In figure 1.73, \(\angle ABC = \angle DCB = 90^\circ\), AB = 6, DC = 8 then \(\dfrac{A(\triangle ABC)}{A(\triangle DCB)} = ?\)
Figure: A and D are on the same side of seg BC; AB ⊥ BC at B and DC ⊥ BC at C.
Answer: \(\dfrac{A(\triangle ABC)}{A(\triangle DCB)} = \dfrac{3}{4}\)
Both triangles have the same base BC. Their heights on this base are AB = 6 (since AB ⊥ BC) and DC = 8 (since DC ⊥ BC).
Areas of triangles with equal bases are proportional to their heights:
\[\frac{A(\triangle ABC)}{A(\triangle DCB)} = \frac{AB}{DC} = \frac{6}{8} = \frac{3}{4}\]Log in or create a free account to answer this question.
