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Circle Test 1
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Circle Test 1
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  • Question 1/10
    1 / -0

    In the adjoining figure, a smaller circle touches a larger circle internally and passes through the centre O of the larger circle. If the area of the smaller circle is 200 cm2, the area of the larger circle in sq cm is

    Solutions

    Let the radius of the larger circle be R, then radius of smaller circle = 

    ∴ π  = 200 ⇒ πR2 = 4 × 200

                            πR2 = 800

    Here, area of larger circle = 800 cm2

  • Question 2/10
    1 / -0

    The values of x and y in the figure are measure of angles, then x + y is equal to

    Solutions

    As, ∠B + ∠D = 180° and ∠A + ∠C = 180°

    x + 10 + 5y + 5 = 180°

    x + 5y = 165°

    2x + 4 + 4y – 4 = 180°

    2x + 4y = 180°

    Solving, x and y are 40° and 25°

    X + y = 40° + 25° = 65°

  • Question 3/10
    1 / -0

    The incircle of a ∆ABC touches the sides AB, BC and AC at the points P, Q, R, respectively, then which of the following statements is/are correct?

    I. AP + BQ + CR = PB + QC + RA

    II. AP + BQ + CR = 1 half (perimeter of ∆ABC)

    III. AP + BQ + CR = 3(AB + BC + CA)

    Solutions

    As, the tangents drawn from an external point to a circle are equal.

    ∴         AP = AR, BQ = BP

    And     CR = QC

    ∴ AP + BQ + CR = BP + QC + RA and perimeter of

    ∆ ABC = AB + BC + CA

    = (AP + PB) + (BQ + QC) + (OR + RA)

    = (AP + BQ) + (BQ + CR) + CR + AP)

    = 2(AP + BQ + CR)

    ∴ AP + BQ + CR =  (perimeter of ∆ ABC)

  • Question 4/10
    1 / -0

    If O is the centre of the circle, then ‘x’ is

    Solutions

    ∠ACB = ∠ADB = 32°

    ∠ACD = ∠ABD = 50°

    ∴         x = ∠BCD = ∠ACB + ∠ACD = 82°

  • Question 5/10
    1 / -0

    In the adjoining figure, POQ is the diameter of the circle, R and S are any two points on the circles. Then,

    Solutions

    ∠PRQ = ∠PSQ = 90°                      (Each angle in semi-circle)

  • Question 6/10
    1 / -0

    It is not possible to draw a circle having its centre on a fixed straight line l and passing through two points A and B not on l, if

    Solutions

    Let P be a fixed point on l. Then, a circle can be drawn through the points P, A, B only when PA = PB. When l ⊥ AB and ldoes not bisect AB, then PA ≠ PB, so in this case, the circle cannot be drawn to pass through P, A, B.

  • Question 7/10
    1 / -0

    Diameter AB and CD of a circle intersect at O. If m ∠BOD = 50°, then m OD is

    Solutions

    m ∠BOD = 50°

    ⇒                    ∠BOD = 50°

    ⇒                    ∠AOD = 180° – ∠BOD = 130°

    ⇒        m OD = 130°

  • Question 8/10
    1 / -0

    In the given figure, OM and ON are the perpendiculars drawn on the chords PQ and RS. If OM = ON = 6 cm. Then,

    Solutions

    PQ = RS as equal chords of circle are equidistant from the distance.

  • Question 9/10
    1 / -0

    If two circles C1 and C2 have three points in common then

    Solutions

    Clearly, two circle can intersect at two points only.

  • Question 10/10
    1 / -0

     

    An equilateral ∆ABC is inscribed in a circle with centre O. Then, ∠BOC is equal to

    Solutions

    Method I. ∠BOC = 90° +  ∠A = 90° +  60°

    ⇒        ∠BOC = 120°

    Method II. ∠BOC = 2∠A = 2 × 60°

    ⇒        ∠BOC = 120°

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