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NCERT Exemplar · Class 10 Mathematics Circles

44 questions · 44 still being checked

EXERCISE 9.1 1–10 (part 1 of 5)

  1. Choose the correct answer from the given four options:

    Exercise 1

    If radii of two concentric circles are 4\displaystyle 4 cm and 5\displaystyle 5 cm, then the length of each chord of one circle which is tangent to the other circle is
    (A)
    3\displaystyle 3 cm
    (B)
    6\displaystyle 6 cm
    (C)
    9\displaystyle 9 cm
    (D)
    1\displaystyle 1 cm

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    (B)
    (B) \(\displaystyle 6\text{ cm}\) NCERT_Solution_Class10_Maths_Exemplar_Ch9_Ex9-1_Q1 \[OM \perp AB, \quad OM = 4,\ OA = 5 \] \[AM = \sqrt{OA^2-OM^2} = \sqrt{25-16} = 3 \] \[AB = 2\,AM = 6 \]

  2. Exercise 2

    In Fig. 9.3\displaystyle 9.3, if ∠AOB=125∘\displaystyle \angle \mathrm{AOB}=125^{\circ}, then ∠COD\displaystyle \angle \mathrm{COD} is equal to
    (A)
    62.5\displaystyle 5°
    (B)
    45\displaystyle 45°
    (C)
    35\displaystyle 35°
    (D)
    55\displaystyle 55°
    NCERT_Question_Class10_Maths_Exemplar_Ch9_Ex9-1_Q2

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    (D)
    (D) \(\displaystyle 55^{\circ}\) \[AB,\,BC,\,CD,\,DA \text{ tangent to the circle} \;\Rightarrow\; \angle AOB+\angle COD=180^{\circ} \] \[\angle COD = 180^{\circ}-125^{\circ}=55^{\circ} \]

  3. Exercise 3

    In Fig. 9.4\displaystyle 9.4, AB is a chord of the circle and AOC is its diameter such that ∠ACB=50∘\displaystyle \angle \mathrm{ACB}=50^{\circ}. If AT is the tangent to the circle at the point A, then ∠BAT\displaystyle \angle \mathrm{BAT} is equal to
    (A)
    65\displaystyle 65°
    (B)
    60\displaystyle 60°
    (C)
    50\displaystyle 50°
    (D)
    40\displaystyle 40°
    NCERT_Question_Class10_Maths_Exemplar_Ch9_Ex9-1_Q3

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    (C)
    (C) \(\displaystyle 50^{\circ}\) \[AOC \text{ diameter} \;\Rightarrow\; \angle ABC = 90^{\circ} \] \[\angle BAC = 180^{\circ}-90^{\circ}-50^{\circ}=40^{\circ} \] \[AT \perp AC \text{ (tangent} \perp \text{radius)} \;\Rightarrow\; \angle CAT = 90^{\circ} \] \[\angle BAT = \angle CAT-\angle BAC = 90^{\circ}-40^{\circ}=50^{\circ} \]

  4. Exercise 4

    From a point P which is at a distance of 13\displaystyle 13 cm from the centre O of a circle of radius 5\displaystyle 5 cm, the pair of tangents PQ and PR to the circle are drawn. Then the area of the quadrilateral PQOR is
    (A)
    60 cm2\displaystyle 60 \mathrm{~cm}^2
    (B)
    65 cm2\displaystyle 65 \mathrm{~cm}^2
    (C)
    30 cm2\displaystyle 30 \mathrm{~cm}^2
    (D)
    32.5 cm2\displaystyle 32.5 \mathrm{~cm}^2

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    (A)
    (A) \(\displaystyle 60\text{ cm}^2\) NCERT_Solution_Class10_Maths_Exemplar_Ch9_Ex9-1_Q4 \[PQ = PR = \sqrt{OP^2-OQ^2} = \sqrt{13^2-5^2} = 12 \] \[\text{ar}(PQOR) = 2\times\tfrac{1}{2}\,OQ\cdot PQ = OQ\cdot PQ = 5\times12 = 60 \]

  5. Exercise 5

    At one end A of a diameter AB of a circle of radius 5\displaystyle 5 cm, tangent XAY is drawn to the circle. The length of the chord CD parallel to XY and at a distance 8\displaystyle 8 cm from A is
    (A)
    4\displaystyle 4 cm
    (B)
    5\displaystyle 5 cm
    (C)
    6\displaystyle 6 cm
    (D)
    8\displaystyle 8 cm

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    NCERT’s answer
    (D)
    (D) \(\displaystyle 8\text{ cm}\) NCERT_Solution_Class10_Maths_Exemplar_Ch9_Ex9-1_Q5 \[OA=5,\quad OA \perp XY \] \[\text{dist}(O,CD)=8-5=3 \] \[\tfrac{1}{2}CD=\sqrt{OA^2-3^2}=\sqrt{25-9}=4 \] \[CD=8 \]

  6. Exercise 6

    In Fig. 9.5\displaystyle 9.5, AT is a tangent to the circle with centre O such that OT =4 cm\displaystyle =4 \mathrm{~cm} and ∠OTA=30∘\displaystyle \angle \mathrm{OTA}=30^{\circ}. Then AT is equal to
    (A)
    4\displaystyle 4 cm
    (B)
    2\displaystyle 2 cm
    (C)
    23 cm\displaystyle 2 \sqrt{3} \mathrm{~cm}
    (D)
    43 cm\displaystyle 4 \sqrt{3} \mathrm{~cm}
    NCERT_Question_Class10_Maths_Exemplar_Ch9_Ex9-1_Q6

