MathsCBSE Class 10 · Basic · Board exam 2024, set 430/3/1 · 5 marks · Long answer

CBSE Class 10 Mathematics Basic 2024: question 33(a)

Set in the CBSE Class 10 board exam of 2024, paper 430/3/1, 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

In the given figure, $A B$ is chord of length $6$ cm of a circle of radius $5$ cm. The tangents at $A$ and $B$ intersect at a point $P$ . Find the length of $P B$ .

Figure in the paper: A circle with centre O and radius 5 cm. Chord AB of length 6 cm is drawn. Tangents at B and A meet at external point P. The radius OB and OA are joined, and the dashed line OP is drawn. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 430/3/1.

Hints

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

Hint 1

OP is the perpendicular bisector of chord AB; find OQ first.

Hint 2

There are two right triangles with PB as a side: $△ P QB$ and $△ O B P$ (right angle at B).

Hint 3

Let $P Q = x$ , $P B = y$ , write Pythagoras in both triangles and eliminate $y^{2}$ .

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 OP meet AB at Q. Since $P A = P B$ [tangents from an external point] and $O A = O B$ [radii], OP is the perpendicular bisector of AB. So $O Q ⊥ A B$ and $A Q = QB = 3$ cm.
1 mark
In right $△ O QB$ : $O Q = O B^{2} - Q B^{2} = 25 - 9 = 4$ cm.
½ mark
Let $P Q = x$ cm and $P B = y$ cm. In right $△ P QB$ : $x^{2} + 3^{2} = y^{2}$ ...(i)
1 mark
$O B ⊥ P B$ [tangent ⟂ radius], so in right $△ O B P$ : $O P^{2} = O B^{2} + P B^{2} \Rightarrow (x + 4)^{2} = 25 + y^{2}$ ...(ii)
1 mark
Substituting (i) in (ii): $x^{2} + 8 x + 16 = 25 + x^{2} + 9 \Rightarrow 8 x = 18 \Rightarrow x = \frac{9}{4}$ cm
1 mark
From (i): $y^{2} = \frac{81}{16} + 9 = \frac{225}{16} \Rightarrow y = \frac{15}{4}$ . So $P B = \frac{15}{4}$ cm $= 3.75$ cm.
½ mark

Answer: $P B = \frac{15}{4}$ cm = 3.75 cm

Where the marks are

OP bisects AB at right angles, so $A Q = QB = 3$ cm (1 mark)
$O Q = 4$ cm by Pythagoras (½ mark)
Equation $x^{2} + 9 = y^{2}$ from $△ P QB$ (1 mark)
Equation $(x + 4)^{2} = y^{2} + 25$ from $△ O B P$ (1 mark)
Solves to get $x = \frac{9}{4}$ (1 mark)
Gets $P B = \frac{15}{4}$ cm (½ mark)

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 paper2024 board paper, set 430/3/1 (Basic): every question, by section.

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CBSE Class 10 Mathematics Basic 2024: question 33(a)

Hints

Step-by-step solution

Revise the chapter

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