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NCERT Exemplar · Class 9 Mathematics Circles

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EXERCISE 10.2 1–10 (part 2 of 6)

  1. Write True or False and justify your answer in each of the following:

    Exercise 1

    Two chords AB and CD of a circle are each at distances 4\displaystyle 4 cm from the centre. Then AB=CD\displaystyle \mathrm{AB}=\mathrm{CD}.

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    NCERT’s answer
    True. Because the distances from the centre of two chords are equal.
    True.NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-2_Q1\[OM \perp AB, \quad ON \perp CD, \quad OM = ON = 4\text{ cm} \] \[AM = \sqrt{OA^{2}-OM^{2}}, \quad CN = \sqrt{OC^{2}-ON^{2}} \] \[OA = OC \quad \text{(radii)} \ \Longrightarrow\ AM = CN \] \[AB = 2AM = 2CN = CD \]Equal distances from the centre give equal chords.

  2. Exercise 2

    Two chords AB and AC of a circle with centre O are on the opposite sides of OA. Then ∠OAB=∠OAC\displaystyle \angle \mathrm{OAB}=\angle \mathrm{OAC}.

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    NCERT’s answer
    False. The angles will be equal only if $\displaystyle \mathrm{AB}=\mathrm{AC}$.
    False.NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-2_Q2\[OA = OB = OC \quad \text{(radii)} \] \[\angle OAB = \angle OBA, \quad \angle OAC = \angle OCA \quad \text{(isosceles triangles)} \] \[\angle OAB = \tfrac12(180^\circ-\angle AOB), \quad \angle OAC = \tfrac12(180^\circ-\angle AOC) \] \[\angle OAB = \angle OAC \iff \angle AOB = \angle AOC \iff AB = AC \]Nothing forces AB = AC, so the two base angles at A need not match.

  3. Exercise 3

    Two congruent circles with centres O and O' intersect at two points A and B. Then ∠AOB=∠AO′B\displaystyle \angle \mathrm{AOB}=\angle \mathrm{AO}^{\prime} \mathrm{B}.

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    NCERT’s answer
    True. Because equal chords of congruent circles subtend equal angles at the respective centres.
    True.NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-2_Q3\[OA = OB = O'A = O'B \quad \text{(congruent circles, equal radii)} \] \[AB = AB \quad \text{(common side)} \] \[\triangle OAB \cong \triangle O'AB \quad \text{(SSS)} \] \[\angle AOB = \angle AO'B \]Congruent circles give equal radii on both sides, so the triangles are congruent.

  4. Exercise 4

    Through three collinear points a circle can be drawn.

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    NCERT’s answer
    False. Because a circle through two points cannot pass through a point which is collinear to these two points.
    False.NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-2_Q4\[DE \perp AB, \quad FG \perp BC \quad \text{(perpendicular bisectors, through midpoints } M, N\text{)} \] \[DE \parallel FG \quad \text{(both perpendicular to the same line } ABC\text{)} \]The centre would have to lie on both bisectors, but two parallel lines never meet.

  5. Exercise 5

    A circle of radius 3\displaystyle 3 cm can be drawn through two points A, B such that AB=6 cm\displaystyle \mathrm{AB}=6 \mathrm{~cm}.

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    NCERT’s answer
    True. Because AB will be the diameter.
    True.NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-2_Q5\[AB = 6\text{ cm} = 2\times 3\text{ cm} = 2r \] \[AB = 2r \ \Longrightarrow\ AB \text{ is a diameter} \] \[O = \text{midpoint of } AB, \quad OA = OB = 3\text{ cm} \]AB equals the diameter, so a circle of radius $\displaystyle 3$ cm centred at its midpoint passes through both points.

  6. Exercise 6

    If AOB is a diameter of a circle and C is a point on the circle, then AC2+BC2=\displaystyle \mathrm{AC}^{2}+\mathrm{BC}^{2}= AB2\displaystyle \mathrm{AB}^{2}.

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    NCERT’s answer
    True. As $\displaystyle \angle \mathrm{C}$ is right angle, $\displaystyle \mathrm{AC}^{2}+\mathrm{BC}^{2}=\mathrm{AB}^{2}$.
    True.NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-2_Q6\[AOB \text{ diameter} \ \Longrightarrow\ \angle ACB = 90^\circ \quad \text{(angle in a semicircle)} \] \[AC^{2}+BC^{2}=AB^{2} \quad \text{(Pythagoras, } \angle C = 90^\circ\text{)} \]The angle in a semicircle is a right angle, so Pythagoras applies directly to triangle ACB.

