Q1. [1]
In the given figure, in $\triangle ABC$, $AD \perp BC$ and $\angle BAC = 90^\circ$. If $BC = 16$ cm and $DC = 4$ cm, then the value of $x$ (= AC) is:
- A $4$ cm
- B $5$ cm
- C $8$ cm
- D $3$ cm
Previously asked in CBSE board exam
2025 30/3/1 Q18
Generated by claude-sonnet-4-6 · 2026-06-15 10:22 · grounding rag
Model Answer
Option C: 8 cm
Using the geometric mean relation in a right triangle with altitude: $AC^2 = BC \times DC = 16 \times 4 = 64$, so $x = AC = 8$ cm.
Explanation
In △ABC with ∠BAC = 90° and AD ⊥ BC, triangles ABC and ACD are similar (AA criterion). This gives the relation $AC^2 = BC \times DC$ (geometric mean). Substituting: $x^2 = 16 \times 4 = 64$, so $x = 8$ cm. This is a standard result from Chapter 6 (Similarity of Triangles).
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