BQ = BP (length of tangents drawn from an external point to a circle are equal) ...(i)
CP = CR                                                                                                                ...(ii)
and AQ = AR
\(\therefore\) 2AQ = AQ + AR
= (AB + BQ) + (AC + CR)
= (AB + BP) + (AC + CP)
                  [using Eqs. (i) and (ii)]
\(\Rightarrow\)  2AQ = AB + AC + (BP + CP)
\(\Rightarrow\) 2AQ = AB + AC + BC
\(\begin{array}{ll} \Rightarrow & A Q=\frac{1}{2}(A B+B C+A C) \end{array}\)
\(\begin{array}{ll} \Rightarrow & A Q=\frac{1}{2} \text { Perimeter of } \triangle A B C \end{array}\)

(i) Let number of white balls be n.
Total number of balls = 6 + 4 + n = 10 + n
Let E be the event of drawing a white ball.
Given, \(P(E)=\frac{1}{3}\)
\(\begin{aligned} \Rightarrow \frac{\text { Number of outcomes favourable to } E}{\text { Total number of outcomes }}=\frac{1}{3} \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad \frac{n}{10+n}=\frac{1}{3} \Rightarrow n=5 \end{aligned}\)
(ii) Now, total number of balls = 10 + n = 10 + 5 = 15
Let x red balls are removed from the bag.
\(\therefore\)  Total number of balls in the bag = 15 - x
Now, \(\begin{gathered} P(E)=\frac{1}{2} \end{gathered}\)
\(\begin{gathered} \Rightarrow \frac{5}{15-x}=\frac{1}{2} \end{gathered}\)
\(\Rightarrow\)    x = 5

\(LHS=\sqrt { \frac { \sec { \theta } -1 }{ \sec { \theta } +1 } } +\sqrt { \frac { \sec { \theta } +1 }{ \sec { \theta } -1 } } \)
\(=\frac { \left( \sec { \theta } -1 \right) +\left( \sec { \theta } +1 \right) }{ \sqrt { \left( \sec { \theta } +1 \right) \left( \sec { \theta } -1 \right) } } \)
\(=\frac { 2\sec { \theta } }{ \sqrt { \sec ^{ 2 }{ \theta } .1 } } =\frac { 2\sec { \theta } }{ \tan { \theta } } \)
\(\left( \because \quad \tan ^{ 2 }{ \theta } =\sec ^{ 2 }{ \theta } -1 \right) \)
\(=2\times \frac { 1 }{ \cos { \theta } } \times \frac { \cos { \theta } }{ \sin { \theta } } \)
\(=2\times \frac { 1 }{ \sin { \theta } } \)
\(=2cosec\theta \)
= RHS