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    NCERT’s answer
    (C)
    (C) \(\displaystyle 2\sqrt3\text{ cm}\) OA \(\displaystyle \perp\) AT (radius \(\displaystyle \perp\) tangent at the point of contact). \[\cos 30^\circ = \frac{AT}{OT} \] \[AT = OT\cos 30^\circ = 4\cdot\frac{\sqrt3}{2} = 2\sqrt3\text{ cm} \] NCERT_Solution_Class10_Maths_Exemplar_Ch9_Ex9-1_Q6

  7. Exercise 7

    In Fig. 9.6\displaystyle 9.6, if O is the centre of a circle, PQ is a chord and the tangent PR at P makes an angle of 50∘\displaystyle 50^{\circ} with PQ, then ∠POQ\displaystyle \angle \mathrm{POQ} is equal to
    (A)
    100\displaystyle 100°
    (B)
    80\displaystyle 80°
    (C)
    90\displaystyle 90°
    (D)
    75∘\displaystyle 75^{\circ}
    NCERT_Question_Class10_Maths_Exemplar_Ch9_Ex9-1_Q7

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    NCERT’s answer
    (A)
    (A) $\displaystyle 100$° \[\angle OPQ = 90^\circ - 50^\circ = 40^\circ \quad \text{(OP \(\displaystyle \perp\) PR, radius \(\displaystyle \perp\) tangent)} \] \[\angle OQP = \angle OPQ = 40^\circ \quad \text{(OP = OQ, radii)} \] \[\angle POQ = 180^\circ - 40^\circ - 40^\circ = 100^\circ \] NCERT_Solution_Class10_Maths_Exemplar_Ch9_Ex9-1_Q7

  8. Exercise 8

    In Fig. 9.7\displaystyle 9.7, if PA and PB are tangents to the circle with centre O such that ∠APB=50∘\displaystyle \angle \mathrm{APB}=50^{\circ}, then ∠OAB\displaystyle \angle \mathrm{OAB} is equal to
    (A)
    25\displaystyle 25°
    (B)
    30\displaystyle 30°
    (C)
    40\displaystyle 40°
    (D)
    50\displaystyle 50°
    NCERT_Question_Class10_Maths_Exemplar_Ch9_Ex9-1_Q8

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    NCERT’s answer
    (A)
    (A) $\displaystyle 25$° \[\angle PAB = \angle PBA = \frac{180^\circ-50^\circ}{2} = 65^\circ \quad \text{(PA = PB, tangents from P)} \] \[\angle OAP = 90^\circ \quad \text{(radius \(\displaystyle \perp\) tangent)} \] \[\angle OAB = \angle OAP - \angle PAB = 90^\circ - 65^\circ = 25^\circ \] NCERT_Solution_Class10_Maths_Exemplar_Ch9_Ex9-1_Q8

  9. Exercise 9

    If two tangents inclined at an angle 60\displaystyle 60° are drawn to a circle of radius 3\displaystyle 3 cm, then length of each tangent is equal to
    (A)
    323 cm\displaystyle \frac{3}{2} \sqrt{3} \mathrm{~cm}
    (B)
    6\displaystyle 6 cm
    (C)
    3\displaystyle 3 cm
    (D)
    33 cm\displaystyle 3 \sqrt{3} \mathrm{~cm}

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    NCERT’s answer
    (D)
    (D) \(\displaystyle 3\sqrt3\text{ cm}\) OP bisects \(\displaystyle \angle APB\), so \(\displaystyle \angle OPA = 30^\circ\); OA \(\displaystyle \perp\) PA. \[\tan 30^\circ = \frac{OA}{PA} \] \[PA = \frac{OA}{\tan 30^\circ} = \frac{3}{1/\sqrt3} = 3\sqrt3\text{ cm} \] NCERT_Solution_Class10_Maths_Exemplar_Ch9_Ex9-1_Q9

  10. Exercise 10

    In Fig. 9.8\displaystyle 9.8, if PQR is the tangent to a circle at Q whose centre is O, AB is a chord parallel to PR and ∠BQR=70∘\displaystyle \angle \mathrm{BQR}=70^{\circ}, then ∠AQB\displaystyle \angle \mathrm{AQB} is equal to
    (A)
    20\displaystyle 20°
    (B)
    40\displaystyle 40°
    (C)
    35\displaystyle 35°
    (D)
    45\displaystyle 45°
    NCERT_Question_Class10_Maths_Exemplar_Ch9_Ex9-1_Q10

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    NCERT’s answer
    (B)
    (B) $\displaystyle 40$° \[OQ \perp PR,\ AB \parallel PR \ \Rightarrow\ OQ \perp AB \ \Rightarrow\ QA = QB \] \[\angle QAB = \angle BQR = 70^\circ \quad \text{(tangent\textendash chord = angle in alternate segment)} \] \[\angle QBA = \angle QAB = 70^\circ \quad (QA=QB) \] \[\angle AQB = 180^\circ - 70^\circ - 70^\circ = 40^\circ \] NCERT_Solution_Class10_Maths_Exemplar_Ch9_Ex9-1_Q10