In the given figure, △ABC is right angled triangle with ∠ A = 90°. AD is perpendicular to BC. Prove that: (i) △DBA ~ △DAC (ii) DA² = DB × DC (iii) Find the area of △ABC when DB = 9 cm and DC = 16 cm.

MathsCBSE Class 10 · Basic · Board exam 2026, set 430/5/1 · 5 marks · Long answer

CBSE Class 10 Mathematics Basic 2026: question 32(a)

Set in the CBSE Class 10 board exam of 2026, paper 430/5/1, section D, question 32(a), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q32(a) · 5 marks · Long answerTriangles

In the given figure, $△$ ABC is right angled triangle with $∠ A = 9 0^{\circ}$ . AD is perpendicular to BC. Prove that: (i) $△$ DBA $\sim$ $△$ DAC (ii) $D A^{2} = D B \times D C$ (iii) Find the area of $△$ ABC when DB = 9 cm and DC = 16 cm.

Figure in the paper: Right triangle ABC with the right angle at A (marked). A is at the top, B at the right and C at the lower left on the base. D lies on BC with AD drawn perpendicular to BC (right-angle mark at D). See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2026 board paper, set 430/5/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

Both triangles have a right angle at D; find one more pair of equal angles using the angle sum property.

Hint 2

From the similarity write $\frac{D B}{D A} = \frac{D A}{D C}$ ; for the area use BC as base and AD as height.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

(i) In $△ D B A$ and $△ D A C$ : $∠ B D A = ∠ A D C = 9 0^{\circ}$ [ $A D ⊥ B C$ ]
1 mark
In $△ A B D$ , $∠ B A D = 9 0^{\circ} - ∠ B$ ; in $△ A B C$ , $∠ A C B = 9 0^{\circ} - ∠ B$ [ $∠ A = 9 0^{\circ}$ ]. So $∠ D A B = ∠ D C A$ . Hence $△ D B A \sim △ D A C$ (AA similarity).
1 mark
(ii) Corresponding sides of similar triangles are proportional: $\frac{D B}{D A} = \frac{D A}{D C} \Rightarrow D A^{2} = D B \times D C$
1 mark
(iii) $D A^{2} = 9 \times 16 = 144 \Rightarrow D A = 12$ cm
1 mark
$B C = D B + D C = 25$ cm. Area of $△ A B C = \frac{1}{2} \times B C \times A D = \frac{1}{2} \times 25 \times 12 = 150 cm^{2}$
1 mark

Answer: (i), (ii) proved; (iii) Area of $△ A B C$ = 150 cm²

Where the marks are

Right angles at D equal (1 mark)
Second pair of equal angles and AA similarity (1 mark)
Proportion of sides giving $D A^{2} = D B \times D C$ (1 mark)
DA = 12 cm (1 mark)
Area = 150 cm² (1 mark)

Revise the chapter

Chapter: Triangles (notes, formulas, where marks are lost and every past question)
Topic: Problems on Similar Triangles

Practise this chapterBoard-style questions from Triangles, one at a time, marked against the answer key and marking points.Back to the full paper2026 board paper, set 430/5/1 (Basic): every question, by section.

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CBSE Class 10 Mathematics Basic 2026: question 32(a)

Hints

Step-by-step solution

Revise the chapter

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