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NCERT Exemplar · Class 9 Mathematics Circles

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EXERCISE 10.3 11–20 (part 4 of 6)

  1. Exercise 11

    If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral so formed is cyclic.

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    \[AB = AC \implies \angle ABC = \angle ACB \quad \text{(isosceles triangle)} \] \[DE \parallel BC \implies \angle ADE = \angle ABC \quad \text{(corresponding angles)} \] NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-3_Q11 \(\displaystyle \angle BDE\) and \(\displaystyle \angle ADE\) are a linear pair on line AB; \(\displaystyle \angle BCE=\angle ACB\) since E lies on AC. \[\angle BDE = 180^{\circ} - \angle ADE = 180^{\circ}-\angle ABC = 180^{\circ}-\angle ACB \] \[\angle BDE + \angle BCE = 180^{\circ} \] So BDEC is cyclic (opposite angles supplementary). Answer: Quadrilateral BDEC is cyclic.

  2. Exercise 12

    If a pair of opposite sides of a cyclic quadrilateral are equal, prove that its diagonals are also equal.

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    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    \[AB = CD \implies \text{arc } AB = \text{arc } CD \quad \text{(equal chords, equal arcs)} \] NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-3_Q12 \[\text{arc } AB + \text{arc } BC = \text{arc } CD + \text{arc } BC \] \[\text{arc } ABC = \text{arc } BCD \implies AC = BD \quad \text{(equal arcs, equal chords)} \] Answer: Diagonals AC = BD.

  3. Exercise 13

    The circumcentre of the triangle ABC is O. Prove that ∠OBC+∠BAC=90∘\displaystyle \angle \mathrm{OBC}+\angle \mathrm{BAC}=90^{\circ}.

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    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    \[OA=OB=OC \quad \text{(circumradius)} \] \[\angle BOC = 2\angle BAC \quad \text{(angle at centre = twice angle at circumference, same arc } BC\text{)} \] NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-3_Q13 \(\displaystyle \triangle OBC\) is isosceles since \(\displaystyle OB=OC\). \[\angle OBC = \angle OCB = \frac{180^{\circ}-\angle BOC}{2} = 90^{\circ}-\angle BAC \] \[\angle OBC + \angle BAC = 90^{\circ} \] Answer: \(\displaystyle \angle OBC + \angle BAC = 90^{\circ}\)

  4. Exercise 14

    A chord of a circle is equal to its radius. Find the angle subtended by this chord at a point in major segment.

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    NCERT’s answer
    $\displaystyle 30^{\circ}$
    \[OA = OB = AB \quad \text{(chord = radius, given)} \] \[\triangle OAB \text{ equilateral} \implies \angle AOB = 60^{\circ} \] NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-3_Q14 P lies on the major arc, so it sees the minor arc AB. \[\angle APB = \tfrac{1}{2}\angle AOB = 30^{\circ} \quad \text{(angle at centre = twice angle at circumference)} \] Answer: \(\displaystyle 30^{\circ}\)

  5. Exercise 15

    In Fig.10.13, ∠ADC=130∘\displaystyle \angle \mathrm{ADC}=130^{\circ} and chord BC=\displaystyle \mathrm{BC}= chord BE . Find ∠CBE\displaystyle \angle \mathrm{CBE}. NCERT_Question_Class9_Maths_Exemplar_Ch10_Ex10-3_Q15

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    NCERT’s answer
    $\displaystyle 100^{\circ}$
    \[\angle ADC + \angle ABC = 180^{\circ} \quad (ADCB \text{ cyclic quadrilateral}) \] \[\angle ABC = 180^{\circ}-130^{\circ} = 50^{\circ} \] NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-3_Q15 AB is a diameter, so \(\displaystyle \angle ACB=\angle AEB=90^{\circ}\) (angle in semicircle). \[\angle BAC = 180^{\circ}-90^{\circ}-50^{\circ} = 40^{\circ} \quad (\triangle ABC) \] \[BC = BE \implies \angle BAE = \angle BAC = 40^{\circ} \quad \text{(equal chords, equal angles)} \] \[\angle ABE = 180^{\circ}-90^{\circ}-40^{\circ}=50^{\circ} \quad (\triangle ABE) \] \[\angle CBE = \angle ABC+\angle ABE = 50^{\circ}+50^{\circ} = 100^{\circ} \quad \text{(C, E opposite sides of AB)} \] Answer: \(\displaystyle \angle CBE = 100^{\circ}\)

  6. Exercise 16

    In Fig.10.14, ∠ACB=40∘\displaystyle \angle \mathrm{ACB}=40^{\circ}. Find ∠OAB\displaystyle \angle \mathrm{OAB}. NCERT_Question_Class9_Maths_Exemplar_Ch10_Ex10-3_Q16

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    NCERT’s answer
    $\displaystyle 50$°
    \[\angle AOB = 2\angle ACB \quad \text{(angle at centre = twice angle at circumference, same arc } AB\text{)} \] \[\angle AOB = 2\times40^{\circ}=80^{\circ} \] NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-3_Q16 \(\displaystyle OA=OB\) (radii), so \(\displaystyle \triangle OAB\) is isosceles. \[\angle OAB=\angle OBA=\frac{180^{\circ}-80^{\circ}}{2}=50^{\circ} \] Answer: \(\displaystyle \angle OAB = 50^{\circ}\)

  7. Exercise 17

    A quadrilateral ABCD is inscribed in a circle such that AB is a diameter and ∠ADC=130∘\displaystyle \angle \mathrm{ADC}=130^{\circ}. Find ∠BAC\displaystyle \angle \mathrm{BAC}.

