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NCERT Exemplar · Class 10 Mathematics Circles

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EXERCISE 9.4 1–10 (part 4 of 5)

  1. Exercise 1

    If a hexagon ABCDEF circumscribe a circle, prove that AB+CD+EF=BC+DE+FA\displaystyle \mathrm{AB}+\mathrm{CD}+\mathrm{EF}=\mathrm{BC}+\mathrm{DE}+\mathrm{FA}.

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    Since the circle touches AB, BC, CD, DE, EF, FA at P, Q, R, S, T, U respectively, tangents from each vertex are equal: \[AP=AU,\quad BP=BQ,\quad CQ=CR,\quad DR=DS,\quad ES=ET,\quad FT=FU \] \[AB+CD+EF=(AP+PB)+(CR+RD)+(ET+TF) \] \[=(AU+BQ)+(CQ+DS)+(ES+FU) \] \[=(BQ+CQ)+(DS+ES)+(FU+AU) \] \[=BC+DE+FA \] NCERT_Solution_Class10_Maths_Exemplar_Ch9_Ex9-4_Q1 Answer: \(\displaystyle AB+CD+EF=BC+DE+FA\)

  2. Exercise 2

    Let s\displaystyle s denote the semi-perimeter of a triangle ABC in which BC=a,CA=b\displaystyle \mathrm{BC}=a, \mathrm{CA}=b, AB=c\displaystyle \mathrm{AB}=c. If a circle touches the sides BC, CA, AB at D, E, F, respectively, prove that BD=s−b\displaystyle \mathrm{BD}=s-b.

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    Let the equal tangent lengths from each vertex be \(\displaystyle AF=AE=x,\ BF=BD=y,\ CD=CE=z\). \[c=AB=x+y,\quad a=BC=y+z,\quad b=CA=z+x \] Adding all three, \[a+b+c=2(x+y+z)\ \Rightarrow\ x+y+z=s \] \[BD=y=s-(x+z)=s-b \] NCERT_Solution_Class10_Maths_Exemplar_Ch9_Ex9-4_Q2 Answer: \(\displaystyle BD=s-b\)

  3. Exercise 3

    From an external point P, two tangents, PA and PB are drawn to a circle with centre O. At one point E on the circle tangent is drawn which intersects PA and PB at C and D, respectively. If PA=10 cm\displaystyle \mathrm{PA}=10 \mathrm{~cm}, find the the perimeter of the triangle PCD.

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    NCERT’s answer
    $\displaystyle 20$ cm
    Since PA, PB and the tangent CD (touching at E) are tangents to the same circle, \[CA=CE,\quad DB=DE \] \[\text{Perimeter}(\triangle PCD)=PC+CD+DP=PC+(CE+ED)+DP \] \[=(PC+CA)+(DB+DP)=PA+PB \] Since \(\displaystyle PA=PB=10\text{ cm}\) (tangents from P), \[\text{Perimeter}(\triangle PCD)=10+10=20\text{ cm} \] NCERT_Solution_Class10_Maths_Exemplar_Ch9_Ex9-4_Q3 Answer: $\displaystyle 20$ cm

  4. Exercise 4

    If AB is a chord of a circle with centre O, AOC is a diameter and AT is the tangent at A as shown in Fig. 9.17. Prove that ∠BAT=∠ACB\angle \mathrm{BAT}=\angle \mathrm{ACB} NCERT_Question_Class10_Maths_Exemplar_Ch9_Ex9-4_Q4

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    Since AOC is a diameter, \(\displaystyle \angle ABC=90^\circ\) (angle in a semicircle). \[\angle ACB+\angle BAC=90^\circ \] Since AT is tangent at A and AC is a diameter, \(\displaystyle AC\perp AT\): \[\angle BAT+\angle BAC=\angle CAT=90^\circ \] \[\therefore\ \angle BAT=90^\circ-\angle BAC=\angle ACB \] NCERT_Solution_Class10_Maths_Exemplar_Ch9_Ex9-4_Q4 Answer: \(\displaystyle \angle BAT=\angle ACB\) (proved)

  5. Exercise 5

    Two circles with centres O and O′\displaystyle \mathrm{O}^{\prime} of radii 3\displaystyle 3 cm and 4\displaystyle 4 cm, respectively intersect at two points P and Q such that OP and O′P\displaystyle \mathrm{O}^{\prime} \mathrm{P} are tangents to the two circles. Find the length of the common chord PQ.

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    NCERT’s answer
    4.$\displaystyle 8$ cm
    Since OP is tangent to the circle centred \(\displaystyle O'\) at P, \(\displaystyle OP\perp O'P\), so \(\displaystyle \triangle OPO'\) is right-angled at P. \[OO'=\sqrt{OP^2+O'P^2}=\sqrt{3^2+4^2}=5\text{ cm} \] \(\displaystyle OO'\) is the perpendicular bisector of the common chord PQ, so half of PQ equals the altitude from P to \(\displaystyle OO'\): \[\tfrac12 PQ=\dfrac{OP\cdot O'P}{OO'}=\dfrac{3\times4}{5}=2.4\text{ cm} \] \[PQ=2\times2.4=4.8\text{ cm} \] NCERT_Solution_Class10_Maths_Exemplar_Ch9_Ex9-4_Q5 Answer: PQ = $\displaystyle 4.8$ cm

  6. Exercise 6

    In a right triangle ABC in which ∠B=90∘\displaystyle \angle \mathrm{B}=90^{\circ}, a circle is drawn with AB as diameter intersecting the hypotenuse AC and P. Prove that the tangent to the circle at P bisects BC.

