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NCERT Exemplar · Class 9 Mathematics Circles

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EXERCISE 10.3 1–10 (part 3 of 6)

  1. Exercise 1

    If arcs AXB and CYD of a circle are congruent, find the ratio of AB and CD.

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    NCERT’s answer
    $\displaystyle 1$:$\displaystyle 1$
    NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-3_Q1\[\operatorname{arc}\, AXB \cong \operatorname{arc}\, CYD \quad \text{(given)} \] \[\Rightarrow AB = CD \quad \text{(congruent arcs subtend equal chords)} \]Answer: \(\displaystyle AB : CD = 1 : 1 \).

  2. Exercise 2

    If the perpendicular bisector of a chord AB of a circle PXAQBY intersects the circle at P and Q , prove that arc⁡PXA≅Arc⁡PYB\displaystyle \operatorname{arc} \mathrm{PXA} \cong \operatorname{Arc} \mathrm{PYB}.

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    NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-3_Q2\[P \text{ on perp. bisector of } AB \ \Longrightarrow\ PA = PB \quad \text{(equidistant from } A, B\text{)} \] \[PA = PB \ \Longrightarrow\ \operatorname{arc}\, PXA \cong \operatorname{arc}\, PYB \quad \text{(equal chords cut off equal arcs)} \]Answer: \(\displaystyle \operatorname{arc}\, PXA \cong \operatorname{arc}\, PYB \).

  3. Exercise 3

    A,B\displaystyle \mathrm{A}, \mathrm{B} and C are three points on a circle. Prove that the perpendicular bisectors of AB, BC and CA are concurrent.

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    NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-3_Q3\[OA = OB = OC \quad \text{(radii of the same circle)} \] \[OA = OB \Rightarrow O \text{ lies on the perpendicular bisector of } AB \] \[OB = OC \Rightarrow O \text{ lies on the perpendicular bisector of } BC \] \[OC = OA \Rightarrow O \text{ lies on the perpendicular bisector of } CA \]Answer: the perpendicular bisectors of AB, BC, CA are concurrent, at the centre O.

  4. Exercise 4

    AB and AC are two equal chords of a circle. Prove that the bisector of the angle BAC passes through the centre of the circle.

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    NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-3_Q4\[AB = AC \quad \text{(given)} \] \[OA = OA, \quad OB = OC \quad \text{(radii of the same circle)} \] \[\Rightarrow \triangle OAB \cong \triangle OAC \quad \text{(SSS)} \] \[\Rightarrow \angle OAB = \angle OAC \]So \(\displaystyle AO\) bisects the angle and passes through the centre.Answer: the bisector of \(\displaystyle \angle BAC \) is the line \(\displaystyle AO \), through the centre.

  5. Exercise 5

    If a line segment joining mid-points of two chords of a circle passes through the centre of the circle, prove that the two chords are parallel.

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    NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-3_Q5\[OM \perp PQ, \quad ON \perp RS \quad \text{(line from the centre to the mid-point of a chord is perpendicular to it)} \] \[M,\, O,\, N \text{ collinear} \quad \text{(given)} \] \[\Rightarrow PQ \perp MN \text{ and } RS \perp MN \] \[\Rightarrow PQ \parallel RS \quad \text{(both perpendicular to } MN\text{)} \]Answer: \(\displaystyle PQ \parallel RS \).

  6. Exercise 6

    ABCD is such a quadrilateral that A is the centre of the circle passing through B, C and D. Prove that ∠CBD+∠CDB=12∠BAD\angle \mathrm{CBD}+\angle \mathrm{CDB}=\frac{1}{2} \angle \mathrm{BAD}

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    NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-3_Q6\[\text{reflex}\angle BAD = 2\,\angle BCD \quad \text{(angle at centre = 2}\times\text{angle at circumference, arc }BD\text{)} \] \[\text{reflex}\angle BAD = 360^\circ - \angle BAD \] \[\Rightarrow \angle BCD = 180^\circ - \tfrac12\angle BAD \] \[\angle CBD + \angle CDB = 180^\circ - \angle BCD \quad \text{(angle sum, } \triangle BCD\text{)} \] \[\Rightarrow \angle CBD + \angle CDB = \tfrac12\angle BAD \]Answer: \(\displaystyle \angle CBD + \angle CDB = \tfrac12\angle BAD \).

  7. Exercise 7

    O is the circumcentre of the triangle ABC and D is the mid-point of the base BC. Prove that ∠BOD=∠A\displaystyle \angle \mathrm{BOD}=\angle \mathrm{A}.

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    NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-3_Q7\[OB = OC \quad \text{(circumradius)} \] \[D \text{ mid-point of } BC \;\Rightarrow\; OD \text{ bisects } \angle BOC \quad \text{(median = bisector in isosceles } \triangle OBC\text{)} \] \[\angle BOD = \tfrac12 \angle BOC \] \[\angle BOC = 2\angle A \quad \text{(angle at centre = 2}\times\text{angle at circumference)} \] \[\Rightarrow \angle BOD = \tfrac12(2\angle A) = \angle A \]Answer: \(\displaystyle \angle BOD = \angle A \).

  8. Exercise 8

    On a common hypotenuse AB, two right triangles ACB and ADB are situated on opposite sides. Prove that ∠BAC=∠BDC\displaystyle \angle \mathrm{BAC}=\angle \mathrm{BDC}.

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    NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-3_Q8\[\angle ACB = \angle ADB = 90^\circ \] \[\Rightarrow A, C, B, D \text{ lie on a circle with } AB \text{ as diameter} \quad \text{(converse, angle in a semicircle)} \] \[\angle BAC, \angle BDC \text{ stand on the same arc } BC \] \[\Rightarrow \angle BAC = \angle BDC \quad \text{(angles in the same segment)} \]Answer: \(\displaystyle \angle BAC = \angle BDC \).

  9. Exercise 9

    Two chords AB and AC of a circle subtends angles equal to 90∘\displaystyle 90^{\circ} and 150∘\displaystyle 150^{\circ}, respectively at the centre. Find ∠BAC\displaystyle \angle \mathrm{BAC}, if AB and AC lie on the opposite sides of the centre.

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    NCERT’s answer
    $\displaystyle 60$°
    \[OA = OB = OC \quad \text{(radii)} \] \[\angle AOB = 90^{\circ}, \ \angle AOC = 150^{\circ} \quad \text{(given)} \] \[\angle OAB = \angle OBA = \frac{180^{\circ}-90^{\circ}}{2} = 45^{\circ} \quad (\triangle OAB \text{ isosceles}) \] \[\angle OAC = \angle OCA = \frac{180^{\circ}-150^{\circ}}{2} = 15^{\circ} \quad (\triangle OAC \text{ isosceles}) \] NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-3_Q9 AB and AC lie on opposite sides of O, so O lies inside \(\displaystyle \angle BAC\). \[\angle BAC = \angle OAB + \angle OAC = 45^{\circ}+15^{\circ} = 60^{\circ} \] Answer: \(\displaystyle \angle BAC = 60^{\circ}\)

  10. Exercise 10

    If BM and CN are the perpendiculars drawn on the sides AC and AB of the triangle ABC, prove that the points B, C, M and N are concyclic.

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    \[BM \perp AC \implies \angle BMC = 90^{\circ} \] \[CN \perp AB \implies \angle BNC = 90^{\circ} \] NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-3_Q10 M and N both view BC at a right angle. \[\angle BMC = \angle BNC = 90^{\circ} \implies B, C, M, N \text{ lie on the circle with diameter } BC \] Answer: B, C, M, N are concyclic, on the circle with diameter BC.