\(\Sigma fx=630\) , \(\Sigma f=40\)
Mean  = \(\frac { 630 }{ 40 } =15.75\)

Area of the sector =\(\frac{1}{2}\times\)(length of the corresponding are) \(\times\)radius
\(=\frac { 1 }{ 2 } \times l\times r\)
\(=\frac { 1 }{ 2 } \times 12\times 6\)
= 36 cm

\(\angle QOR+\angle QPR={ 180 }^{ ° }\)
(Supplementary angle)
\(\Rightarrow \quad \angle QOR+{ 46 }^{ ° }={ 180 }^{ ° }\)
\(\Rightarrow \quad \angle QOR={ 180 }^{ ° }-{ 46 }^{ ° }={ 134 }^{ ° }\)

Given, \(\angle CBA=90^0,\angle BRS=90^0,\angle BSC=90^0,\) BC = 7.5 cm, RS = 5 cm, BR = 6 cm and AB = 18 cm
Then, AR = AB - RB = 18 - 6 = 12 cm

In \(\triangle ABC,\)  by using Pythagoras theorem, we get  
AS² = AR² + RS² = (12)² + (5)²
AS² = 144 + 25 = 169
\(\Rightarrow\) AS = \(\sqrt 169\)
\(\therefore\)  AS = 13 cm [since, side cannot be negative]
\((i)\quad In\quad \triangle BRS,\ tan{ x }^{ 0 }=\frac { P }{ B } =\frac { BR }{ RS } =\frac { 6 }{ 5 } \quad \)
\((ii)\quad In\quad \triangle ARS,\quad sin{ y }^{ 0 }=\frac { P }{ H } =\frac { SR }{ AS } =\frac { 5 }{ 13 } \)
\((iii)\quad In\quad \triangle ARS,\quad cos{ y }^{ 0 }=\frac { AR }{ AS } =\frac { 12 }{ 13 } \)

k=7

Let a be the first term and d be the common of given AP.
Then, a₁ + a₅ + a₁₀ + a₁₅ + a₂₀ + a₂₄ = 225
\(\Rightarrow \quad a+(a+4d)+(a+9d)+(a+14d)+(a+19d)+(a+23d)=225\)
\(\\ \Rightarrow \quad 6a+69d=225\ \)
\(\ \Rightarrow \quad 2a+23d=75\)
Now, \(S_{ 24 }=\frac { 24 }{ 2 } \left[ 2a+(24-1)d \right] \)
\(=12\left[ 2a+23d \right] =12\times 75\)
= 900

As c is negative, find such factors of 10 whose difference is 3.(-5,2)

6 cm

Total number of cards = 1000 
(i) Perfect square greater than 500 = 529, 576, 625,676,729,784,841,900,961 
= 9 outcomes only 
∴ Required Probability = \(\frac { 9 }{ 1000 } =0.009\)
(ii) After first player, who has won the prize the number of perfect squares greater than 500 will be 9 -1 i.e., 8 
∴ Required probability = \(\frac{8}{999}\)

Yes

70^∘

Let PQ be the building of height h m . Let R and S br  the two positions on the ground such that
ㄥPRQ=45^o, ㄥPSQ=60^o
and RS=100 m
Now in rt ΔSQP, we obtain 
\(\frac { PQ }{ SQ } \)=tan 60^o
⇒ \(\frac { PQ }{ \sqrt { 3 } } =SQ\ or\ \frac { h }{ \sqrt { 3 } } =SQ\)
Also, in rt . ΔRQP, we obtain
\(\frac { PQ }{ RQ } \)=tan 450 ⇒ PQ=RQ
⇒ h=100+SQ
⇒ h=100+\(\frac { h }{ \sqrt { 3 } } \)
⇒ h\(\left( 1-\frac { 1 }{ \sqrt { 3 } } \right) \)=100
h=\(\frac { 100\sqrt { 3 } }{ \sqrt { 3 } -1 } \times \frac { \sqrt { 3 } +1 }{ \sqrt { 3 } +1 } \)
=50(3+\(\sqrt { 3 } \))m
Thus, height of the building is 50(3+ \(\sqrt { 3 } \))m
(i) Some applications of trigonometry
(ii) Kindness, charity and sacrifice.

(-2, 5)

\(k =\pm \ 2\)

Here, a = 2, d = 7 – 2 = 5 and n = 10.
We have a_n = a + (n – 1) d
So,  a₁₀ = 2 + (10 – 1) × 5 = 2 + 45 = 47
Therefore, the 10th term of the given AP is 47.