Given, CD = 6 cm, BD = 8 cm and radius = 4 cm

Join OC, OA and OB.
Let the circle touches the other sides AB and AC at points E and F, respectively.
We know that tangents drawn from an external point to the circle are equal in length.
\(\therefore\) CD = CF = 6 cm            [\(\because\) C is an external point]
BD = BE = 8 cm                [\(\because\) B is an external point]
and AF = AE = x cm (say  [\(\because\) A is an external point]
Area of \(\Delta\)OCB, \(A_1=\frac{1}{2} \times \text { Base } \times \text { Height }\)
\(\begin{aligned} & =\frac{1}{2} \times C B \times O D \\ \end{aligned}\)
\(\begin{aligned} & =\frac{1}{2} \times 14 \times 4=28 \mathrm{~cm}^2 \\ \end{aligned}\)
\(\begin{aligned} & \quad[\because C B=C D+B D=6+8=14] \end{aligned}\)
Area of \(\Delta\)OCA,
\(\begin{aligned} A_2 & =\frac{1}{2} \times A C \times O F \end{aligned}\)
\(\begin{aligned} =\frac{1}{2}(6+x) \times 4=(12+2 x) \mathrm{cm}^2 \end{aligned}\)
and area of \(\Delta\)OBA,
\(A_3=\frac{1}{2} \times A B \times O E=\frac{1}{2}(8+x) \times 4=(16+2 x) \mathrm{cm}^2\)
Thus, area of  \(\Delta\)ABC
= A₁ + A₂ + A₃ = [28 + (12 + 2x) + (16+ 2x)]
= (56 + 4x) cm²     ...(i)
Now, semi-perimeter of \(\Delta\)ABC=\(\frac{1}{2}\)(AB + BC + CA)
\(\Rightarrow \quad s=\frac{1}{2}(x+8+14+6+x)\)
\(\Rightarrow\) s = (14 + x) cm
Using Heron's formula,
area of \(\Delta\)ABC = \(\begin{aligned} & =\sqrt{s(s-a)(s-b)(s-c)} \end{aligned}\)
\(\begin{aligned} =\sqrt{(14+x)(14+x-14)(14+x-x-6)(14+x-x-8)} \end{aligned}\)
\(\begin{aligned} & =\sqrt{(14+x) \times x \times 8 \times 6} \end{aligned}\)
\(\begin{aligned} =\sqrt{(14+x) 48 x} \end{aligned}\)     ...(ii)
From Eqs. (i) and (ii), we get
\(\sqrt{(14+x) 48 x}=56+4 x=4(14+x)\)
On squaring both sides, we get
(14 + x) 48 x = 4² (14 + x)²
\(\Rightarrow\) 3x = 14 + x
\(\Rightarrow\) 2x = 14
\(\Rightarrow\) x = 7
\(\therefore\)  Length of AC = 6 + x = 6 + 7 = 13 cm
and length of AB = 8 + x = 8 + 7 = 15 cm

Given, one gulabjamun is a combination of a cylinder and two hemispheres. Here, total length of one gulabjamun = 5 cm and diameter = 2.8 cm

\(\therefore\) Radius of cylindrical part = Radius of hemispherical part
\(=r=\frac{2.8}{2}=1.4 \mathrm{~cm}\)
Height of cylindrical part,
h = PQ - (PR + SQ) = 5 - (1.4 + 1.4)
= 5 - 2.8 = 2.2 cm
\(\therefore\) Volume of one gulabjamun = 2 \(\times\) Volume of hemispherical part + Volume of cylindrical part
\(\begin{aligned} & =2 \times\left[\frac{2}{3} \pi r^{3}\right]+\pi r^2 h=\frac{4}{3} \pi r^{3}+\pi r^2 h \\ \end{aligned}\)
\(\begin{aligned} & =\pi r^2\left[\frac{4 r}{3}+h\right]=\frac{22}{7} \times 1.4 \times 1.4\left[\frac{4}{3} \times 1.4+2.2\right] \\ \end{aligned}\)
\(\begin{aligned} & =\frac{22}{7} \times \frac{14}{10} \times \frac{14}{10}\left[\frac{4}{3} \times \frac{14}{10}+\frac{22}{10}\right] \\ \end{aligned}\)
\(\begin{aligned} & =22 \times \frac{1}{5} \times \frac{7}{5}\left[\frac{28}{15}+\frac{11}{5}\right]=\frac{154}{25} \times \frac{61}{15}=\frac{9394}{375} \mathrm{~cm}^3 \end{aligned}\)
Now, volume of 45 gulabjamuns
\(=45 \times \frac{9394}{375}=\frac{3 \times 9394}{25}=\frac{28182}{25}=1127.28 \mathrm{~cm}^3\)
Since, one gulabjamun contains sugar syrup upto about 30% of its volume.
Hence, quantity of syrup found in 45 gulabjamuns
\(=1127.28 \times \frac{30}{100}=1127.28 \times \frac{3}{10}=338.184 \approx 338 \mathrm{~cm}^3\)

