\(\frac { 5 }{ 16 } \)

2(3)²+3k=0⇒k=-21

192.5 cm²

Perimeter of a sector OAB = 16.4 cm
⇒ OA+ AB + 0B = 16.4cm
⇒ 5.2 cm + 5.2 cm + AB = 16.4 cm
⇒10.4 cm + AB = 16.4 cm
⇒ AB = (16.4 - 10.4) cm
Length of the arc = 6.0 cm
Area of the sector =\({lr\over 2}={6.0\times5.2\over 2}\)
= 6.0 x 2.6 cm² = 15.6 cm² 

We know that in 1 hour (i.e., 60 minutes), the minute hand rotates 360°.
In 5 minutes, minute hand will rotate = 360^@/60xx5 = 30^@
Therefore, the area swept by the minute hand in 5 minutes will be the area of a sector of 30° in a circle of 14 cm radius.
Area of sector of angle θ = \(\frac{\theta}{360^{\circ}} \times \pi r^{2}\)
Area of sector of 30° \(=\frac{30^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 14 \times 14\)
\(\begin{array}{l} =\frac{22}{12} \times 2 \times 14 \\ =\frac{11 \times 14}{3} \end{array}\)
=154/3 cm²
Therefore, the area swept by the minute hand in 5 minutes is 154/3 cm²

a = 9, d = 10
a₁₂= a + 11d
= 9 + 11 x 10 = 119

Given capacities of two tankers are 850 Land 680 L.
Here,   850 > 680
Now,   850 = (680xl) +170
 Here, remainder = 170 \( \neq\) 0.
So, new dividend is
680 and divisor is 170.
Now,      680 = (170 x  4)+0
 [by Euclid's division lemma]
Here, remainder is zero and divisor is 170.
So, the HCF of 850 and 680 is 170.
Hence, the maximum capacity of the required container is 170 L

3x²+8x+4;-2 and \(-\frac { 2 }{ 3 } \)

Let P(x)=x²-5
For finding a zeroes of p(x), p(x)=0
x²-5=0
⇒ x=±5
Let α=5 and β=-5
Now, 1+a=1+5=6
and 1-a=1-5=4
Thus, 6 and -4 are the zeroes of new quadratic polynomial.
Therefore the new quadratic polynomial will be (x-6)(x+4) or x²-2x-24

Here,  OA = OB = r = 12cm, \(\angle \)AOB = 120°
In \(\triangle\)AOB, AO = BO = 12 cm, draw OD ⊥ AB
Since \(\triangle\)AOB is an isosceles and OD ⊥ AB
ஃ OD is the angle bisector as well as median

Now, in rt.  \(\triangle\)ADO, \(\angle \)D = 90° , \(\angle \)AOD = 60°
∴ \(OD\over AO\) = cos 60°
= 12 x 1/2 =6 cm
and  \(AD\over AO\) = sin 60°

⇒ AD = AO, sin 60° = 12 x \(\sqrt3 \over2\) = 6\(\sqrt3\) cm
∴ AB = 2 AD = 2\(\sqrt3\) x = 12\(\sqrt3\) cm
Area of minor segment 
= Area of sector AOB - Area of \(\triangle\)AOB
= \(\theta\over360°\) x \(\pi\)r² - \(1\over2\) x AB x OD
= \(120° \over360°\) x 3.14 x 12 x 12 - \(1\over2\) x 12\(\sqrt3\) x 6
= 150.72 - 62.28 = 88.44 cm².

Given: A circle with centre O. AB is the diameter of this circle. I is tangent to the circle. AH and BK are perpendicular to I from A and B at H  and K respectively. 
To prove: AH + BK = AB 
Proof: AH and HP are tangents from the external point H 
ஃ  AH = HP        .....(i)
and BK, KP are tangent from the external point K
ஃ   BK = KP      ......(ii)
Adding (i) and (ii) we get
AH + BK = HP + PK = HK ......(iii)
AB ⊥  AH
AB ⊥ BK
[Tangent makes 90° angle with radius at the point of contact]
⇒ \(\angle \)1 = \(\angle \)2 = 90°
Given that AH ⊥ i ⇒ \(\angle \)3 = 90°
and BK ⊥ i ⇒ \(\angle \)4 = 90°
∵ \(\angle \)1 = \(\angle \)2 = \(\angle \)3 = \(\angle \)4 = 90°​​​​​​​
⇒ AHKB is a rectangle
⇒  AB = HK        ......(iv)
[Opposite sides of a rectangle are equal]
From (iii) and (iv) ⇒ PH + PK = AB
AH + BK = AB from (i) and (ii) Hence proved.

