NCERT Solutions for Class 9 Maths Chapter 10: Circles

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    Q. Two circles of radius 10 cm and 8 cm intersect and the length of the common chord is 12 cm. Find the distance between their centres.

    Ans. Let O and O' be the centres of the circles.

    Two circles

    $$\text{We have OP = 10 cm, O’P = 8 cm and PQ = 12 cm (their common chord)}\\PL =\frac{1}{2}PQ = 6 cm\\ \text{[Since perpendicular from centre bisect the chord] In right triangle OLP , we have}\\ OP^2 = OL^2 + LP^2\\ ⇒ OL =\sqrt{\mathstrut OP^2LP^2} =\sqrt{\mathstrut 10^2-6^2}=\sqrt{\mathstrut 64}=8 cm\\ \text{In right DO’LP, we have}\\ O’P^2 = O’L^2 + LP^2\\ ⇒ O’L =\sqrt{\mathstrut O’P^2-O’L^2}\\ =\sqrt{\mathstrut 8^2-6^2}\\ =\sqrt{\mathstrut 28}\\ = 5.29 cm\\ OO’ = OL + LO’\\ = 8 + 5.29\\ = 13.29 cm\\\text{Hence, the distance between the centres of the two circles is 13.29 cm.}$$

    Q. Two circles of radius 10 cm and 8 cm intersect and the length of the common chord is 12 cm. Find the distance between their centres.

    circles

    Ans. Given : A circle with centre O.
    ∠OAB = 30°
    ∠OCB = 57°
    Now, in DOAB
    OA = OB = radius of circle
    ∠OAB = ∠OBA = 30°
    And in DBOC,
    ∠OCB = ∠OBC = 57° [... OB = OC]
    ∠OCB + ∠OBC + ∠BOC = 180°
    ⇒ 57° + 57° + ∠BOC = 180°
    ⇒ ∠BOC = 180° – 114°
    = 66°
    Now , in DOAB
    ∠OAB + ∠OBA + ∠AOB = 180°
    ⇒ ∠AOB = 180° – (30° + 30°) = 120°
    ∠AOC = ∠AOB – ∠BOC
    = 120° – 66°
    = 54°
    Hence, m∠BOC = 66° and m∠AOC = 54°.

    Q. A circular ground of radius 40 m is situated in a sport club stadium. Three boys Ankur, Amit and Anand are standing at equal distance on its boundary, each having strings in their hands to form a triangle. Find the total length of the string used.

    circular ground

    Ans.

    $$\text{According to the given situation, we have}\\\widehat{AB}=\widehat{BC}=\widehat{CA}\\ ⇒ chord AB = chord BC = chord CA\\ ⇒ \text{△ABC is an equilateral triangle.}\\ \text{Centroid O of DABC coincides with its circumcentre.}\\ ⇒ AO = OB = OC = 40 m\\ \text{Since, O divides AD in the ratio 2 : 1}\\ \text{Therefore,}\frac{OA}{OD}=\frac{2}{1}\\ ⇒\frac{40}{OD}=\frac{2}{1}⇒ OD = 20 m\\ In △BOD, we have\\ BO^2 = BD^2 + OD^2\\ ⇒ (40)^2 = BD^2 + (20)^2\\ BD =\sqrt{\mathstrut (40-20)(40+20)}$$

    Q. Two equal chords AB and CD of a circle when produced intersect at point P. Prove that PB = PD.

    Ans. Given : AB and CD are equal two chords of a circle
    C(O, r).
    These chords produced meet at point P .
    Two equal chords

    $$\text{To prove} : PB = PD\\ \text{Construction : Join OP, and draw perpendicular from O to AB at Q and OR at CD.}\\ Proof : Since, AB = CD (Given)\\ ⇒ OQ = OR\\ \text{[Equal chords are equidistant from the centre]}\\ \text{In △OQP and △ORP}\\ OQ = OR \space\text{[Proved above]}\\ \text{OP is common.}\\ ∠OQP = ∠ORP = 90°\\ △OQP ≅ △ORP \text{(By RHS congruency)}\\ ⇒ QP = RP (By C.P.C.T) …(i)\\ Now,\space AB = CD \space(Given)\\ ⇒\frac{1}{2}AB=\frac{1}{2}CD\\ ⇒ QB = RD …(ii)\\ \text{(Since, perpendicular from centre bisects the chord) Subtracting (ii) from (i), we get}\\ QP – QB = RP – RD\\ ⇒ BP = PD \space Hance$$

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