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    NCERT’s answer
    $\displaystyle 40^{\circ}$
    Since \(\displaystyle ABCD\) is a cyclic quadrilateral, \[\angle ADC + \angle ABC = 180^\circ \quad \text{(opposite angles of a cyclic quadrilateral)} \] \[\angle ABC = 180^\circ - 130^\circ = 50^\circ \] NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-3_Q17 Since \(\displaystyle AB\) is a diameter, \[\angle ACB = 90^\circ \quad \text{(angle in a semicircle)} \] In \(\displaystyle \triangle ABC\), \[\angle BAC + \angle ABC + \angle ACB = 180^\circ \] \[\angle BAC = 180^\circ - 50^\circ - 90^\circ = 40^\circ \] Answer: \(\displaystyle \angle BAC = 40^\circ\)

  8. Exercise 18

    Two circles with centres O and O′\displaystyle \mathrm{O}^{\prime} intersect at two points A and B . A line PQ is drawn parallel to OO′\displaystyle \mathrm{OO}^{\prime} through A (or B ) intersecting the circles at P and Q . Prove that PQ=2OO′\displaystyle \mathrm{PQ}=2 \mathrm{OO}^{\prime}.

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    Draw \(\displaystyle OM \perp PQ\) and \(\displaystyle O'N \perp PQ\), feet \(\displaystyle M\), \(\displaystyle N\) on line \(\displaystyle PQ\).NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-3_Q18 \[OM \perp PA \Rightarrow PM = MA \quad \text{(perpendicular from centre bisects a chord)} \] \[O'N \perp AQ \Rightarrow AN = NQ \quad \text{(perpendicular from centre bisects a chord)} \] \(\displaystyle OM \parallel O'N\) (both \(\displaystyle \perp PQ\)) and \(\displaystyle OO' \parallel PQ\), so \(\displaystyle OMNO'\) is a rectangle: \[MN = OO' \] \[PQ = PM + MA + AN + NQ = 2(MA+AN) = 2MN \] \[PQ = 2\,OO' \]Answer: \(\displaystyle PQ = 2\,OO' \).

  9. Exercise 19

    In Fig.10.15, AOB is a diameter of the circle and C, D, E are any three points on the semi-circle. Find the value of ∠ACD+∠BED\displaystyle \angle \mathrm{ACD}+\angle \mathrm{BED}. NCERT_Question_Class9_Maths_Exemplar_Ch10_Ex10-3_Q19

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    Since \(\displaystyle AB\) is a diameter, \[\angle ADB = 90^\circ \quad \text{(angle in a semicircle)} \] NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-3_Q19 \(\displaystyle ACDB\) is a cyclic quadrilateral: \[\angle ACD + \angle ABD = 180^\circ \] \(\displaystyle ADEB\) is a cyclic quadrilateral: \[\angle DAB + \angle DEB = 180^\circ \] Adding, and using \(\displaystyle \angle DAB + \angle ABD = 180^\circ - \angle ADB = 90^\circ\) in \(\displaystyle \triangle ADB\): \[\angle ACD + \angle DEB = 360^\circ - (\angle DAB+\angle ABD) = 360^\circ - 90^\circ = 270^\circ \]Answer: \(\displaystyle \angle ACD + \angle BED = 270^\circ\).NCERT prints: $\displaystyle 278$° — a misprint: the working gives $\displaystyle 360$° − ∠ADB = $\displaystyle 360$° − $\displaystyle 90$° = $\displaystyle 270$°.

  10. Exercise 20

    In Fig. 10.16\displaystyle 10.16, ∠OAB=30∘\displaystyle \angle \mathrm{OAB}=30^{\circ} and ∠OCB=57∘\displaystyle \angle \mathrm{OCB}=57^{\circ}. Find ∠BOC\displaystyle \angle \mathrm{BOC} and ∠AOC\displaystyle \angle \mathrm{AOC}. NCERT_Question_Class9_Maths_Exemplar_Ch10_Ex10-3_Q20

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    NCERT’s answer
    $\displaystyle \angle \mathrm{BOC}=66^{\circ}, \angle \mathrm{AOC}=54^{\circ}$
    \(\displaystyle OA = OB\) (radii), so \[\angle OBA = \angle OAB = 30^\circ \] \[\angle AOB = 180^\circ - 30^\circ - 30^\circ = 120^\circ \] NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-3_Q20 \(\displaystyle OC = OB\) (radii), so \[\angle OBC = \angle OCB = 57^\circ \] \[\angle BOC = 180^\circ - 57^\circ - 57^\circ = 66^\circ \] \[\angle AOC = \angle AOB - \angle BOC = 120^\circ - 66^\circ = 54^\circ \] Answer: \(\displaystyle \angle BOC = 66^\circ,\ \angle AOC = 54^\circ\)