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In the following figure, seg AE \(\perp\) seg BC, seg DF \(\perp\) line BC, AE = 4, DF = 6, then find \(\dfrac{A(\Delta ABC)}{A(\Delta DBC)}\).[Figure: B, E, C, F are points in that order on a line; A is above E with AE \(\perp\) BC and D is above F…

Q.1 (B)(ii) In the following figure, seg AE \(\perp\) seg BC, seg DF \(\perp\) line BC, AE = 4, DF = 6, then find \(\dfrac{A(\Delta ABC)}{A(\Delta DBC)}\).

[Figure: B, E, C, F are points in that order on a line; A is above E with AE \(\perp\) BC and D is above F with DF \(\perp\) line BC; segments AB, AC, DB, DC are drawn; right-angle marks at E and F.]

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Answer: \(\dfrac{A(\Delta ABC)}{A(\Delta DBC)} = \dfrac{2}{3}\)

\(\Delta ABC\) and \(\Delta DBC\) have the same base BC. Their heights are AE and DF respectively.

\(\dfrac{A(\Delta ABC)}{A(\Delta DBC)} = \dfrac{\frac{1}{2} \times BC \times AE}{\frac{1}{2} \times BC \times DF} = \dfrac{AE}{DF} = \dfrac{4}{6} = \dfrac{2}{3}\).

Doubtly AI · 25 Oct 2026 AI-verified · Oct 2026 Report
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