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NCERT Exemplar · Class 9 Mathematics Circles

54 questions · 54 still being checked

EXERCISE 10.4 11–14 (part 6 of 6)

  1. Exercise 11

    Two equal chords AB and CD of a circle when produced intersect at a point P. Prove that PB=PD\displaystyle \mathrm{PB}=\mathrm{PD}.

    Not cross-checked

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

    Let \(\displaystyle M,N\) be the feet of the perpendiculars from centre \(\displaystyle O\) to \(\displaystyle AB, CD\). NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-4_Q11 \[AM=MB=\tfrac12AB,\qquad CN=ND=\tfrac12CD \quad \text{(perpendicular from centre bisects chord)} \] \[AB=CD \ \Rightarrow\ MB=ND \qquad (i) \] \[OM=ON \quad \text{(equal chords are equidistant from centre)} \] \[\angle OMP=\angle ONP=90^\circ,\ OP=OP \ \Rightarrow\ \triangle OMP\cong\triangle ONP \quad \text{(RHS)} \] \[\Rightarrow PM=PN \qquad (ii) \] \[PB=PM-MB,\qquad PD=PN-ND \] \[\Rightarrow PB=PD \quad \text{[by (i), (ii)]} \]Answer: \(\displaystyle PB = PD \).

  2. Exercise 12

    AB and AC are two chords of a circle of radius r\displaystyle r such that AB=2AC\displaystyle \mathrm{AB}=2 \mathrm{AC}. If p\displaystyle p and q\displaystyle q are the distances of AB and AC from the centre, prove that 4q2=p2+3r2\displaystyle 4 q^{2}=p^{2}+3 r^{2}.

    Not cross-checked

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

    Let \(\displaystyle M,N\) be the midpoints of \(\displaystyle AB, AC\), so \(\displaystyle OM=p,\ ON=q\) (perpendicular from the centre bisects a chord). Let \(\displaystyle AC=a\), so \(\displaystyle AB=2a\). NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-4_Q12 \[OM^2+\left(\tfrac{AB}2\right)^2=OB^2 \ \Rightarrow\ p^2+a^2=r^2 \qquad (i) \] \[ON^2+\left(\tfrac{AC}2\right)^2=OC^2 \ \Rightarrow\ q^2+\tfrac{a^2}4=r^2 \qquad (ii) \] \[\text{From (ii):}\ a^2=4(r^2-q^2) \] \[\text{Substituting in (i):}\ p^2+4r^2-4q^2=r^2 \] \[\Rightarrow 4q^2=p^2+3r^2 \]Answer: \(\displaystyle 4q^2 = p^2 + 3r^2 \).

  3. Exercise 13

    In Fig. 10.20\displaystyle 10.20, O is the centre of the circle, ∠BCO=30∘\displaystyle \angle \mathrm{BCO}=30^{\circ}. Find x\displaystyle x and y\displaystyle y. NCERT_Question_Class9_Maths_Exemplar_Ch10_Ex10-4_Q13

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    NCERT’s answer
    $\displaystyle x=30^{\circ}, y=15^{\circ}$
    \[OB = OC \quad \text{(radii)} \Rightarrow \angle OBC = \angle OCB = 30^\circ \] \[\angle BOC = 180^\circ - 30^\circ - 30^\circ = 120^\circ \] NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-4_Q13 \(\displaystyle OA\) extended meets \(\displaystyle BC\) at its midpoint \(\displaystyle M\), at right angles. \[\angle MOB = \angle MOC = \tfrac{1}{2}\angle BOC = 60^\circ \quad \text{(OM bisects the apex angle of isosceles } \triangle OBC\text{)} \] \[\angle AOB = \angle AOC = 180^\circ - 60^\circ = 120^\circ \quad (A, O, M \text{ collinear}) \] \[\triangle OAB,\ OA = OB \Rightarrow x = \angle OAB = \tfrac{180^\circ - 120^\circ}{2} = 30^\circ \] \[\angle DOC = \angle AOC - \angle AOD = 120^\circ - 90^\circ = 30^\circ \] \[y = \angle DBC = \tfrac{1}{2}\angle DOC = 15^\circ \quad \text{(angle at centre = twice at circle)} \] Answer: \(\displaystyle x = 30^\circ,\ y = 15^\circ \)

  4. Exercise 14

    In Fig. 10.21\displaystyle 10.21, O is the centre of the circle, BD=OD\displaystyle \mathrm{BD}=\mathrm{OD} and CD⊥AB\displaystyle \mathrm{CD} \perp \mathrm{AB}. Find ∠CAB\displaystyle \angle \mathrm{CAB}. NCERT_Question_Class9_Maths_Exemplar_Ch10_Ex10-4_Q14

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    $\displaystyle 30^{\circ}$
    \[OB = OD \quad \text{(radii)}, \quad BD = OD \quad \text{(given)} \Rightarrow OB = OD = BD \] \[\triangle OBD \text{ equilateral} \Rightarrow \angle BOD = 60^\circ \] NCERT_Solution_Class9_Maths_Exemplar_Ch10_Ex10-4_Q14 Chord \(\displaystyle CD \perp AB\) at \(\displaystyle N\); the diameter \(\displaystyle AB\) reflects \(\displaystyle C\) onto \(\displaystyle D\). \[\angle BOC = \angle BOD = 60^\circ \] \[\angle CAB = \tfrac{1}{2}\angle BOC = 30^\circ \quad \text{(angle at centre = twice at circle)} \] Answer: \(\displaystyle \angle CAB = 30^\circ \)