Each angle of regular hexagon \(=\frac{\text { Sum of all angles }}{\text { Number of sides }}\)
\(=\frac{(6-2) \times 180^{\circ}}{6}=120^{\circ}\)
\(\therefore \angle A=\angle B=\angle C=\angle D=\angle E=\angle \mathrm{F}=120^{\circ}\)
 Area of shaded region = 6 \(\times\) Area of sector with \(\angle\)A.
Ans. 2\(\pi\)r²

Steps of Construction:
1. Draw a line segment AB of 7 cm
2. Taking A and B as centre draw two circle of 3 cm and 2 cm radius

3. Bisect the line AB. Let mid-point of AB is C.
4. Taking C as centre draw a circle of radius AC which will intersect the circle at point P, Q, R and S.
5. Join BP, BQ, AS and AR. These are the required tangents.

Since, ABC is an isosceles right angled triangle.
So, AB = BC  ...(i)

By Pythagoras theorem,
AC² = AB² + BC²
\(\Rightarrow \)  AC² = 2 AB²     [From Eq. (i)]
\(\Rightarrow \) \(\left( 5\sqrt { 2 } \right) ^{ 2 }\) = 2 AB²
\(\Rightarrow \) \(\frac{50}{2}\)  =  AB²
\(\Rightarrow \) AB² = 25  \(\Rightarrow \)  AB = 5
Hence, the length of the equal sides of a triangle is 5 cm.

Let C(-1, 1) divides AB in the ratio k:1.

Then, by using section formula, we get
\(C(-1,1)=C\left( \frac { k-3 }{ k+1 } ,\frac { 3k-1 }{ k+1 } \right) \)
On equating x-coordinate from both sides, we get
\(-1=\frac { k-3 }{ k+1 } \\ \Rightarrow -k-1=k-3\\ \Rightarrow -2k=-3+1 \ \Rightarrow -2k=-2\\ \Rightarrow k=1\)
On equating y-coordinate from both sides, we get=
\(1=\frac { 3k-1 }{ k+1 } \\ \Rightarrow k+1=3k-1\\ \Rightarrow 2k=2 \ \\ \Rightarrow k=1\)
Since in both cases value of k is same. So, C divides AB in the ratio of 1:1. i.e. C is the mid-point of AB.
Hence, A, B and C are collinear.

(i)  \(\frac { 3 }{ 8 } \)
(ii) \(\frac { 1 }{ 2 } \)
(iii)\(\frac { 7 }{ 8 } \)

Let four parts be a - 3d, a - d, a + d and a + 3d
a - 3d + a - d + a + d + a +3d = 56
4a = 56
a = 14
A.T.Q
\(\frac { \left( a-3d \right) \left( a+3d \right) }{ (a-d)(a+d) } =\frac { 5 }{ 6 } \)
\(\frac { { a }^{ 2 }-{ ad }^{ 2 } }{ { a }^{ 2 }-{ d }^{ 2 } } =\frac { 5 }{ -6 } \)
\(\Rightarrow \frac { { \left( 14 \right) }^{ 2 }-9{ d }^{ 2 } }{ { \left( 14 \right) }^{ 2 }-{ d }^{ 2 } } =\frac { 5 }{ 6 } \Rightarrow \frac { 196-9{ d }^{ 2 } }{ 196-{ d }^{ 2 } } =\frac { 5 }{ 6 } \)
\(\Rightarrow 1176-54{ d }^{ 2 }=980-5{ d }^{ 2 }\)
\(\Rightarrow 196=4{ d }^{ 2 }\)
\(\Rightarrow { d }^{ 2 }=4\Rightarrow d=\pm 2\)
when a = 14, d = 2
four parts are
14 - 3 x 2, 14 - 2, 14 + 2, 14 + 3 x 2,
i.e. 8,12,16,20
when a = 14, d = -2
four parts are 20,16,12 and 8

Let AB is statue, BC is pedestal and BC=x m, CD=y m.
In right ΔBCD, ΔBCD, \(\frac { BC }{ CD } \)=tan 45^o
⇒, \(\frac { x }{ y } \)=1 ⇒ x=y     .....(i)
In right ΔACD, \(\frac { AC }{ CD } \)=tan 600
⇒ \(\frac { x+1.46 }{ y } =\sqrt { 3 } \)
⇒ \(\frac { x+1.46 }{ x } \)=1.73    [Using (i)]
⇒ x+1.46=1.73x ⇒ 0.73x=1.46 ⇒ x=2

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.

\(15\tan ^{ 2 }{ \theta } +4\sec ^{ 2 }{ \theta } =23\)
\(15\tan ^{ 2 }{ \theta } +4\left( \tan ^{ 2 }{ \theta } +1 \right) =23\)
\(\left( \because \sec ^{ 2 }{ \theta } =1+\tan ^{ 2 }{ \theta } \quad \right) \)
\(\Rightarrow \quad 15\tan ^{ 2 }{ \theta } +4\tan ^{ 2 }{ \theta } +4=23\)
\(\Rightarrow \quad 19\tan ^{ 2 }{ \theta } =19\)
\(\Rightarrow \quad \tan { \theta } =1=\tan { { 45 }^{ ° } } \)
\(\therefore \quad \theta ={ 45 }^{ ° }\)
Now, \(\left( \sec { \theta } +cosec\theta \right) ^{ 2 }-\sin ^{ 2 }{ \theta } \)
\(=\left( \sec { { 45 }^{ ° } } +cosec{ 45 }^{ ° } \right) ^{ 2 }-\sin ^{ 2 }{ { 45 }^{ ° } } \)
\(={ \left( \sqrt { 2 } +\sqrt { 2 } \right) }^{ 2 }-{ \left( \frac { 1 }{ \sqrt { 2 } } \right) }^{ 2 }\)
\(=\left( 2\sqrt { 2 } \right) ^{ 2 }-\frac { 1 }{ 2 } \)
\(=8-\frac{1}{2}=\frac{15}{2}\)

