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NCERT Exemplar · Class 9 Mathematics Circles

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EXERCISE 10.1 1–10 (part 1 of 6)

  1. Exercise 1

    AD is a diameter of a circle and AB is a chord. If AD=34 cm,AB=30 cm\displaystyle \mathrm{AD}=34 \mathrm{~cm}, \mathrm{AB}=30 \mathrm{~cm}, the distance of AB from the centre of the circle is : (A) 17\displaystyle 17 cm (B) 15\displaystyle 15 cm (C) 4\displaystyle 4 cm (D) 8\displaystyle 8 cm

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    (D)
    (D) $\displaystyle 8$ cm.NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-1_Q1\[OA = \frac{AD}{2} = 17 \text{ cm}, \qquad AM = \frac{AB}{2} = 15 \text{ cm} \quad \text{(perpendicular from centre bisects chord)} \] \[OM = \sqrt{OA^2 - AM^2} = \sqrt{17^2 - 15^2} = \sqrt{64} = 8 \text{ cm} \]

  2. Exercise 2

    In Fig. 10.3\displaystyle 10.3, if OA=5 cm,AB=8 cm\displaystyle \mathrm{OA}=5 \mathrm{~cm}, \mathrm{AB}=8 \mathrm{~cm} and OD is perpendicular to AB , then CD is equal to: (A) 2\displaystyle 2 cm (B) 3\displaystyle 3 cm (C) 4\displaystyle 4 cm (D) 5\displaystyle 5 cm NCERT_Question_Class9_Maths_Exemplar_Ch10_Ex10-1_Q2

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    (A)
    (A) $\displaystyle 2$ cm.NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-1_Q2\[AC = \frac{AB}{2} = \frac{8}{2} = 4 \text{ cm} \quad \text{(perpendicular from centre bisects the chord)} \] \[OC = \sqrt{OA^2 - AC^2} = \sqrt{5^2 - 4^2} = 3 \text{ cm} \] \[CD = OD - OC = 5 - 3 = 2 \text{ cm} \quad (OD = OA, \text{ radii}) \]

  3. Exercise 3

    If AB=12 cm,BC=16 cm\displaystyle \mathrm{AB}=12 \mathrm{~cm}, \mathrm{BC}=16 \mathrm{~cm} and AB is perpendicular to BC, then the radius of the circle passing through the points A,B\displaystyle \mathrm{A}, \mathrm{B} and C is : (A) 6\displaystyle 6 cm (B) 8\displaystyle 8 cm (C) 10\displaystyle 10 cm (D) 12\displaystyle 12 cm

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    (C)
    (C) $\displaystyle 10$ cm.NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-1_Q3\[\angle ABC = 90^\circ \Rightarrow AC \text{ is a diameter} \quad \text{(converse of angle in semicircle)} \] \[AC = \sqrt{AB^2 + BC^2} = \sqrt{12^2 + 16^2} = \sqrt{400} = 20 \text{ cm} \] \[r = \frac{AC}{2} = 10 \text{ cm} \]

  4. Exercise 4

    In Fig. 10.4\displaystyle 10.4, if ∠ABC=20∘\displaystyle \angle \mathrm{ABC}=20^{\circ}, then ∠AOC\displaystyle \angle \mathrm{AOC} is equal to: (A) 20\displaystyle 20° (B) 40\displaystyle 40° (C) 60\displaystyle 60° (D) 10\displaystyle 10° NCERT_Question_Class9_Maths_Exemplar_Ch10_Ex10-1_Q4

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    (B)
    (B) $\displaystyle 40$°.NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-1_Q4\[\angle AOC = 2\angle ABC \quad \text{(angle at centre = twice angle at circumference, same arc AC)} \] \[\angle AOC = 2 \times 20^\circ = 40^\circ \]

  5. Exercise 5

    In Fig.10.5, if AOB is a diameter of the circle and AC=BC\displaystyle \mathrm{AC}=\mathrm{BC}, then ∠CAB\displaystyle \angle \mathrm{CAB} is equal to: (A) 30\displaystyle 30° (B) 60\displaystyle 60° (C) 90\displaystyle 90° (D) 45\displaystyle 45° NCERT_Question_Class9_Maths_Exemplar_Ch10_Ex10-1_Q5

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    (D)
    (D) $\displaystyle 45$°.NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-1_Q5\[\angle ACB = 90^\circ \quad \text{(angle in a semicircle, AOB a diameter)} \] \[AC = BC \Rightarrow \angle CAB = \angle CBA = \frac{180^\circ - 90^\circ}{2} = 45^\circ \]

