In the adjoining figure, △ CAB is a right triangle, right angled at A and AD ⊥ BC. Prove that △ ADB ~ △ CDA. Further, if BC = 10 cm and CD = 2 cm, find the length of AD.

MathsCBSE Class 10 · Standard · Board exam 2025, set 30/6/1 · 5 marks · Long answer

CBSE Class 10 Mathematics Standard 2025: question 33(b)

Set in the CBSE Class 10 board exam of 2025, paper 30/6/1, section D, question 33(b), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q33(b) · 5 marks · Long answerTriangles

In the adjoining figure, $△ C A B$ is a right triangle, right angled at A and $A D ⊥ B C$ . Prove that $△ A D B \sim △ C D A$ . Further, if $B C = 10$ cm and $C D = 2$ cm, find the length of AD.

Figure in the paper: Right triangle CAB with the right angle at A (A at bottom left, C at top left, B at bottom right). D lies on hypotenuse CB, and AD is drawn perpendicular to CB, marked with a right-angle symbol at D. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 30/6/1.

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.

Hint 2

Use the fact that the acute angles of a right triangle add to $9 0^{\circ}$ ; then from the similarity, $A D^{2} = B D \cdot C D$ with $B D = B C - C D$ .

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.

In $△ A D B$ and $△ C D A$ : $∠ A D B = ∠ C D A = 9 0^{\circ}$ (given $A D ⊥ B C$ ).
½ mark
In right $△ C A B$ , $∠ B + ∠ C = 9 0^{\circ}$ ; in right $△ A D B$ , $∠ D A B + ∠ B = 9 0^{\circ}$ . Hence $∠ D A B = ∠ C = ∠ D C A$ .
1 mark
So $△ A D B \sim △ C D A$ [AA similarity]. Hence proved.
1 mark
Corresponding sides are proportional: $\frac{A D}{C D} = \frac{B D}{A D}$
½ mark
$\Rightarrow A D^{2} = B D \times C D$
½ mark
$B D = B C - C D = 10 - 2 = 8$ cm, so $A D^{2} = 8 \times 2 = 16$
1 mark
$A D = 4$ cm
½ mark

Answer: $△ A D B \sim △ C D A$ ; $A D = 4$ cm

Where the marks are

Right angles at D in both triangles (½ mark)
Shows $∠ D A B = ∠ D C A$ (1 mark)
Concludes similarity by AA (1 mark)
Writes proportion $A D / C D = B D / A D$ (½ mark)
$A D^{2} = B D \times C D$ (½ mark)
$B D = 8$ cm and $A D^{2} = 16$ (1 mark)
$A D = 4$ cm (½ 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 paper2025 board paper, set 30/6/1 (Standard): every question, by section.

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CBSE Class 10 Mathematics Standard 2025: question 33(b)

Hints

Step-by-step solution

Revise the chapter

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