Q9) In Fig 2, ABCD is a parallelogram in which DC = 10 cm and BC = 43 cm. AP is perpendicular to DC. If ADC = 60^o, find the length of AP.  (Marks 2)

Ans9) In rt. APD, AP/AD = sin 60^o
= AP/BC = sin 60^o   = AP/43 = 3/2
= AP = 43 x 3/2   = AP = 4 x 3/2 = 6 cm

Q10) ABC is right angled at B. On the side AC, a point D is taken such that AD = DC and AB = BD. Find the measure of CAB.  (Marks 2)
Ans10)  

The vertices of a rt. touch a circle with diameter equal to the hypotenuse.
Since AC is the diameter and AD = DC (given)
.^.. D is the centre of the circle.
.^.. AD = DC = BD  (radii of the same circle)
But AB = BD  (given)
.^.. AB = BD = AD    ...(i)
.^.. ABD is an equilateral
.^.. DAB = 60^o   ...(In equilateral all angles are equal to 60^o)
or CAB = 60^o

Q11) ) In Fig. 3, ABCD is a quadrilateral in which AD = BC and ADC = BCD. Show that the points A, B, C and D lie on a circle.  (Marks 2)

Ans11)  Proof 2 = 3   ...(i) (given)
Also AD = BC   ...(given)
.^.. AB || DC
Now 1 + 2 = 180^o    ...(consecutive interior angles)
= 1 + 3 = 180^o 
Since the sum of opposite angle of the given quadrilateral is 180^o
.^.. It is a cyclic quadrilateral 
or, Points A, B, C and D lie of the circle.

Q12) D is a point on the side BC of ABC such that ADC and BAC are equal. Prove that CA² = DC x CB.  (Marks 2)
Ans12) Given : D is a point on the side of BC of a ABD such that ADC = BAC

To prove : CA² = DC x CB
Proof : In ABC and DAC,
BAC = ADC    ...(given) 
C = C   ...(common)
.^.. ABC = DAC
.^.. s ABC and DAC are equiangular and hence similar 
.^.. BC/AC = AC/DC    = AC² = DC . BC

Q13) If the mean of n observations
x₁, x₂, x₃,  ..., x_n is , prove that 
(x₁ - ) + (x₂ - ) + (x₃ - ) + ... + (x_n - ) = 0  (Marks 2)
Ans13)  (x₁ - ) + (x₂ - ) + (x₃ - ) + ... + (x_n - )
= (x₁ + x₂ + x₃ + ... + x_n) - n
= x - n
= n - n = 0 

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