  6. Exercise 6

    In Fig. 10.6\displaystyle 10.6, if ∠OAB=40∘\displaystyle \angle \mathrm{OAB}=40^{\circ}, then ∠ACB\displaystyle \angle \mathrm{ACB} is equal to : (A) 50\displaystyle 50° (B) 40\displaystyle 40° (C) 60\displaystyle 60° (D) 70\displaystyle 70° NCERT_Question_Class9_Maths_Exemplar_Ch10_Ex10-1_Q6

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    (A)
    (A) $\displaystyle 50$°.NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-1_Q6\[OA = OB \Rightarrow \angle OBA = \angle OAB = 40^\circ \quad \text{(isosceles, radii)} \] \[\angle AOB = 180^\circ - 2\times 40^\circ = 100^\circ \] \[\angle ACB = \tfrac{1}{2}\angle AOB = 50^\circ \quad \text{(angle at centre)} \]

  7. Exercise 7

    In Fig. 10.7\displaystyle 10.7, if ∠DAB=60∘,∠ABD=50∘\displaystyle \angle \mathrm{DAB}=60^{\circ}, \angle \mathrm{ABD}=50^{\circ}, then ∠ACB\displaystyle \angle \mathrm{ACB} is equal to: (A) 60\displaystyle 60° (B) 50\displaystyle 50° (C) 70\displaystyle 70° (D) 80\displaystyle 80° NCERT_Question_Class9_Maths_Exemplar_Ch10_Ex10-1_Q7

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    (C)
    (C) $\displaystyle 70$°.NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-1_Q7\[\angle ADB = 180^\circ - \angle DAB - \angle ABD = 180^\circ - 60^\circ - 50^\circ = 70^\circ \] \[\angle ACB = \angle ADB \quad \text{(angles in the same segment on AB)} \] \[\angle ACB = 70^\circ \]

  8. Exercise 8

    ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and ∠ADC=140∘\displaystyle \angle \mathrm{ADC}=140^{\circ}, then ∠BAC\displaystyle \angle \mathrm{BAC} is equal to: (A) 80\displaystyle 80° (B) 50\displaystyle 50° (C) 40\displaystyle 40° (D) 30\displaystyle 30°

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    (B)
    (B) $\displaystyle 50$°.NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-1_Q8\[\angle ABC = 180^\circ - \angle ADC = 180^\circ - 140^\circ = 40^\circ \quad \text{(opposite angles of cyclic quad.)} \] \[\angle ACB = 90^\circ \quad \text{(angle in semicircle, AB a diameter)} \] \[\angle BAC = 180^\circ - 90^\circ - 40^\circ = 50^\circ \]

  9. Exercise 9

    In Fig. 10.8\displaystyle 10.8, BC is a diameter of the circle and ∠BAO=60∘\displaystyle \angle \mathrm{BAO}=60^{\circ}. Then ∠ADC\displaystyle \angle \mathrm{ADC} is equal to : (A) 30\displaystyle 30° (B) 45\displaystyle 45° (C) 60\displaystyle 60° (D) 120\displaystyle 120° NCERT_Question_Class9_Maths_Exemplar_Ch10_Ex10-1_Q9

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    (C)
    (C) $\displaystyle 60$°.NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-1_Q9\[OA = OB \Rightarrow \angle ABC = \angle OAB = 60^\circ \quad \text{(isosceles, radii; O on diameter BC)} \] \[\angle ADC = \angle ABC = 60^\circ \quad \text{(same segment, chord AC)} \]

  10. Exercise 10

    In Fig. 10.9\displaystyle 10.9, ∠AOB=90∘\displaystyle \angle \mathrm{AOB}=90^{\circ} and ∠ABC=30∘\displaystyle \angle \mathrm{ABC}=30^{\circ}, then ∠CAO\displaystyle \angle \mathrm{CAO} is equal to: (A) 30\displaystyle 30° (B) 45\displaystyle 45° (C) 90\displaystyle 90° (D) 60\displaystyle 60° NCERT_Question_Class9_Maths_Exemplar_Ch10_Ex10-1_Q10

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    (D)
    (D) $\displaystyle 60$°.NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-1_Q10\[OA = OB \Rightarrow \angle OAB = \angle OBA = \frac{180^\circ - 90^\circ}{2} = 45^\circ \quad \text{(isosceles, radii)} \] \[\angle ACB = \tfrac{1}{2}\angle AOB = 45^\circ \quad \text{(angle at centre)} \] \[\angle BAC = 180^\circ - \angle ABC - \angle ACB = 180^\circ - 30^\circ - 45^\circ = 105^\circ \] \[\angle CAO = \angle BAC - \angle OAB = 105^\circ - 45^\circ = 60^\circ \]