CTET Solved Paper - CTET September 2016 Paper2
- 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
- Option : C
- Explanation : ΔABC ≈ ΔPRQ
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- 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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- Option : B
- Explanation :
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- 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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