  7. Exercise 7

    ABCD is a cyclic quadrilateral such that ∠A=90∘,∠B=70∘,∠C=95∘\displaystyle \angle \mathrm{A}=90^{\circ}, \angle \mathrm{B}=70^{\circ}, \angle \mathrm{C}=95^{\circ} and ∠D=105∘\displaystyle \angle \mathrm{D}=105^{\circ}.

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    NCERT’s answer
    False, as $\displaystyle \angle \mathrm{A}+\angle \mathrm{C}=90^{\circ}+95^{\circ}=185^{\circ} \neq 180^{\circ}$.
    False.NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-2_Q7\[\angle A + \angle C = 90^\circ + 95^\circ = 185^\circ \] \[\angle B + \angle D = 70^\circ + 105^\circ = 175^\circ \] \[\text{cyclic quadrilateral} \ \Longrightarrow\ \angle A+\angle C = \angle B+\angle D = 180^\circ \]Opposite angles of a cyclic quadrilateral must sum to \(\displaystyle 180^\circ\); here neither pair does.

  8. Exercise 8

    If A,B,C,D\displaystyle \mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D} are four points such that ∠BAC=30∘\displaystyle \angle \mathrm{BAC}=30^{\circ} and ∠BDC=60∘\displaystyle \angle \mathrm{BDC}=60^{\circ}, then D is the centre of the circle through A, B and C.

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    NCERT’s answer
    False, because there can be many points D such that $\displaystyle \angle \mathrm{BDC}=60^{\circ}$ and each such point cannot be the centre of the circle through A,B,C.
    False.NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-2_Q8\[\angle BOC = 2\angle BAC = 60^\circ \quad \text{(central = 2}\times\text{inscribed)} \] \[\angle BDC = \angle BOC = 60^\circ \Rightarrow B, D, O, C \text{ concyclic} \quad \text{(equal angles, same side of } BC\text{)} \] \[OB = OC \text{ (radii)} \Rightarrow O \text{ is the only point of arc } BOC \text{ on the perpendicular bisector of } BC \] \[D \neq O \Rightarrow DB \neq DC \]Answer: False — the angle condition only places D on that arc, not at its centre.

  9. Exercise 9

    If A, B, C and D are four points such that ∠BAC=45∘\displaystyle \angle \mathrm{BAC}=45^{\circ} and ∠BDC=45∘\displaystyle \angle \mathrm{BDC}=45^{\circ}, then A, B, C, D are concyclic.

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    NCERT’s answer
    True. Angles in the same segment.
    True.NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-2_Q9\[\angle BAC = \angle BDC = 45^\circ \quad \text{(given; A and D taken on the same side of } BC\text{)} \] \[\angle BAC = \angle BDC \ \Longrightarrow\ A, B, C, D \text{ concyclic} \quad \text{(angles in the same segment, converse)} \]

  10. Exercise 10

    In Fig. 10.10\displaystyle 10.10, if AOB is a diameter and ∠ADC=120∘\displaystyle \angle \mathrm{ADC}=120^{\circ}, then ∠CAB=30∘\displaystyle \angle \mathrm{CAB}=30^{\circ}. NCERT_Question_Class9_Maths_Exemplar_Ch10_Ex10-2_Q10 NCERT_Question_Class9_Maths_Exemplar_Ch10_Ex10-2_Q10

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    NCERT’s answer
    True. $\displaystyle \angle \mathrm{B}=180^{\circ}-120^{\circ}=60^{\circ}, \angle \mathrm{CAB}=90^{\circ}-60^{\circ}=30^{\circ}$.
    True.NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-2_Q10\[\angle ADC + \angle ABC = 180^\circ \quad \text{(opp. angles, cyclic quad. } ADCB\text{)} \] \[\angle ABC = 180^\circ - 120^\circ = 60^\circ \] \[\angle ACB = 90^\circ \quad \text{(angle in semicircle, } AB \text{ diameter)} \] \[\angle CAB = 180^\circ - 90^\circ - 60^\circ = 30^\circ \]