We have to prove, \(\frac { a^{ 2 } }{ \beta ^{ 2 } } +\frac { \beta ^{ 2 } }{ a^{ 2 } } =\frac { p^{ 4 } }{ q^{ 2 } } -\frac { 4p^{ 2 } }{ q } +2\)
Given α and β are the zeroes of the quadratic polynomial f(x)=x²-px+q
∴ Sum of zeroes, α+β=\(\frac { -(-p) }{ 1 } =p\) and product of zeroes, \(a\beta =\frac { q }{ 1 } =q\)
Now, LHS\(=\frac { a^{ 2 } }{ \beta ^{ 2 } } +\frac { \beta ^{ 2 } }{ a^{ 2 } } \frac { a^{ 4 }+\beta ^{ 4 } }{ a^{ 2 }\beta ^{ 2 } } =\frac { (a^{ 2 })^{ 2 }(\beta ^{ 2 })^{ 2 } }{ a^{ 2 }\beta ^{ 2 } } \)
\(=\frac { (a^{ 2 }+\beta ^{ 2 })^{ 2 }-2a^{ 2 }\beta ^{ 2 } }{ a^{ 2 }\beta ^{ 2 } } \)
\(\left[ \because \quad a^{ 2 }+b^{ 2 }=(a+b)^{ 2 }-2ab \right] \)
\(=\frac { \left[ (a+\beta )^{ 2 }-2a\beta \right] ^{ 2 }-2\left( a\beta \right) ^{ 2 } }{ \left( a\beta \right) ^{ 2 } } \)
On substituting α+β=p and αβ=q, we get
\(LHS=\frac { \left[ p^{ 2 }-2q \right] ^{ 2 }-2q^{ 2 } }{ q^{ 2 } } \)
\(=\frac { (p^{ 2 })^{ 2 }+(2q)^{ 2 }-2\times p^{ 2 }\times 2q-2q^{ 2 } }{ q^{ 2 } } \)
\(\left[ \because \quad (a-b)^{ 2 }=a^{ 2 }+b^{ 2 }-2ab \right] \)
\(=\frac { p^{ 4 }+4q^{ 2 }-4p^{ 2 }q-2q^{ 2 } }{ q^{ 2 } } \)
\(=\frac { p^{ 4 }+2q^{ 2 }-4p^{ 2 }q }{ q^{ 2 } } =\frac { p^{ 4 } }{ q^{ 2 } } -\frac { 4p^{ 2 } }{ q } +2\)
=RHS
\(\therefore \) LHS=RHS

77 m²

Let AB= hm be height of the hill. BC = x km and BD = (x+1) km As ∠ACB = 60⁰ and ∠ADB = 30⁰ Consider rt. △ABC, we have
\(\frac{AB}{BC}\) =tan 60^o
\(\frac{h}{x}=\sqrt{3}\)
⇒ h=\(\sqrt{3}x\)  --- (i)

Consider rt. △ABD, we have
\(\frac{AB}{BD} \)=tan 30^o
⇒  \(\sqrt{3}\)h=x+1
⇒  \(\sqrt{3}\)ｘ\(\sqrt{3}\)x=x+1  [using(i)]
⇒ 3x=x+1
2x=1 ⇒ \(x=\frac{1}{2}\)
Putting the value of x in eq.(i), we have
h=\(\sqrt{3}\)ｘ\(\frac{1}{2}\)=\(\frac{1.732}{2}\)
=0.866 km
=866 metre
∴ Height of the hill = 866 metre.

Volume of the wood 
= External volume - internal volume
= (18 x 10 x 6 -17 x 9  x 5 ) cm³
= 315 cm³
Weight of box = 15 kg
Weight of 1 cubic cm of wood 
\(=\frac { 15 }{ 315 } =\frac { 1 }{ 21 } kg\)
Weight of box with sand = 100 kg
 Weight of sand = 100 - 15 = 85 kg
Voluhle of sand = volume of the box 
\(\Rightarrow \) Weight of 1 cu cm of stand 
=  \(\frac { 85 }{ 315 } =0.27\ kg\)

Since 4 + 7 = 11points are to be located on AX  at equal distantes, so B is joined to last point, A₁₁.