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    \[\angle APB=90^\circ \quad\text{(angle in semicircle, AB diameter)} \] \[\therefore\ BP\perp AC,\qquad \angle BPC=90^\circ \] Since \(\displaystyle \angle ABC=90^\circ\), \(\displaystyle BC\perp AB\), so BC is tangent to the circle at B. Let the tangent at P meet BC at Q. \[QB=QP \quad\text{(tangents from Q)} \] \[\angle QPB=\angle QBP=\angle PBC \] \[\angle QPC=90^\circ-\angle QPB=90^\circ-\angle PBC=\angle PCB \quad\text{(}\angle PBC+\angle PCB=90^\circ\text{ in }\triangle PBC\text{)} \] \[\therefore\ QP=QC \] \[QB=QP=QC \] NCERT_Solution_Class10_Maths_Exemplar_Ch9_Ex9-4_Q6 Answer: The tangent at P bisects BC (proved)

  7. Exercise 7

    In Fig. 9.18\displaystyle 9.18, tangents PQ and PR are drawn to a circle such that ∠RPQ=30∘\displaystyle \angle \mathrm{RPQ}=30^{\circ}. A chord RS is drawn parallel to the tangent PQ. Find the ∠RQS\displaystyle \angle \mathrm{RQS}. [Hint: Draw a line through Q and perpendicular to QP.] NCERT_Question_Class10_Maths_Exemplar_Ch9_Ex9-4_Q7

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    NCERT’s answer
    \(\displaystyle 30^{\circ}\)
    \[\angle PQR=\angle PRQ=\dfrac{180^\circ-30^\circ}{2}=75^\circ \quad\text{(}PQ=PR\text{, tangents from P)} \] \[\angle QRS=\angle PQR=75^\circ \quad\text{(alternate angles, }RS\parallel PQ\text{)} \] \[\angle QSR=\angle PQR=75^\circ \quad\text{(tangent-chord angle = angle in alternate segment)} \] \[\angle RQS=180^\circ-\angle QRS-\angle QSR=180^\circ-75^\circ-75^\circ=30^\circ \] NCERT_Solution_Class10_Maths_Exemplar_Ch9_Ex9-4_Q7 Answer: \(\displaystyle \angle RQS=30^\circ\)

  8. Exercise 8

    AB is a diameter and AC is a chord of a circle with centre O such that ∠BAC=30∘\displaystyle \angle \mathrm{BAC}=30^{\circ}. The tangent at C intersects extended AB at a point D. Prove that BC=BD\displaystyle \mathrm{BC}=\mathrm{BD}.

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    \[\angle ACB = 90^\circ \quad \text{(angle in a semicircle, } AB \text{ diameter)} \] \[\angle ABC = 180^\circ-90^\circ-30^\circ = 60^\circ \quad \text{(angle sum, } \triangle ABC\text{)} \] \[\angle CBD = 180^\circ-\angle ABC = 120^\circ \quad \text{(linear pair, } D \text{ on } AB \text{ extended)} \] \[\angle BCD = \angle BAC = 30^\circ \quad \text{(tangent–chord} = \text{alternate segment)} \] \[\angle BDC = 180^\circ-120^\circ-30^\circ = 30^\circ \quad \text{(angle sum, } \triangle BCD\text{)} \] \[BC = BD \quad (\angle BCD=\angle BDC \Rightarrow \text{sides opposite equal angles are equal}) \]NCERT_Solution_Class10_Maths_Exemplar_Ch9_Ex9-4_Q8

  9. Exercise 9

    Prove that the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc.

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    \[MA = MB \quad \text{(equal arcs, equal chords)} \] \[OA = OB \quad \text{(radii)} \] \[OM \perp AB \quad \text{(O and M are each equidistant from A and B)} \] \[OM \perp PQ \quad \text{(tangent} \perp \text{radius at the point of contact } M) \] \[\therefore AB \parallel PQ \quad \text{(both} \perp OM) \]NCERT_Solution_Class10_Maths_Exemplar_Ch9_Ex9-4_Q9

  10. Exercise 10

    In Fig. 9.19\displaystyle 9.19, the common tangent, AB and CD to two circles with centres O and O′\displaystyle \mathrm{O}^{\prime} intersect at E. Prove that the points O,E,O′\displaystyle \mathrm{O}, \mathrm{E}, \mathrm{O}^{\prime} are collinear. NCERT_Question_Class10_Maths_Exemplar_Ch9_Ex9-4_Q10

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    \[EA = EC \quad (\text{tangents from } E \text{ to circle } O) \] \[\triangle OAE \cong \triangle OCE \quad (OA=OC,\ OE=OE,\ EA=EC) \Rightarrow \angle OEA = \angle OEC \] \[EB = ED \quad (\text{tangents from } E \text{ to circle } O') \] \[\triangle O'BE \cong \triangle O'DE \quad (O'B=O'D,\ O'E=O'E,\ EB=ED) \Rightarrow \angle O'EB = \angle O'ED \] \[\angle AEC = \angle BED \quad (\text{vertically opposite angles}) \] \[\angle OEA = \tfrac12\angle AEC = \tfrac12\angle BED = \angle O'EB \] \[\angle AEO+\angle OEB = 180^\circ \quad (A,E,B \text{ collinear}) \] \[\Rightarrow \angle OEB+\angle BEO' = 180^\circ \] \[\therefore O,\,E,\,O' \text{ are collinear} \]NCERT_Solution_Class10_Maths_Exemplar_Ch9_Ex9-4_Q10