The area of the following trapezoid is 24 cm^{2}. If 5 times its longer base is equal to 7 times its shorter base, then what is the difference between the area of ∆DOC and the area of∆ AOB?
cm^{2}
We label the sections as shown. Since 7AB = 5CD, we have
.
Since the height of ABC and ACD are the same, the ratio of their areas is the ratio of their bases, so
.
Thus, the ratio of ABC to ABCD is
,
so its area is 24 × 5/12 = 10 cm^{2}. ADC = ABCD-ABC, so ADC = 24 - 10 = 14 cm^{2}. We see that ADC and BDC have the same area because they have the same base and height. Thus,
ADC-DOC = BDC-DOC,
so (3) = (4) . Therefore, we find the difference of DOC and AOB to be
(1) - (2) = [ (1) + (4) ]-[ (2) + (3) ] = 14 - 10 = 4 cm^{2}.