Plane Geometry


  1. In a triangle ABC, the side BC is extended up to D. Such that CD = AC, if ∠BAD = 109° and ∠ACB = 72° then the value of ∠ABC is









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    On the basis of given in question , we draw a figure triangle ABC ,

    ∠ACD = 180° – ∠ACB (Linear Pair)
    ∠ACD = 180° – 72° = 108°

    ∠CAD = ∠ADC =
    72
    = 36 cm
    2

    Correct Option: A

    On the basis of given in question , we draw a figure triangle ABC ,

    ∠ACD = 180° – ∠ACB (Linear Pair)
    ∠ACD = 180° – 72° = 108°

    ∠CAD = ∠ADC =
    72
    = 36 cm
    2

    ∴ ∠ABC = 180° – 109° – 36° = 35°


  1. ABC is a triangle. The bisectors of the internal angle ∠B and external angle ∠C intersect at D. If ∠BDC= 50°, then ∠A is









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    As per the given in question , we draw a figure of triangle ABC in which bisectors of the internal angle ∠B and external angle ∠C intersect at point D ,

    [Exterior angle is sum of opp. interior angles]
    ∠ACE = ∠BAC + ∠ABC
    ⇒ 2y = ∠A + 2x
    Similarly ,
    ∠DCE = ∠DBC + ∠BDC

    Correct Option: A

    As per the given in question , we draw a figure of triangle ABC in which bisectors of the internal angle ∠B and external angle ∠C intersect at point D ,

    [Exterior angle is sum of opp. interior angles]
    ∠ACE = ∠BAC + ∠ABC
    ⇒ 2y = ∠A + 2x
    Similarly ,
    ∠DCE = ∠DBC + ∠BDC
    ⇒ y = 50° + x
    ⇒ ∠A = 2y – 2x = 100 + 2x – 2x = 100°



  1. If the length of the sides of a triangle are in the ratio 4 : 5 : 6 and the inradius of the triangle is 3 cm, then the altitude of the triangle corresponding to the largest side as base is :









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    On the basis of given in question , we draw a figure a triangle ABC ,

    The ratio of the sides of a triangle = 4 : 5 : 6
    Let AB = 4y , BC = 5y , CA = 6y
    ∆OBA + ∆BOC + ∆AOC = ∆ABC

    1
    × 4y ×3 +
    1
    × 5y × 3 +
    1
    × 6y ×3 =
    1
    × 6y × h
    2222

    ⇒ 6y +
    15y
    + 9y = 3yh
    2

    Correct Option: A

    On the basis of given in question , we draw a figure a triangle ABC ,

    The ratio of the sides of a triangle = 4 : 5 : 6
    Let AB = 4y , BC = 5y , CA = 6y
    ∆OBA + ∆BOC + ∆AOC = ∆ABC

    1
    × 4y ×3 +
    1
    × 5y × 3 +
    1
    × 6y ×3 =
    1
    × 6y × h
    2222

    ⇒ 6y +
    15y
    + 9y = 3yh
    2

    ⇒ 12 + 15 + 18 = 6h
    ⇒ 45 = 6h
    ⇒ h =
    15
    = 7.5 cm
    2


  1. Taking any three of the line segments out of segments of length 2 cm, 3 cm, 5 cm and 6 cm, the number of triangles that can be formed is :









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    According to question ,
    The sum of two sides of a triangle should be greater than the third side.

    Correct Option: B

    According to question ,
    The sum of two sides of a triangle should be greater than the third side.
    (3, 5, 6) and (2, 5, 6)



  1. If the circumcentre of a triangle lies outside it, then the triangleis









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    As we know that the right bisectors of the sides of a triangle meet at a point. The point of intersection is called circum-centre.

    Correct Option: D

    As we know that the right bisectors of the sides of a triangle meet at a point. The point of intersection is called circum-centre. For an obtuse angled triangle, circum-centre lies outside the triangle.