Additional Questions
Q. If the mean of x and 1/x is M then find the mean of x² and 1/x².
Sol
Mean of x and 1/x
Mean = 1/2[x+1/x]
M = 1/2[x+1/x]
[x+1/x] = 2M ........(1)
Mean of x² and 1 / x²
= 1/2[x² + 1/x² + 2–2] adding and subtracting 2 in x² + 1/x²
= 1/2[ {(x)² +(1/x)² + 2} –2]
= 1/2[ (x + 1/x)² –2]
placing the value of x + 1/x
= 1/2[ (2M)² –2]
= 1/2[ 4M² –2]
= 1/2× 4M² –1/2×2]
= 2M² –1
Q. In the given figure ABCD is a quadrilateral and AO and DO are bisector of ∠A and ∠D, find the value of x.
Sol.
In quadrilateral ABCD
∠A+∠B+∠C+∠D = 360° (BY ASPOQ)
2n + 70° + 120° + 2m = 360°
2m + 2n + 190° = 360°
2m + 2n = 360°– 190°
2m + 2n = 170°] ÷ 2
m + n = 85° ...... (1)
In ∆ AOD
∠A+∠O+∠D = 180° (BY ASPOT)
n+ x + m = 180°
m + n + x = 180°
(m + n) + x = 180°
85° + x = 180° [From equation (1)]
x = 180° – 85°
x = 95° Ans.
[Note
ASPOQ => Angle Sum Property Of Quadrilatera (A.S.P.O.Q)
ASPOT => Angle Sum Property Of Triangle (A.S.P.O.T)]
Q. If the surface area and the volume of a cylinder are equal, find the diameter of the cylinder.
Sol.
ATQ
Surface Area of cylinder = volume of cylinder
2πrh = πr²h
2 = πr²h/πrh
2 = r
Now
d = 2.r
d = 2×2 = 4
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