Given XY and X'Y' are two parallel tangents. Another tangent AB touches the circle at C and intersect XY at A and X'Y'at B.
To prove \(\angle\)AOB = 90°
Proof We know that, tangents drawn from an external point to a circle are equal in length.
\(\therefore\)   AP = AC   [\(\because\) A is an external point] ...(i)
Thus, in \(\Delta\)APO and \(\Delta\)ACO, AP = AC   [from Eq. (i)]
AO = AO  [common sides]
OP = OC [radii of circle]
\(\Delta\)APO \(\cong\)\(\Delta\)ACO  [by SSS congruence rule]
Then, \(\angle\)OAP = \(\angle\)OAC    [by CPCT]        ...(ii)
\(\Rightarrow\) \(\angle\)PAC = 2 \(\angle\)CAO                             ....(iii)
Similarly, we can prove that \(\angle\)CBO = \(\angle\)OBQ
\(\Rightarrow\)   \(\angle\)CBQ = 2 \(\angle\)CBO                           ...(iv)
since, XY || X'Y'      [given]
\(\therefore\) \(\angle\)PAC +  \(\angle\)QBC = 180°
[\(\because\) sum of interior angles on the same side of transversal is 180°]
\(\Rightarrow\) 2 \(\angle\)CAO + 2 \(\angle\)CBO = 180°    [from Eqs. (iii) and (iv)]
\(\Rightarrow\)   \(\angle\)CAO + \(\angle\)CBO = 90°                    ...(v)
Now, in \(\Delta\)AOB, \(\angle\)CAO + \(\angle\) CBO + \(\angle\)AOB = 180°
[by angle sum property of triangle]
\(\Rightarrow\) \(\angle\)CAO + \(\angle\)CBO = 180° - \(\angle\)AOB    ...(vi)
From Eqs. (v) and (vi), we get
180° - \(\angle\)AOB = 90° \(\Rightarrow\) \(\angle\)AOB = 90° Hence proved.

\(LHS=\frac { \left( \sin ^{ 4 }{ \theta } +\cos ^{ 4 }{ \theta } \right) }{ 1-2\sin ^{ 2 }{ \theta } \cos ^{ 2 }{ \theta } } \)
\(=\frac { { \left( \sin ^{ 2 }{ \theta } \right) }^{ 2 }+{ \left( \cos ^{ 2 }{ \theta } \right) }^{ 2 } }{ 1-2\sin ^{ 2 }{ \theta } \cos ^{ 2 }{ \theta } } \)
\(=\frac { { \left( \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } \right) }^{ 2 }-2\sin ^{ 2 }{ \theta } \cos ^{ 2 }{ \theta } }{ 1-2\sin ^{ 2 }{ \theta } \cos ^{ 2 }{ \theta } } \)
\(=\frac { 1-2\sin ^{ 2 }{ \theta } \cos ^{ 2 }{ \theta } }{ 1-2\sin ^{ 2 }{ \theta } \cos ^{ 2 }{ \theta } } =1=RHS\)

LHS=\((\tan { \theta } +2)(2\tan { \theta } +1)\)
\(=2\tan ^{ 2 }{ \theta } +4\tan { \theta } +\tan { \theta } +2\)
\(=2\tan ^{ 2 }{ \theta } +2+5\tan { \theta } \)
\(=2(\tan ^{ 2 }{ \theta } +1)+5\tan { \theta } \)
\(=2\sec ^{ 2 }{ \theta } +5\tan { \theta } \quad \left[ \because 1+\tan ^{ 2 }{ \theta } =\sec ^{ 2 }{ \theta } \right] \)
\(=5\tan { \theta } +2\sec ^{ 2 }{ \theta } \)=RHS
Hence proved.

Let h be height of the cylinder, and r the common radius of the cylinder and hemisphere. Then, the total surface area of the bird-bath = CSA of cylinder + CSA of hemisphere
= 2\(\pi\)rh + 2\(\pi\)r² = 2\(\pi\)r(h + r)
\(=2 \times \frac{22}{7} \times 30(145+30) \mathrm{cm}^2\)
= 33000 cm² = 3.3 m²

Total no.of cards = 46
Total no.of ways to select a card = 46
(i) Prime no.less than 10 in these cards are 5, 7
\(\therefore\)  No.of ways to select a prime no.less than 10 = 2
\(\therefore\)  Probability that the number on the card is prime = \(\frac{2}{46}=\frac{1}{23}\)
(ii) No. which is a perfect square, i.e. 9, 16, 25, 36, 49
No. of ways to select a card with perfect square = 5
\(\therefore\)   Probability = \(\frac{5}{46}\)

