A backyard is in the shape of a triangle ABC with right angle at B. AB = 7 m and BC = 15 m. A circular pit was dug inside it such that it touches the walls AC, BC and AB at P, Q and R respectively such that AP = x m. (i) Find the length of AR in terms of x. (ii) Write the type of quadrilateral BQOR. (iii) (a) Find the length PC in terms of x and hence find the value of x. (iii) (b) Find x and hence find the radius r of circle.

MathsCBSE Class 10 · Standard · Board exam 2024, set 30/1/1 · 4 marks · Case-based

CBSE Class 10 Mathematics Standard 2024: question 38

Set in the CBSE Class 10 board exam of 2024, paper 30/1/1, section E, question 38, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q38 · 4 marks · Case-basedCompetency-basedCircles

A backyard is in the shape of a triangle ABC with right angle at B. AB = 7 m and BC = 15 m. A circular pit was dug inside it such that it touches the walls AC, BC and AB at P, Q and R respectively such that AP = $x$ m.

(i)
Find the length of AR in terms of $x$ .
1 mark
(ii)
Write the type of quadrilateral BQOR.
1 mark
(iii)
(a) Find the length PC in terms of $x$ and hence find the value of $x$ .
2 marks
OR
(iii)
(b) Find $x$ and hence find the radius $r$ of circle.
2 marks

Figure in the paper: Right triangle ABC with right angle at B; AB = 7 m vertical, BC = 15 m horizontal. A circle with centre O and radius r is inscribed, touching AC at P, BC at Q and AB at R. OR, OQ are drawn to the touching points. Labels: 7 m on AB, 15 m on BC. A photograph of a triangular backyard with a circular pit is shown alongside. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 30/1/1.

Hints

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

Hint 1

Tangents drawn to a circle from the same outside point are equal in length.

Hint 2

BR = BQ = r, and BR = 7 − x. Use this to get CQ, and hence PC.

Hint 3

AC = AP + PC = 8 + 2x. Equate its square to 7² + 15² using Pythagoras.

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) Tangents drawn from an external point to a circle are equal. So from A: $A R = A P = x$ m.
1 mark
(ii) In BQOR:
- $∠ B = 9 0^{\circ}$ [given]
- $∠ O R B = ∠ O QB = 9 0^{\circ}$ [tangent ⟂ radius]
So all four angles are $9 0^{\circ}$ . Also $O R = O Q = r$ [radii], and $B R = B Q$ [tangents from B].
Hence BQOR is a square.
1 mark
(iii)(a) $B R = A B - A R = 7 - x$ . Since BQOR is a square, $B Q = B R = 7 - x$ .
$C Q = B C - B Q = 15 - (7 - x) = 8 + x$
$P C = C Q$ [tangents from C], so $P C = 8 + x$ .
½ mark
(iii)(a) $A C = A P + P C = x + 8 + x = 8 + 2 x$ .
By Pythagoras in $△ A B C$ :
$A C^{2} = A B^{2} + B C^{2} = 49 + 225 = 274$
$\Rightarrow (8 + 2 x)^{2} = 274 \Rightarrow 8 + 2 x = 274$
1 mark
(iii)(a) $x = \frac{274 - 8}{2} \approx \frac{16.55 - 8}{2} \approx 4.28$ m
½ mark
(iii)(b) As in (iii)(a): $A R = x$ , $B Q = B R = 7 - x$ , and $P C = C Q = 8 + x$ , so $A C = 8 + 2 x$ .
By Pythagoras: $(8 + 2 x)^{2} = 7^{2} + 1 5^{2} = 274$
$\Rightarrow 8 + 2 x = 274$
1 mark
(iii)(b) $x = \frac{274 - 8}{2} \approx 4.28$ m
½ mark
(iii)(b) BQOR is a square, so $r = B R = 7 - x$ :
$r = 7 - (\frac{274}{2} - 4) = 11 - \frac{274}{2} \approx 2.72$ m
½ mark

Answer: (i) AR = x m; (ii) square; (iii)(a) PC = (8 + x) m, x = (√274 − 8)/2 ≈ 4.28 m; (iii)(b) x ≈ 4.28 m, r = 11 − √274/2 ≈ 2.72 m

Where the marks are

(i) Uses equal tangents from A: $A R = x$ m (1 mark)
(ii) Identifies BQOR as a square (1 mark)
(iii)(a) Gets $P C = 8 + x$ (½ mark)
(iii)(a) Uses Pythagoras: $(8 + 2 x)^{2} = 274$ (1 mark)
(iii)(a) Finds $x = (274 - 8) /2 \approx 4.28$ (½ mark)
(iii)(b) Forms $(8 + 2 x)^{2} = 274$ (1 mark)
(iii)(b) Finds $x \approx 4.28$ (½ mark)
(iii)(b) Finds $r = 7 - x \approx 2.72$ m (½ 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 paper2024 board paper, set 30/1/1 (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 2024: question 38

Hints

Step-by-step solution

Revise the chapter

More questions from Circles