If the quadratic equation has two roots, then find the value of p.
The given quadratic equation has two equal roots.
therefore,
by using the discriminant method
⇒ 20p^{2}  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.
Find the relation between x and y if the points A (x, y), B (5, 7) and C (4, 5) are collinear.
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.
In an AP, if S_{5} + S_{7} = 167 and S_{10} = 235, then find the AP, Where S_{n} denotes the sum of its first n terms.
Let a and d be the first term and the common difference of the AP, respectively
In the given figure, a triangle ABC is drawn to circumscribe of radius 3 cm, such that the segments RD and DC are respectively of length 6 cm and 9 cm. If the area of ΔABC is 54 cm^{2}, then find lengths of sides AB and AC.
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
Solve the following quadratic equation for x,
4x^{2} + 4bx  (a^{2}  b^{2}) = 0
Given equation,
4x^{2} + 4bx  (a^{2}  b^{2}) = 0
⇒ 4x^{2 } +4bx  ( a + b) (a  b) = 0
⇒ 4x^{2} + 2[(a + b)  (a  b)] x  (a + b)(ab) = 0
⇒ 4x^{2} +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
In the given figure, two tangents RQ and RP are drawn from an external point R to the circle with centre O. If ∠PRQ = 120^{o}, then prove that OR = PR + RQ.
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,
The points A (4, 7), B (p, 3) and C (7, 3) are the vertices a right triangle, rightangled at B find the value of p.
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}] + [(33)^{2} +(7  p)^{2}] = [(37)^{2} + (7  4)^{2}]
⇒ (p  4)^{2} + (7  p)^{2} = 9
⇒ p^{2} + 16 8p + 49 + p^{2}  14p = 9
⇒ 2p^{2}  22p + 56 = 0
⇒ p^{2 } 11p + 28 = 0
⇒ p^{2}  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
In the given figure, PQ is a chord of a circle with centre O and PT is tangent. If ∠QPT = 60^{o}, find ∠ PRQ.
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}.
The different dice are tossed together. Find the probability that the product of the two number on the top of the dice is 6.
When two dice are thrown simultaneously, the possible outcomes can be listed as:

1 
2 
3 
4 
5 
6 
1 
(1, 1) 
(1, 2) 
(1, 3) 
(1, 4) 
( 1, 5) 
(1, 6) 
2 
(2, 1) 
(2, 2) 
(2, 3) 
(2, 4) 
(2, 5) 
(2, 6) 
3 
(3, 1) 
(3, 2) 
(3, 3) 
(3, 4) 
(3, 5) 
(3, 6) 
4 
(4, 1) 
(4, 2) 
(4, 3) 
(4, 4) 
(4, 5) 
(4, 6) 
5 
(5, 1) 
(5, 2) 
(5, 3) 
(5, 4) 
(5, 5) 
(5, 6) 
6 
(6, 1) 
(6, 2) 
(6, 3) 
(6, 4) 
(6, 5) 
(6, 6) 
∴ 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
In Given figure, a tower AB is 20 m high and BC, its shadow on the ground, is m long. Find the sun's altitude.
Let the sun's altitude be θ.
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