Given, umbrella to be a flat circle. So, the central angle of an umbrella is 360°.
Since, umbrella has 8 ribs.
\(\therefore\) Angle between two ribs.
\(=\frac{360^{\circ}}{8}=45^{\circ}\)
Area between two ribs = Area of one sector of the umbrella
\(=\frac{\theta}{360^{\circ}} \times \pi r^2=\frac{45^{\circ}}{360^{\circ}} \times \frac{22}{7} \times(45)^2[\because r=45, \text { given }]\)
\(\begin{aligned} & =\frac{22}{7 \times 8}(45)^2 \\ \end{aligned}\)
\(\begin{aligned} & =\frac{22275}{28} \mathrm{~cm}^2 \end{aligned}\)

Given: PQ is chord of length 16 cm, TP and TQ are the tangents to a circle with centre O, radius 10 cm. 
To find: TP. 
Solution: Join OP and OQ. 
In triangles OTP and OTQ, OT is common 
OP = OQ (radii) 
TP = TQ [length of the tangents drawn from a point outside the circle to the circle are equal]   
ஃ  \(\triangle\)OPT ≅ \(\triangle\)OQT (SSS cong. rule) 
ஃ  \(\angle \)POT = \(\angle \)QOT       .....(i) 
Consider, triangles OPR and OQR; 
OP = OQ (radii); OR is common 
\(\angle \)POR = \(\angle \)QOR    [from (i)] 
ஃ \(\triangle\)OPR ≅ \(\triangle\)OQR (SAS cong. rule) 
\(\angle \)ஃ PR = RQ =  \(1\over2\)x16 = 8 cm  ...(ii) 
\(\angle \)ORP = \(\angle \)ORQ = 90^o     ....(iii) 
In right angled triangle TRP,
TR² = TP² - (8)² = TP² - 64  [from (iii) .... (iv)]
Also    OT² = TP² - (10)²
 (TR+6)² = TP² + 100
[∵ OR = \(\sqrt{100-64}\)=6]
TR² + 12TR + 36 = TP² + 100
TP² - 64 + 12TR + 36 = TP² + 100
12TR = 128 ⇒ TR=\(32\over3\)cm
Form (iv) \({ \left( \frac { 32 }{ 3 } \right) }^{ 2 }\)=TP² - 64
⇒ TP² = \({1024\over9}+64={1024+576\over9}={1600\over9}\)
⇒ TP = \(40\over3\) cm

Given: A circle touches the side BC of \(\triangle\)ABC at P and touch AB and AC produce at Q and R respectively and AQ = 5 cm.
To find: Perimeter of \(\triangle\)ABC.Sol: AQ and AR are tangent from the same point AQ = AR = 5 cm ...(i)
[Tangent from the same external points are equal]
BQ and BP are tangent from the same point  ....(ii)
⇒   BQ = BP
CP and CR are also  tangent from the same point
⇒  CP = CR   .....(iii)
In ABC
Perimeter of \(\triangle\)ABC = AB + BC + AC = AB + BP + CP + AC AB + BQ + CR + AC =  AQ + AR  [From (ii) and (iii)]
= 5 cm + 5 cm = 10 cm  [From (i)]
Permeter \(\triangle\)ABC = 10 cm

Hint \(L H S =\frac{1}{1+\sin \theta}+\frac{1}{1-\sin \theta}=\frac{1-\sin \theta+1+\sin \theta}{(1+\sin \theta)(1-\sin \theta)}\)
\(=\frac{2}{\left(1-\sin ^2 \theta\right)}=\frac{2}{\cos ^2 \theta} \quad\left[\because 1-\sin ^2 A=\cos ^2 A\right] \)
\( =2 \sec ^2 \theta=\text { RHS } \quad \text { Hence proved. }\)

