, since DC is perpendicular to AF, AF is also the height for triangle ADC, so we can calculate the area ADC.