Segment bd is an altitude of triangle abc. find the area of the triangle. triangle abc with altitude bd is shown. point a is at 2, 1. point b is at negative 1, 4. point c is at 2, 6. point d is at 2, 4.

Where's the "triangle with alt. BD?"  This problem can be solved without the diagram, but the solution would be easier with it.

BD is the altitude.  Find the length of BD by finding the dist. between (-1,4) and (2,4); it is 2-(-1), or 3.  |BD| = 3.

I've graphed the triangle myself and have found that the "base" of the triangle is the vertical line thru (2,1) and (2,6); its length is 6-1, or 5.

Thus, the area of this triangle is  A = (b)(h) / 2, or  A = (5)(3) / 2 = 10/3 square inches.

444(2)4-2(7)2+40=? 8(4)-14(2)+40=? 32-28+40=? 4+40=? 44