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  1. In a triangle ABC, AB + BC = 12 cm, BC + CA = 14 cm and CA + AB = 18 cm. Find the radius of the circle (in cm) which has the same perimeter as the triangle.








  1. View Hint View Answer Discuss in Forum

    Given that , AB + BC = 12
    BC + CA = 14
    CA + AB = 18
    On adding , we get
    ∴ 2(AB + BC + CA) = 12 + 14 + 18 = 44
    ⇒ AB + BC + CA = 22
    ∴ 2πr = 22

    ⇒ 2 ×
    22
    		× r = 22
    7

    Correct Option: B

    Given that , AB + BC = 12
    BC + CA = 14
    CA + AB = 18
    On adding , we get
    ∴ 2(AB + BC + CA) = 12 + 14 + 18 = 44
    ⇒ AB + BC + CA = 22
    ∴ 2πr = 22

    ⇒ 2 ×
    22
    		× r = 22
    7

    ⇒ r =
    7
    		cm
    2

  1. If I be the incentre of ∆ ABC and ∠B = 70° and ∠C = 50°, then the magnitude of ∠BIC is








  1. View Hint View Answer Discuss in Forum

    Firstly , We draw a figure of triangle ABC whose I is incentre ,

    Here , ∠B = 70° and ∠C = 50°

    ∠IBC =
    70°
    		 = 35° ;
    2

    ∠ICB =
    50°
    		 = 25° ;
    2

    Correct Option: C

    Firstly , We draw a figure of triangle ABC whose I is incentre ,

    Here , ∠B = 70° and ∠C = 50°

    ∠IBC =
    70°
    		 = 35° ;
    2

    ∠ICB =
    50°
    		 = 25° ;
    2

    We know that , ∠BIC + ∠ICB + ∠IBC = 180°
    ∴ ∠BIC = 180° – 35° – 25°
    ∠BIC = 180° – 60° = 120°

  1. For a triangle ABC, D, E, F are the mid-points of its sides. If ∆ABC = 24 sq. units then ∆DEF is








  1. View Hint View Answer Discuss in Forum

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

    Here , ∆ABC = 24 sq. units
    From figure , we have

    ∆DEF =
    1
    		 ∆ ABC
    4

    Correct Option: B

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

    Here , ∆ABC = 24 sq. units
    From figure , we have

    ∆DEF =
    1
    		 ∆ ABC
    4

    ∆DEF =
    1
    		 × 24 = 6 Sq. unit
    4

  1. Angle between sss ∠B is








  1. View Hint View Answer Discuss in Forum

    As per the given in question , we draw a figure of triangle ABC

    ∠BPC = 120°
    ∴ ∠PBC + ∠PCB = 180°– 120°= 60°

    Correct Option: D

    As per the given in question , we draw a figure of triangle ABC

    ∠BPC = 120°
    ∴ ∠PBC + ∠PCB = 180°– 120°= 60°
    ∴ ∠ABC + ∠ACB = 2 × 60° = 120°
    ∴ ∠A = 180 – 120 = 60°

  1. In a ∆ABC, ∠A + ∠B = 70° and ∠B + ∠C = 130°, value of ∠A is








  1. View Hint View Answer Discuss in Forum

    In ∆ ABC,
    As we know that , ∠A + ∠B + ∠C = 180°
    Given , ∠A + ∠B = 70° and ∠B + ∠C = 130°

    Correct Option: B

    In ∆ ABC,
    As we know that , ∠A + ∠B + ∠C = 180°
    Given , ∠A + ∠B = 70° and ∠B + ∠C = 130°
    ∴ ∠A = (∠A + ∠B + ∠C) – (∠B + ∠C)
    ∠A = 180– 130°= 50°