MathsCBSE Class 10 · Standard · Board exam 2023, set 30/5/1 · 5 marks · Long answer

CBSE Class 10 Mathematics Standard 2023: question 33(a)

Set in the CBSE Class 10 board exam of 2023, paper 30/5/1, section D, question 33(a), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q33(a) · 5 marks · Long answerCircles

A triangle 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 area $△ A B C = 90$ cm $^{2}$ .

Figure in the paper: Triangle ABC with incircle centre O of radius 4 cm. The incircle touches BC at D; OD is drawn vertically with length 4 cm. BD = 10 cm and DC = 8 cm are marked on base BC. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2023 board paper, set 30/5/1.

Hints

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

Hint 1

Tangents from the same external point are equal: use this at B, C and A.

Hint 2

Join O to A, B, C; each of the three small triangles has height equal to the radius 4 cm.

Hint 3

Set the sum of the three areas equal to 90 and solve for the unknown tangent length from A.

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.

Let the incircle touch AB at F and AC at E. Join OA, OB, OC, and draw $O E ⊥ A C$ , $O F ⊥ A B$ (radius ⟂ tangent), so $O D = O E = O F = 4$ cm.
Tangents from an external point are equal: $B F = B D = 10$ cm, $C E = C D = 8$ cm; let $A F = A E = x$ cm.
1½ marks
ar $△ A B C$ = ar $△ B O C$ + ar $△ C O A$ + ar $△ A O B$ $= \frac{1}{2} (4) (B C) + \frac{1}{2} (4) (C A) + \frac{1}{2} (4) (A B) = \frac{1}{2} \cdot 4 (B C + C A + A B)$
1½ marks
$B C = 18$ , $C A = 8 + x$ , $A B = 10 + x$ . So $90 = 2 (18 + 8 + x + 10 + x) = 2 (36 + 2 x) = 4 (18 + x)$ $\Rightarrow 18 + x = 22.5 \Rightarrow x = 4.5$
1 mark
$A B = 10 + 4.5 = 14.5$ cm and $A C = 8 + 4.5 = 12.5$ cm.
1 mark

Answer: AB = 14.5 cm, AC = 12.5 cm

Where the marks are

Draws radii to points of contact, uses equal tangents BF = 10, CE = 8, AF = AE = x (1½ marks)
Splits area of ΔABC into ar BOC + ar COA + ar AOB with heights 4 cm (1½ marks)
Forms 90 = 2(36 + 2x) and gets x = 4.5 (1 mark)
States AB = 14.5 cm and AC = 12.5 cm (1 mark)

Revise the chapter

Chapter: Circles (notes, formulas, where marks are lost and every past question)
Topic: Lengths of Tangents from an External Point

Practise this chapterBoard-style questions from Circles, one at a time, marked against the answer key and marking points.Back to the full paper2023 board paper, set 30/5/1 (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 2023: question 33(a)

Hints

Step-by-step solution

Revise the chapter

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