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Geometry Test 40
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Geometry Test 40

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  • Question 1/10
    1 / -0
    In ∆ABC, AB = AC, A circle drawn through B touches AC at D and interest AB at P. If D is the mid point of AC and AP = 2.5 cm, then AB is equal to :
    Solutions

    D is mid point ⇒ AD = DC = x

    ⇒ AC = 2x and AB = 2x

    AC is acting as tangent

    ⇒ AD2 = AP×AB

    ⇒ x2 = 2.5×2x

    ⇒ x = 5 cm

    AB = 2x = 10cm

  • Question 2/10
    1 / -0
    In the given diagram of ∆ABC, ∠B = 800, ∠C = 300 BF and CF are the angel bisectors of ∠CBD and ∠BCE respectively. Find the value of ∠BFC:
    Solutions
    ∠BFC = 900 – 1/2 (∠A)
    = 900 – 1/2 (1800 - (800 + 300)] = 550
  • Question 3/10
    1 / -0
    The circumference of a circle is 88 cm, find its diameter?
    Solutions
    Circumference of a circle = 2π r
    2 π r = 88
    ⇒ r = 88 × (7/22) × (1/2)
    = 14 cm
    ∴ Diameter of the circle = 2 r = 28 cm
  • Question 4/10
    1 / -0
    What is the reflection of the point (2, 3) in the line y = 4?
    Solutions
    Given point is (2,3) and the line is y = 4
    Y = 4 is a horizontal line parallel to x axis,
    So reflected point = (2, 2 × 4 - 3) = (2,5)
  • Question 5/10
    1 / -0
    ABC ~ NLM and ar(ABC): ar(LMN) = 4 : 9. If AB = 6 cm, BC = 8 cm and AC = 12 cm, then ML is equal to:
    Solutions

    ∆ABC similar to ∆NLM (Given)

    From the property of similar triangles

    = 3/2

    3/2

  • Question 6/10
    1 / -0
    A regular hexagon is inscribed in a circle. What is the ratio of the area of the circle to that of its portion not covered by the hexagon?
    Solutions

    Let the side of hexagon be a.

    Let O be the centre and  is the radius of the circle.

    Area of hexagon

    Area of circle

    The area of portion that is not covered by hexagon

    Required ratio

    .

  • Question 7/10
    1 / -0
    ABC is an isosceles triangle such that AB = AC = 30 cm and BC = 48 cm. AD is a median to base BC. What is the length (in cm) of AD?
    Solutions
    Since AD is the median
    Therefore,
    BD = DC
    BD = 24 cm
    By Pythagoras we get,

    Therefore putting the values we get,



  • Question 8/10
    1 / -0
    A(7, - 8) and C(1,4) are vertices of a square ABCD. Find equation of diagonal BD ?
    Solutions
    Slope of diagonal AC =
    As BD is perpendicular to AC then slope of AC × Slope of BD = - 1
    Slope of BD = 1/2
    Also BD goes through midpoint of AC
    So midpoint of AC, = (4, - 2)
    As we know that y = mx + c and mid point of AC are the mid point of BD also
    Here m = 1/2 and y = - 2
    - 2 = 4/2 + c
    C = - 4
    So equation of line y = mx + c
    y = x × 1/2 + ( - 4)
    2y = x - 8
    x - 2y = 8
  • Question 9/10
    1 / -0
    Two circles of radii 10 cm and 8 cm intersect each other and hence the length of the common chord is 12 cm. Then the distance between their centers is approximately how much (in cms)?
    Solutions

    AB= 12cm
    AC= BC= 6cm
    OA= 10cm









  • Question 10/10
    1 / -0
    In triangle PQR, A is the point of intersection of all the altitudes and B is the point of intersection of all the angle bisectors of the triangle. If ∠PBR = 105 °, then what is the value of ∠PAR (in degrees)?
    Solutions

    Here A is orthocentre and B is incentre
    ∠PBR = 105 °
    Also, in triangle PQR
    ∠PQR + ∠QPR + ∠QRP = 180°
    In triangle PBR
    ∠PBR + ∠QPR/2 + ∠QRP/2 = 180°
    105° +(∠QPR + ∠QRP)/2 = 180°
    (∠QPR + ∠QRP)/2 = 75°
    ∠QPR + ∠QRP = 150°
    180° - ∠PQR = 150°
    ∠PQR = 30°
    In Quadrilateral AMQN,
    ∠MAN + ∠ANQ + ∠NQM + ∠QNA = 360°
    ∠PAR + 90° + 30° + 90° = 360°         ......(∠MAN = ∠PAR ....vertically opposite angles)
    ∠PAR = 150°
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