Figure 1-9 shows a rhomboid, ABCD. If you drop a perpendicular, CF, from C to AD, and project another from A to BC, you will create two right triangles, DAEB and DCFD, and the rectangle DECF. It can be shown geometrically that the right triangles are similar and equal. You can see that the area of the rectangle AECF equals the product of AF x FC. The area of the triangle CFD equals 1 /2(FD)(FC). Because the triangle AEB is equal and similar to CFD, the area of that triangle also equals 1/2(FD)(FC). Therefore, the total area of both triangles equals (FD)(FC). The total area of the rhomboid equals the area of the rectangle AECF + the total area of both triangles.

The total area of the rhomboid equals (AF)(FC) + (FD)(FC), or (AF + FD)(FC). But AF + FD equals AD, the base. FC equals the altitude. Therefore, the formula for the area of a rhomboid is A = bh. Here again you must

remember that h in a rectangle corresponds to the length of one of the sides, but h in a rhombus or rhomboid does not.

Figure 1-10 shows a trapezoid, ABCD. If you drop perpendiculars BE and CF from points B and C, respectively, you create the right triangles AEB and DFC and the rectangle EBCF between them. The area of the trapezoid obviously equals the sum of the areas of these figures.

The area of DAEB equals 1/2(AE)(FC), the area of DDFC equals 1/2(FD)(FC), and the area of EBCF equals (EF)(FC). There-fore, the area of the trapezoid ABCD equals l/2(AE)(FC) + (EF)(FC) + 1/2(FD)(FC), or

However, 2EF = EF + BC. Therefore, the area of the trapezoid equals

But AE + FD + EF = AD. Therefore, the area of the trapezoid equals

Stated in words, the area of a trapezoid is equal to one-half the sum of its bases times its altitude.