In trapezoid ABCD as shown in a figure, the point O is an intersection of a diagonal line.
The area of △ AOB is 10 cm2 and △ BOC is 25 cm2.
Find the area of trapezoid ABCD.
Answer
Moreover, since the area of △ABD and △ACD is equal, it is P + S = R + S.
Thus, it is found that R = P = 10 cm2.
Furthermore, since the area ratio of S and R turns into a ratio of a base, it is 2 : 5.
Thus, the area of S is 10 cm2 × 2/5 = 4 cm2.
Therefore, the area of trapezoid ABCD is 10 + 25 + 10 + 4 = 49 cm2.
49 cm2
Solution
In △AOB (P) and △BOC(Q), the ratio of triangular base is equal to the ratio of area of triangle, it is AO : CO = 10 : 25 = 2 : 5.
Moreover, since the area of △ABD and △ACD is equal, it is P + S = R + S.
Thus, it is found that R = P = 10 cm2.
Furthermore, since the area ratio of S and R turns into a ratio of a base, it is 2 : 5.
Thus, the area of S is 10 cm2 × 2/5 = 4 cm2.
Therefore, the area of trapezoid ABCD is 10 + 25 + 10 + 4 = 49 cm2.