r₁ = 5 cm, r₂ = ?,
AB = 24 cm
∵ AB is tangent to circle
C(C, r₁) at C
ஃ OC ⊥ AB
In circle C(O, r₂), AB is a chord and OC ⊥ AB
ஃ AC = BC
[∵ Perpendicular from the centre bisects the chord]
In right \(\triangle\)OCA
OC² + AC² = AC²
⇒ 5² + (12)² = (r₂)²
⇒ 25 + 144 = (r₂)² ⇒ (r₂)² = 169
r₂ = 13 cm

Given, a = 9, d = 17 - 9 = 8,
Sn = 636
Sn = \(\frac{n}{2}[2a+(n-1)d]\)
\(\Rightarrow\)    636 = \(\frac{n}{2}[2\times9+(n-1)8]\Rightarrow636\times 2=n[18+8n-8]\)
\(\Rightarrow\)    636 X 2 = n ( 10 + 8n ) \(\Rightarrow\) 636 x 2 = 2n ( 5 + 4n )
\(\Rightarrow\)    \(\frac{636\times2}{2}\) = 5n + 4n²    \(\Rightarrow\) 4n² + 5n - 636 = 0
\(\Rightarrow\)   4n² + 53n + 48n - 636 = 0 \(\Rightarrow\) 4n( n + 53 ) - 48 ( n + 53 ) = 0
\(\Rightarrow\)     ( 4n - 48 ) ( n + 53 ) = 0 \(\Rightarrow\) Either 4n = 48 or n + 53 = 0
\(\Rightarrow\)     n = \(\frac{48}{4}\) or n = -53 (rejected)
\(\Rightarrow\)   n = 12
Hence   n = 12

According to the question,  \(\alpha\) and \(\beta\) are zeroes of the polynomial p(x) = 6x² - 5x + k
So, Sum of zeroes = \(\alpha +\beta =-\left( \frac { -5 }{ 6 } \right) =\frac { 5 }{ 6 } \)            ... (i)
Product of zeroes = \(\alpha \beta =\frac { k }{ 6 } \)
\(\alpha-\beta=\frac{1}{6}\)           (Given)     ...... (ii)
Adding equations (i) and (ii), we get
\(2\alpha=2\)
\(\Rightarrow \quad \alpha=\frac{1}{2}\)
Putting the value of \(\alpha\) in equation (ii), we get
\(\frac{1}{2}-\beta=\frac{1}{6}\)
\(\Rightarrow \quad \frac{1}{2}-\frac{1}{6}=\beta\)
\(\Rightarrow \quad \frac{2}{6}=\frac{1}{3}=\beta\)
\(\therefore \quad \alpha\beta=\frac{k}{6}=\frac{1}{2}\times\frac{1}{3}\)
\(\therefore \quad k=1\)

Let AB, BC, CD and AD touch the circle at points P, Q, R and S respectively, then AS = AP, BQ = BP, CQ = CR and DS = DR on adding,
AS + DS + BQ + CQ = AP + DR + BP + CR \(\Rightarrow\) AD + BC = AB + CD
\(\Rightarrow\) AD = (6 + 4 - 7) cm = 3 cm

Area of the segment AYB = Area of sector OAYB - Area of \(\Delta\)OAB                  (1)
Now, area of the sector OAYB = \(=\frac{120}{360} \times \frac{22}{7} \times 21 \times 21 \mathrm{~cm}^2=462 \mathrm{~cm}^2\) (2)
For finding the area of \(\Delta\)OAB, draw OM \(\perp\) AB as shown in Figure.
Note that OA = OB. Therefore, by RHS congruence, \(\Delta\)AMO \(\cong\) BMO.
So, M is the mid-point of AB and \(\angle\)AOM = \(\angle\)BOM = \(\frac{1}{2} \times 120^{\circ}=60^{\circ}\)
Let  OM = x cm
So, from \(\Delta\)OMA, \(\frac{\mathrm{OM}}{\mathrm{OA}}=\cos 60^{\circ}\)

