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In Δ ABC, ∠ABC = 90° and seg BD ⊥ hypotenuse AC, with A–D–C. If AD = 9 cm and DC = 16 cm, find BD, AB and BC.[Figure: right-angled Δ ABC with the right angle at B; the hypotenuse AC is horizontal at the bottom with A at the left and C at the right; B…

Q.4 (2) In Δ ABC, ∠ABC = 90° and seg BD ⊥ hypotenuse AC, with A–D–C. If AD = 9 cm and DC = 16 cm, find BD, AB and BC.

[Figure: right-angled Δ ABC with the right angle at B; the hypotenuse AC is horizontal at the bottom with A at the left and C at the right; B is above AC; the perpendicular BD from B meets AC at D.]

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Answer: BD = 12 cm, AB = 15 cm, BC = 20 cm

In right Δ ABC, seg BD ⊥ hypotenuse AC.

By the theorem of geometric mean, \(BD^2 = AD \times DC = 9 \times 16 = 144\), so BD = 12 cm.

In right Δ ADB (∠ADB = 90°), by Pythagoras theorem:
\(AB^2 = AD^2 + BD^2 = 81 + 144 = 225\), so AB = 15 cm.

In right Δ CDB (∠CDB = 90°):
\(BC^2 = DC^2 + BD^2 = 256 + 144 = 400\), so BC = 20 cm.

Check: AC = 9 + 16 = 25 and \(AB^2 + BC^2 = 225 + 400 = 625 = 25^2\).

Doubtly AI · 27 Oct 2026 AI-generated · not yet verified Report
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