CBSE Class 10 Mathematics Solved Question Paper 2015

The given quadratic equation has two equal roots.
therefore,

by using the discriminant method




⇒  20p² - 60p = 0

⇒  20p (p - 3) = 0

⇒  p = 0 or p - 3 = 0

⇒  p = 0 or p = 3

p cannot be zero.

Hence, the value of p  is 3. 

It is given that the points A (x, y), B (-5, 7) and C (-4, 5) are collinear.

∴ Area of triangle, ABC = 0

⇒ 1/2 [x (7 - 5) + (-5) (5 - y) + (-4) (y -7)]  = 0

⇒ 2x - 25 + 5y - 4y + 28 = 0

⇒ 2 x + y + 3 = 0

This is the required relation between x and y.

Let a and d be the first term and the common difference of the AP, respectively

Suppose the given circle touch the sides AB and AC of the triangle at point F and E, respectively. Let the length of the line segment AF is x cm.


We know that the length of tangents drawn from an external point to a circle are equal.

In ΔABC,

CE = CD = 9 cm (Tangents of the circle from point C)

BF = BD = 6 cm (Tangents on the circle from point B)

AE = AF + FB = x + 6

BC = BD + DC = 6 + 9 = 15

CA = CE + EA  = 9 + x

Given equation,
4x² + 4bx - (a² - b²) = 0 

⇒ 4x²  +4bx - ( a + b) (a - b) = 0

⇒ 4x² + 2[(a + b) - (a - b)] x - (a + b)(a-b) = 0

⇒ 4x² +2 (a + b)x - 2(a - b)x - (a + b ) (a - b) = 0

⇒ 2x [2x + (a + b)] [(a - b)[2x + (a + b)] = 0

⇒ 2x + (a + b) = 0 or 2x - (a - b) = 0

⇒ x =  - (a + b)/2 OR x = (a - b)/2

Given: RQ and RP are tangents drawn from an external point R  to the circle with centre O such that ∠PRQ = 120^o

To prove, PR + PQ = OR

Construction, Join OP and OQ.



Proof:
∠OPR = ∠OQR = 90^o
(Radius of the circle is perpendicular to the tangent to the circle through the point contact)

We Know that the centre lies on the bisector of the angle between the two tangents.
So, ∠PRO = ∠QRO = 1/2 ∠PRQ = 60^o

Now, In ΔPRO,

Given ,
The vertices of a right triangle, such as,
A (4, 7), B (p, 3) and C (7, 3) 



In right ΔABC, using Pythagoras theorem

(AB)² + (BC)² = (AC)²  

⇒ [(3 - 7)² + (p - 4)²] + [(3-3)² +(7 - p)²] = [(3-7)² + (7 - 4)²]

⇒ (p - 4)² + (7 - p)² = 9

⇒ p² + 16 -8p + 49 + p² - 14p = 9

⇒ 2p² - 22p + 56 = 0

⇒ p² - 11p + 28 = 0

⇒ p² - 4p -7p + 28 =0

⇒ p(p - 4) -7(p - 4)=0

⇒ (p - 7 )(p - 4) = 0

⇒ p - 7 = 0 or p - 4 = 0

⇒ p = 7 or p = 4

Hence, the value of p is 4 or 7 

PQ is the chord of the circle and PT is tangent.

We know that the tangent to a circle is perpendicular to the radius through the point of contact.

∴ ∠ OPT = 90^o

Now, Given

∠ QPT = 60^o 

∴ ∠ OPQ = ∠OPT - ∠QPT

 ⇒ ∠ OPQ = 90^o - 60^o = 30^o

In Δ OPQ,

OP= OR (Radii of the same circle)

∠ OQP = ∠ OPQ = 30^o (in a triangle, equal sides have equal angles opposite to them.)

Now,

∠ OQP + ∠OPQ + ∠POQ = 180^o [Angle sum property]

⇒ 30^o + 30^o + ∠POQ = 180^o 

⇒ ∠POQ = 180^o - 60^o = 120^o

⇒ Reflex ∠POQ = 360^o -120^o = 240^o

We know that the angle subtended by an arc of a circle at the centre is twice the angle subtended by it any point on the remaining part of the circle.

Therefore,

Reflex ∠POQ = 2 ∠PRQ

⇒ 240^o = 2 ∠PRQ

⇒ ∠PRQ = 240 / 2 = 120^o 

Hence, the measure of angle PRQ is 120^o.

When two dice are thrown simultaneously, the possible outcomes can be listed as:

∴ Total number of possible outcomes = 36

The outcomes favourable to the event the product of the two number of the top of the dice is 6 denoted by E are (1, 6), (2, 3), (3, 2) and (6, 1)

∴ Number of favourable outcomes = 4

Let the sun's altitude be θ.

In ΔABC,