\(\frac { { cosec }^{ 2 }\theta }{ { cosec }\theta -1 } -\frac { { cosec }^{ 2 }\theta }{ { cosec }\theta +1 } \)
\(={ cosec }^{ 2 }\theta \left[ \frac { 1 }{ \frac { 1 }{ \sin { \theta } } -1 } -\frac { 1 }{ \frac { 1 }{ \sin { \theta } } +1 } \right] \)
\(={ cosec }^{ 2 }\theta \left[ \frac { \sin { \theta } }{ 1-\sin { \theta } } -\frac { \sin { \theta } }{ 1+\sin { \theta } } \right] \)
\(=\frac { 1\times \sin { \theta } }{ \sin ^{ 2 }{ \theta } } \left[ \frac { \left( 1+\sin { \theta } \right) -\left( 1-\sin { \theta } \right) }{ \left( 1-\sin { \theta } \right) \left( 1+\sin { \theta } \right) } \right] \)
\(=\frac { 1 }{ \sin { \theta } } \left[ \frac { 2\sin { \theta } }{ 1-\sin ^{ 2 }{ \theta } } \right] \)
\(=\frac { 2 }{ \cos ^{ 2 }{ \theta } } =2\sec ^{ 2 }{ \theta } \)
= RHS

2100.78 cm²

Here, Volume of a hemisphere = 2425 \(\frac{1}{2}\) cm³
\(\Rightarrow\)   \(\frac { 2 }{ 3 } \times \frac { 22 }{ 7 } \times { r }^{ 3 }=\frac { 4851 }{ 2 } \)
\(\Rightarrow\)    \({ r }^{ 3 }=\frac { 4851 }{ 2 } \times \frac { 3\times 7 }{ 2\times 22 } \)
\(=\frac { 441\times 21 }{ 2\times 2\times 2 } \)
\(\Rightarrow\)      r = \(\frac{21}{2} cm\)
Now, curved surface area = 2\(\pi\)r²
= 2 x \(\frac { 22 }{ 7 } \times \frac { 21 }{ 2 } \times \frac { 21 }{ 2 } \)
= 693 cm²

(i) Possible outcomes HHH, HHT, HTH,THH, TTH, THT, HTT, TTT
Total number of possible outcomes = 8
Let A =  getting exactly two heads
Favourable outcomes to event A are HHT, HTH to THH.
\(\therefore \ P(A)\ =\ \frac { 3 }{ 8 } \)
(ii) Let B = getting atleast two heads 
Favourable outcomes to B are HHT, HTH, THH or HHH.
\(\therefore P(B)\ =\ \frac { 4 }{ 8 } =\frac { 1 }{ 2 } \)
(iii) Let C = getting at least two tails 
\(\therefore \) Favourable outcomes to event C are TTH, THT, HTT or TTT
\(\therefore \ P(C)\ =\ \frac { 4 }{ 8 } =\frac { 1 }{ 2 } \)

Elernentarv events are
(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)
(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)
(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)​​​​​​​
(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)​​​​​​​
(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)​​​​​​​
(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)​​​​​​​
Total no. of elementary events = 36
(a)Let A = 5 will not collie up on either of thetn.
\(\therefore\) Total no. of eletnentary events favourable to A = 25. Hence, P(A) = \(\frac {25}{36}\) 
(b)B=5 will come up at least one die
 \(\therefore\) total no. of eletnentarv events lhsourable to B = 11. Hence, P(B) =\(\frac {11}{36}\) 
(c)C= 5 will come up at both dice
\(\therefore\)No. of elementary event favourable to C = 1 Hence, P(C)= \(\frac {1}{36}\)​​​​​​​

In OBD
cos 60°=\(OD \over OB\)and sin 60°=\(BD\over OB\)
⇒ \(1\over2\)= \(OD\over6\) and \(\frac { \sqrt { 3 } }{ 2 } \) = \(BD\over6\)
⇒ \(6\over2\)=OD and \(\frac { \sqrt { 3 } }{ 2 } \)x6 = BD
⇒ OD = 3 and BD = 3\(\sqrt3\)
BC = 2BD = 2 x 3\(\sqrt3\)=6\(\sqrt3\)
Area of the shaded region = area of circle - area of \(\triangle\)ABC
= \(2\pi { (6) }^{ 2 }-\frac { \sqrt { 3 } }{ 2 } { \left( 6\sqrt { 3 } \right) }^{ 2 }\)
= 3.14 x 6 x - \(\frac { \sqrt { 3 } }{ 4 } \) x 6\(\sqrt3\) x 6 \(\sqrt3\)
=  113.04 - 27 x 1.732 = 113.04 - 46.76 = 66.28 cm²

Total surface area (TSA) of toy = CSA of hemisphere + CSA of cone
This is because the toy is obtained by joining the plane surfaces of hemisphere and cone.

