MathsCBSE Class 10 · Standard · Board exam 2026, set 30/4/1 · 5 marks · Long answer

CBSE Class 10 Mathematics Standard 2026: question 33

Set in the CBSE Class 10 board exam of 2026, paper 30/4/1, section D, question 33, for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q33 · 5 marks · Long answerCircles

In the given figure, TP and TQ are tangents to a circle with centre M, touching another circle with centre N at A and B respectively. It is given that MQ = 13 cm, NB = 8 cm, BQ = 35 cm and TP = 80 cm.

(i)
Name the quadrilateral MQBN.
1 mark
(ii)
Is MN parallel to PA ? Justify your answer.
1 mark
(iii)
Find length TB.
1 mark
(iv)
Find length MN.
2 marks

Figure in the paper: A larger circle with centre M and a smaller circle with centre N. Two lines from T: tangent line TQ along the bottom touching the big circle at Q and small circle at B; and the other tangent line T-A-P touching the small circle at A and the big circle at P. MQ = 13 (perpendicular to TQ), NB = 8 (perpendicular to TQ), QB = 35 cm, MP and NA are radii perpendicular to the upper tangent. M and N are joined; a dashed horizontal line from M's level toward N is drawn with right angle marks. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2026 board paper, set 30/4/1.

Hints

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

Hint 1

The radius to the point of contact is perpendicular to the tangent; use it to find parallel sides.

Hint 2

Tangents from T to the bigger circle are equal, so TQ = TP.

Hint 3

For MN, drop a perpendicular from N onto MQ and use Pythagoras in the right triangle formed.

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) $M Q ⊥ QB$ and $N B ⊥ QB$ [radius ⟂ tangent], so $M Q ∥ N B$ . Since $M Q = 13 \neq = N B = 8$ , MN is not parallel to QB. Exactly one pair of opposite sides is parallel, so MQBN is a trapezium (a right trapezium).
1 mark
(ii) No. $M P ⊥ P A$ and $N A ⊥ P A$ [radius ⟂ tangent], so $M P ∥ N A$ . If MN were parallel to PA, MPAN would be a rectangle and $M P = N A$ ; but $M P = 13$ cm $\neq = N A = 8$ cm. Hence MN is not parallel to PA.
1 mark
(iii) $T Q = T P = 80$ cm [tangents from an external point are equal]. So $T B = T Q - B Q = 80 - 35 = 45$ cm.
1 mark
(iv) Draw $N S ⊥ M Q$ meeting MQ at S. NSQB is a rectangle, so $N S = B Q = 35$ cm and $S Q = N B = 8$ cm, giving $M S = 13 - 8 = 5$ cm. In right $△ M S N$ : $M N^{2} = N S^{2} + M S^{2} = 3 5^{2} + 5^{2}$
1 mark
$= 1225 + 25 = 1250$
½ mark
$M N = 1250 = 252$ cm.
½ mark

Answer: (i) Trapezium; (ii) No, since MP ≠ NA; (iii) TB = 45 cm; (iv) MN = $252$ cm

Where the marks are

(i) Identifies MQBN as a trapezium with reason (1 mark)
(ii) Says No, justified by $M P \neq = N A$ (1 mark)
(iii) Uses $T Q = T P = 80$ cm to get TB = 45 cm (1 mark)
(iv) Applies Pythagoras with $N S = 35$ , $M S = 5$ (1 mark)
(iv) $M N^{2} = 1250$ (½ mark)
(iv) $M N = 252$ cm (½ mark)

Revise the chapter

Chapter: Circles (notes, formulas, where marks are lost and every past question)
Topic: Solving Problems Using Tangent Properties

Practise this chapterBoard-style questions from Circles, one at a time, marked against the answer key and marking points.Back to the full paper2026 board paper, set 30/4/1 (Standard): every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.

CBSE Class 10 Mathematics Standard 2026: question 33

Hints

Step-by-step solution

Revise the chapter

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