or, \(\begin{aligned} \frac{x}{21} & =\frac{1}{2} \quad\left(\cos 60^{\circ}=\frac{1}{2}\right) \\ \end{aligned}\)
or, \(\begin{aligned} x & =\frac{21}{2} \\ \end{aligned}\)
So, \(\begin{aligned} \mathrm{OM} & =\frac{21}{2} \mathrm{~cm} \\ \end{aligned}\)
Also,\(\begin{aligned} \frac{\mathrm{AM}}{\mathrm{OA}} & =\sin 60^{\circ}=\frac{\sqrt{3}}{2} \\ \end{aligned}\)
So, \(\begin{aligned} \mathrm{AM} & =\frac{21 \sqrt{3}}{2} \mathrm{~cm} \end{aligned}\)
Therefore, \(\mathrm{AB}=2 \mathrm{AM}=\frac{2 \times 21 \sqrt{3}}{2} \mathrm{~cm}=21 \sqrt{3} \mathrm{~cm}\)
So, area of \(\begin{aligned} \Delta \mathrm{OAB} & =\frac{1}{2} \mathrm{AB} \times \mathrm{OM}=\frac{1}{2} \times 21 \sqrt{3} \times \frac{21}{2} \mathrm{~cm}^2 \\ \end{aligned}\)
\(\begin{aligned} =\frac{441}{4} \sqrt{3} \mathrm{~cm}^2 \end{aligned}\)                                                                              (3)
Therefore, are of the segment \(\mathrm{AYB}=\left(462-\frac{441}{4} \sqrt{3}\right) \mathrm{cm}^2\) [From (1), (2) and (3)]
\(=\frac{21}{4}(88-21 \sqrt{3}) \mathrm{cm}^2\)

(i) Number of logs stacked in each row form a sequence 20, 19, 18, 17,...., which is an AP with first term, a= 20 and common difference, d = 19 - 20 = -1.
Suppose number of rows is n, then S_n= 200
\(\begin{aligned} \Rightarrow \frac{n}{2}[2 \times 20+(n-1)(-1)] & =200 \\ \end{aligned}\)
\(\begin{aligned} {\left[\because S_n\right.} & \left.=\frac{n}{2}\{2 a+(n-1) d\}\right] \end{aligned}\)
\(\begin{array}{lr} \Rightarrow & 400=40 n-n^2+n \\ \end{array}\)
\(\begin{array}{lr} \Rightarrow & n^2-41 n+400=0 \\ \end{array}\)
\(\begin{array}{lr} \Rightarrow & n^2-25 n-16 n+400=0 \\ \end{array}\)
\(\begin{array}{lr} \Rightarrow & n(n-25)-16(n-25)=0 \\ \end{array}\)
\(\begin{array}{lr} \Rightarrow & (n-25)(n-16)=0 \\ \end{array}\)
\(\begin{array}{lr} \Rightarrow & n=16 \text { or } n=25 \end{array}\)
Hence, the number of rows is either 25 or 16.
When, n = 16,
a_n= a + (n -1) d = 20 + (16 - 1) ( - 1)
= 20 - 15 = 5
When, n = 25,
a_n = a + (n - 1) d = 20 + (25-1) (-1)
= 20 - 24 = - 4
[\(\because\) number of logs cannot be negative]
Hence, the number of rows is 16 and number of logs in the top row is 5.
(ii) The pattern of logs show space saving creativity, reasoning and balancing.

A.T.Q., S₅ + S₇ = 167
\(\Rightarrow\) \({5\over2}[2a+14d]+{7\over2}[2a+6d]=167\)
\(\Rightarrow\) 5 ( a + 2d ) + 7 ( a + 3d ) = 167
\(\Rightarrow\)  12a + 31d = 167                   ...(i)
and  S₁₀ = 235
\(\Rightarrow\)  \({10\over2}[2a+9d]=235\)
\(\Rightarrow\)  2a + 9d = \({235\over4}=47\)                ...(ii)
Multiplying equation (ii) by 6 and then subtractigfrom (i), we have
12a + 31d = 167
12a + 54d = 282
-     -          -      
     -23d = -115
d= \({-115\over-23}=5\)
\(\therefore\)  From (ii), 2a + 9d = 47
\(\Rightarrow\)   2a + 9 x 5 = 47
\(\Rightarrow\)    2a = 47 - 45
\(\Rightarrow\)   2a = 2 \(\Rightarrow\) a = 1
Hence, S₂₀ = \({20\over2}[2a+19d]\)
= 10 [ 2 x 1 + 19x 5 ]
= 10 [ 2 + 95 ] = 10 x 97 = 970

