CTET Solved Paper - CTET September 2016 Paper2

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41. If one angle of a triangle is 110°, then the angle between the bisectors of the other two angles measures

  • Option : B
  • Explanation : Let the ΔABC, in which AM and CN are angle bisectors.
    ∠A + ∠B + ∠C =180°
    ⇒ ∠A + 110° + ∠C = 180°
    ⇒ ∠A + ∠C=180-110°
    ⇒ ∠A/2 + ∠C/2 = 70°/2
    ⇒ ∠A/2 + ∠C/2 = 35 ... (i)
    Now the angle bisectors meet at O.
    In ΔAOC,
    ∠A/2 + ∠C/2 + ∠AOC =180°
    => 35° + ∠AOC = 180° [From Eq. (i)]
    ⇒ ∠AOC = 180° - 35°
    ∠AOC = 145°
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42. In ΔABC, AB=4cm, AC=5cm and BC=6cm. In ΔPQR, PR=4cm, PQ=5cm and RQ=6cm. ΔABC is congruent to

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43. A tank is in the form of a cuboid. It holds a maximum of 540 m3 water. If the tank is 8 m long and 15 m wide, then how many metres deep must the water be when the tank is 2/3 full?

  • Option : D
  • Explanation : Volume of a cuboid (tank)=lxbxh
    ⇒ 540 m3 = 8 x 15 x h
    ⇒ Height of the tank = 540/8x15
                                            = 4.5 m
    Now, water height in the tank = 2/3 of tank height = 2/3 x 4.5 = 3 m
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44. The ratio of the areas of two equilateral triangles is 16:9. If the perimeter of the smaller triangle is 63cm, then how much larger is a side of the larger triangle than a side of the smaller triangle?

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45. The area of a triangle is equal to the area of a circle whose perimeter is 6Π cm. If the base of the triangle is 8cm, then its corresponding height (in cm) is

  • Option : D
  • Explanation : Perimeter of circle = 2Πr
    ⇒6Π = 2Πr r - 3cm
    Area of circle= Πr2 = Π(3)2
                                           =9Π cm2 ....(i)
    Since Area of Triangle = Area of a circle (given)
    ⇒ 1/2 x base x height = 9Π
    (put the value of Eq(i) i.e., A=9Π)
    ⇒ 1/2 x 8 x h=9Π
    ⇒ h=9Π x 2/8
    ⇒ h=9/4 Π
    ⇒ h=2.25Π
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