In Fig. 12.33, ABC is a quadrant of a circle of radius 14 cm and
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    Areas Related to Circles

    Multiple Choice QuestionsShort Answer Type

    31.

    In Fig. 12.30, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the (i) quadrant OACB, (ii) shaded region. 

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    32.

    In Fig. 12.31, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. (Use π = 3.14).


    Fig. 12.31

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    33.

    AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see Fig. 12.32). If ∠ AOB = 30°, find the area of the shaded region.


    Fig. 12.32

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    Multiple Choice QuestionsLong Answer Type

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    34.

    In Fig. 12.33, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region. 


    Fig. 12.33

    Let points P and Q on arcs BPC and BQC respectively with BC as diameter

    Let points P and Q on arcs BPC and BQC respectively with BC as diamet

    Now in increment ABC, we have
    BC2 = AB2 + AC2 
                  [Using pythagoras theorem]
    rightwards double arrow   BC2 = 142 + 142
    rightwards double arrow   BC  = square root of 196 plus 196 end root
    rightwards double arrow   BC  = 14square root of 2 cm
    Since, BC is a diameter for semicircle BCQB
    therefore Therefore, radius (r) = 7 square root of 2 space cm
    Now, area of semicircle (BCQB)

                 equals space πr squared over 2 equals space space 22 over 7 cross times 1 half cross times 7 square root of 2 cross times 7 square root of 2 equals space 154 space cm squared
    Area of sector (ACPBA)

                equals space πr squared over 2 equals space space 22 over 7 cross times 1 half cross times 14 cross times 14 equals space space 154 space cm squared
    And comma space left parenthesis increment ACB right parenthesis space equals space 1 half cross times AC cross times AB equals space 1 half cross times AC cross times AB equals 1 half cross times 14 cross times 14 space equals space 98 space cm squared

    Hence, required area
    = [ar (∆ABC) + ar (BCQB)] - ar(ACPBA)] cm2
    = [(98 + 154) - (154)]cm2 = 98 cm2.

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    Multiple Choice QuestionsShort Answer Type

    35.

    Calculate the area of the designed region in Fig. 12.34 common between the two quadrants of circles of radius 8 cm each.


    Fig. 12.34

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    36. A chord of a circle of radius 14 cm subtends a right angle at the centre. What is the area of the minor sector?
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    37. A chord of a circle of radius 14 cm subtends 60° at the centre. Find the area of the major sector.
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    38. What is the perimeter of a sector of angle 45° of a circle with radius 7 cm ? [Use π = 22/7].  
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    39. In the following figure, the length of an arc AB = 20Ï€ cm is a sector of a circle, find the radius of the circle.
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    40. An arc of a circle is of length 5Ï€ cm and the sector it bounds has an area 20Ï€ cm2. Find the radius of the circle.
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