(a) Both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

(b) If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

(i) (b) Given, number of students in each subject are Hindi =60, English = 84, Mathematics = 108.
The prime factors of each subject students are
60 =2\(\times\) 2  \(\times\)3 \(\times\)5= 2² \(\times\) 3 \(\times\) 5
84 = 2 \(\times\) 2 \(\times\)3 \(\times\)7 = 2² \(\times\)3 \(\times\)7
108 =2 \(\times\) 2  \(\times\)3  \(\times\)3  \(\times\)3= 2² \(\times\)3³
The maximum number of participants that an accommodated in each room
=HCF (60, 84, 108)
= Product of the smallest power of each common prime factor involved in the numbers
=2² \(\times\)3 = 12
(ii) (d) The minimum number of rooms required during the event = \(\frac{Total \quad number \quad of \quad participants}{12}=\frac{252}{12}=21\)
(iii) (a) LCM of (60, 84, 108)
= Product of the greatest power of each prime factor involved in the numbers with highest power
= 2² \(\times\)3³ \(\times\)5 \(\times\) 7
= 4 \(\times\)27 \(\times\)35
= 3780
(iv) (d) Now, HCF (60, 84, 108) \(\times\)LCM (60, 84, 108) = 12 \(\times\)3780 = 45360
(v) (d) Number 108 can be expressed as a product of its prime as 2²\(\times\) 3³.

(i) (b): Angle made by minute hand in 60 minutes = 360°
\(\therefore\) Angle made by minute hand in 10 minutes
\(=\frac{360^{\circ}}{60} \times 10=60^{\circ}\)
Length of minute hand = 9 cm
Area swept by minute hand in 10 minutes = Area of sector having central angle 60°
\(=\pi r^{2}\left(\frac{60^{\circ}}{360^{\circ}}\right)=\frac{22}{7} \times 9 \times 9 \times \frac{1}{6}\)
\(=\frac{297}{7}=42.42 \mathrm{~cm}^{2}\)
(ii) (b): We have, r = 15 cm

\(\text { and } l=\frac{1}{2}(22)=11 \mathrm{~cm}\)
\(\text { We known that } l=2 \pi r\left(\frac{\theta}{360^{\circ}}\right)\)
\(\Rightarrow \theta=\frac{11 \times 360^{\circ}}{2 \times \frac{22}{7} \times 15}=\frac{90^{\circ} \times 7}{15}=6^{\circ} \times 7=42^{\circ}\)
(iii) (a): Angle made by hour hand in 12 hours = 360°
\(\therefore\) Angle made by hour hand in 10 minutes
\(=\left(\frac{360^{\circ}}{12} \times \frac{1}{6}\right)=5^{\circ}\)
(iv) (d): Angle made by hour hand in 1 hour
\(=\frac{360^{\circ}}{12}=30^{\circ}\)
Also, r = 6 cm
Area swept by hour hand in 1 hour = Area of sector having central angle 30°
(v) (a): Number of hours from 11 a.m. to 5 p.m. = 6
Area swept by hour hand in 1 hour = 9.428 cm²
\(\therefore\) Area swept by hour hand in 6 hours = 9.428 x 6
= 56.568 cm²

Let the numbers on the cards be a, a + d, a + Zd, ...
According to question, We have (a + 5d) + (a + 13d) = -76
\(\Rightarrow\) 2a+18d = -76\(\Rightarrow\)a + 9d= -38 ... (1)
And (a + 7d) + (a + 15d) = -96
\(\Rightarrow\) 2a + 22d = -96 \(\Rightarrow\) a + 11d = -48 ...(2)
From (1) and (2), we get
2d= -10 \(\Rightarrow\) d= -5
From (1), a + 9(-5) = -38 \(\Rightarrow\) a = 7
(i) (b): The difference between the numbers on any two consecutive cards = common difference of the A.P.=-5
(ii) (d): Number on first card = a = 7
(iii) (b): Number on 19th card = a + 18d = 7 + 18(-5) = -83
(iv) (a): Number on 23rd card = a + 22d = 7 + 22( -5) = -103
(v) (d): \(S_{15}=\frac{15}{2}[2(7)+14(-5)]=-420\)