A △ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC are of lengths 10 cm and 8 cm respectively. Find the lengths of the sides AB and AC, if it is given that ar(△ABC) = 90cm² For Visually Impaired candidates: A circle is inscribed in a right-angled triangle ABC, right angled at B. If BC=7cm and AB=24cm, find the radius of the circle

MathsCBSE Class 10 · Standard · Sample paper, session 2025-26 · 2 marks · Very short answer

CBSE Class 10 Mathematics Standard sample paper 2025-26: question 25

From the official CBSE sample question paper for 2025-26, section B, question 25, for 2 marks. Try it first; the hints and our step-by-step solution are below.

Q25 · 2 marks · Very short answerCircles

A $△$ ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC are of lengths 10 cm and 8 cm respectively. Find the lengths of the sides AB and AC, if it is given that ar( $△$ ABC) = 90cm $^{2}$

For Visually Impaired candidates: A circle is inscribed in a right-angled triangle ABC, right angled at B. If BC=7cm and AB=24cm, find the radius of the circle

Figure in the paper: Triangle ABC with an inscribed circle with centre O touching BC at D, CA at E and AB at F. A is at top, C bottom-left, B bottom-right. Lines OA, OB, OC, OD, OE, OF are drawn from centre O. See it in the official paper.

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2025-26.

Hints

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

Hint 1

Tangent lengths from the same vertex are equal. Call the unknown tangent length from A $x$ .

Hint 2

Join O to A, B and C. The three triangles each have height r = 4 cm.

Hint 3

The area of ABC equals ½ × r × perimeter.

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.

Tangents drawn from an external point are equal. So $B F = B D = 10$ cm and $C E = C D = 8$ cm. Let $A E = A F = x$ cm.
Then $A B = x + 10$ , $A C = x + 8$ and $B C = 18$ cm.
½ mark
The radii OD, OE, OF are perpendicular to the sides [tangent ⟂ radius]. Hence
ar( $△ A B C$ ) = ar( $△ A O B$ ) + ar( $△ B O C$ ) + ar( $△ A O C$ ) $= \frac{1}{2} r (A B + B C + A C)$ .
½ mark
$90 = \frac{1}{2} \times 4 (x + 10 + 18 + x + 8) = 2 (2 x + 36) = 4 x + 72$ , so $4 x = 18$ and $x = 4.5$ cm.
½ mark
$A B = 4.5 + 10 = 14.5$ cm and $A C = 4.5 + 8 = 12.5$ cm.
½ mark
For Visually Impaired candidates (alternative question): In right $△ A B C$ with $∠ B = 9 0^{\circ}$ , $A C^{2} = A B^{2} + B C^{2} = 2 4^{2} + 7^{2} = 625$ , so $A C = 25$ cm.
(VI) Area $= \frac{1}{2} \times 7 \times 24 = 84$ cm $^{2}$ … (i)
(VI) Joining the centre to the vertices, area $= \frac{1}{2} r (24 + 25 + 7) = 28 r$ … (ii)
(VI) From (i) and (ii), $28 r = 84$ , so $r = 3$ cm. (This version carries ½ mark for each of these four steps.)

Answer: AB = 14.5 cm, AC = 12.5 cm (VI version: r = 3 cm)

Where the marks are

Uses equal tangents: BF = 10, CE = 8, AE = AF = x (½ mark)
Writes the area of ABC as the sum of areas of AOB, BOC and AOC (½ mark)
Solves $90 = 2 (2 x + 36)$ to get x = 4.5 cm (½ mark)
States AB = 14.5 cm and AC = 12.5 cm (½ mark)
VI: AC = 25 cm (½), area 84 cm² (½), area = 28r (½), r = 3 cm (½)

Revise the chapter

Chapter: Circles (notes, formulas, where marks are lost and every past question)

Practise this chapterBoard-style questions from Circles, one at a time, marked against the answer key and marking points.Back to the full paperSample paper 2025-26 (Standard): every question, by section.

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CBSE Class 10 Mathematics Standard sample paper 2025-26: question 25

Hints

Step-by-step solution

Revise the chapter

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