Asked in: June 2026 board paper (same idea as a 2-mark activity in July 2025) In the following figure, seg AE ⊥ seg BC, seg DF ⊥ line BC, AE = 4, DF = 6, then find (A(Δ ABC))/(A(Δ DBC)). [Figure: Δ ABC and Δ DBC stand on the same base BC; AE and DF are the perpendiculars from A and D to BC.]

Asked in: June 2026 board paper (same idea as a 2-mark activity in July 2025)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: \(\Delta ABC\) and \(…

Asked in: June 2026 board paper (same idea as a 2-mark activity in July 2025)

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: \(\Delta ABC\) and \(\Delta DBC\) stand on the same base BC; AE and DF are the perpendiculars from A and D to BC.]

Doubtly Asked 29 Oct 2026 Practice test ssc 10th Save WhatsApp Report

Best answer

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

The two triangles have the same base BC, so the ratio of their areas equals the ratio of their heights.

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

Doubtly AI · 29 Oct 2026 AI-verified · Nov 2026 Report

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