If Assertion is incorrect but Reason is correct.

(a) Length of the square shaped grass field = 20 m

Area of the square shaped grass field = 20 \(\times\) 20 = 400 m²
(b) (i) Length of the rope = 7 m
Thus, each horse can graze upto 7 m of distance along the side.
The grazed area is making a complete circle by taking all the four grazed parts.
So, area of grazed part = \(\pi\)r²
\(=\frac{22}{7} \times 7^2=22 \times 7=154 \mathrm{~m}^2\)
Therefore, area of the total field in which these horses can graze is 154 m².
(ii) New length of the rope of each horse = 10 m
Area grazed by one horse \(=\frac{1}{4} \pi r^2\)
\(\begin{aligned} & =\frac{1}{4} \times 3.14 \times 10^2 \\ \end{aligned}\)
\(\begin{aligned} & =\frac{314}{4}=78.5 \mathrm{~m}^2 \end{aligned}\)
Therefore, the required area grazed by one horse is 78.5 m².
(c) Length of rope of each horse = 7 m
Area of square shaped grass field = 400 m²
                                                        [from part (a)]
Area of total field grazed by horses = 154 m²
                                                        [from Eq. (ii) (a)]
Area of field left ungrazed = Area of square field - Area of grazed field by horses
= 400 - 154 = 246
Therefore, area of field left ungrazed is 246m².

(i) Total angle = 360°
Angle of white = 120°
\(\therefore \text { Probability }=\frac{120^{\circ}}{360^{\circ}}=\frac{1}{3}\)
(ii) Total angle = 360°
Angle of blue = 60°
Angle of green = 60°
\(\therefore \text { Probability }=\frac{60^{\circ}+60^{\circ}}{360^{\circ}}=\frac{120^{\circ}}{360^{\circ}}=\frac{1}{3}\)
(iii) Let n students participated in the survey
\(\Rightarrow \frac{90^{\circ}}{360^{\circ}} \times n=15 \Rightarrow n=15 \times 4=60\)
\(\therefore\) 60 students participated in the survey.
Or
Total angle = 360°
Angle of red = 30°
Angle of blue = 60°
\(\therefore \text { Probability }=\frac{60^{\circ}+30^{\circ}}{360^{\circ}}=\frac{90^{\circ}}{360^{\circ}}=\frac{1}{4}\)

Given,
Hemisphere Outer radius(R) = 6 cm
and Inner radius (r) = 5 cm
Cylinder (Mallet) Height (h) = 10 cm
and Radius (r') = 2 cm
(i) Volume of material used in making the mallet
= \(\pi\)r²h
= 3.14 \(\times\)2 \(\times\)2\(\times\)10 = 125.6 cm³
(ii) Inner surface area of bowel = LSA of hemisphere
= 2\(\pi\)r² = 2 \(\times\)3.14 \(\times\) 5 \(\times\) 5 = 157 cm²
(iii) Volume of metal used to make the bowl
\(\begin{aligned} & =\frac{2}{3} \pi\left(R^3-r^3\right) \\ \end{aligned}\)
\(\begin{aligned} & =\frac{2}{3} \times 3.14\left(6^3-5^3\right) \\ \end{aligned}\)
\(\begin{aligned} & =\frac{2}{3} \times 3.14 \times 91=190.493 \mathrm{~cm}^3 \end{aligned}\)
Or
Total surface area of mallet = 2\(\pi\)r' (h + r')
= 2 \(\times\) 3.14 \(\times\) 2(10 + 2)
=4 \(\times\)3.14 \(\times\)12 = 150.72 cm²