Given, radius and height = 7 m & 24 m.
Slant height(l)=\(\sqrt { { r }^{ 2 }+{ h }^{ 2 } } =\sqrt { { 7 }^{ 2 }+{ 24 }^{ 2 } } \)
\(=\sqrt { 625 } =25\quad m\)
\(C.S.A=\pi rl\)
\(=\frac { 22 }{ 7 } \times 7\times 25=550\quad { m }^{ 2 }.\)   
Let x m of cloth is required C.S.A.
= area of cloth.
\(\Rightarrow \quad 5x=550\quad \Rightarrow \quad x=\frac { 550 }{ 5 } =110cm\)
110 m of cloth is required.
Cost of cloth = 25 x 110=Rs.2750

Given: AD=1.5 cm, BD=3 cm and AB=AD+BD=1.5+3.0=4.5 cm.
Given, In triangle ADE and ABC,
\(\angle\)A is common and DE || BC
\(\angle\)ADE=\(\angle\)ABC  (Corresponding angles)
\(\triangle\)ADE~\(\triangle\)ABC,  (AA similarity)
\(\frac { ar(\triangle ADE) }{ ar(\triangle ABC) } =\frac { { AD }^{ 2 } }{ { AB }^{ 2 } } =\frac { { (1.5) }^{ 2 } }{ { (4.5) }^{ 2 } } =\frac { 1 }{ 9 } \)
\(\Rightarrow \quad \frac { ar(\triangle ADE) }{ ar(\triangle ABC)-ar(\triangle ADE) } =\frac { 1 }{ 9-1 } \)
\(\Rightarrow \quad \frac { ar(\triangle ADE) }{ ar(trapezium\quad BCED) } =\frac { 1 }{ 8 } \)

We know that,
Centroid of \(\Delta =\left( \frac { { x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 } }{ 3 } ,\frac { { y }_{ 1 }+{ y }_{ 2 }+{ y }_{ 3 } }{ 3 } \right) \)
Given, (0, 0) = \(\left( \frac { a+b+c }{ 3 } ,\frac { b+c+a }{ 3 } \right) \)
On comparing x-coordinates both sides, we get
\(\frac { a+b+c }{ 3 } =0\Rightarrow a+b+c=0\)

\(\frac{11}{12}\)

31p

Total number of ways to draw a card = 52
Number of ways to draw a black queen = 2
\(\therefore\)  Probability of getting a black green = \(\frac{2}{52}=\frac{1}{26}\)

  a = p and d = q
 a_n = a + (n - 1)d
a₁₀ = p + 9q

Yes, 16 : 1 ; unity of nation, fraternity and patriotism.

1:7

Let a be the first term and d be the common difference of the given AP.
Given, sum of 7 terms, \({S}_{7}=49\)
\(\Rightarrow\)              \(\frac {7}{2}[2a+(7-1)d]=49\)
\(\left[ \because { S }_{ n }=\frac { n }{ 2 } \left\{ 2a+\left( n-1 \right) d \right\} \right] \)
\(\Rightarrow\) \(\frac { 7 }{ 2 } \left( 2a+6d \right) =49\)
\(\Rightarrow\) a+3d= 7    ...(i)
and sum of 17 terms,\({ S }_{ 17 }=289\)
\(\Rightarrow\)   \(\frac { 17 }{ 2 } \left[ 2a+(17-1)d \right] =289\)
\(\Rightarrow\) \(\frac { 17 }{ 2 } \left[ (2a+16d) \right] =289\)
\(\Rightarrow\) a + 8d =17 ...(ii)
On subtracting Eq. (i) from Eq. (ii), we get
 5d = 10
\(\Rightarrow\)  d = 2 
On putting d = 2 in Eq. (i), we get a + 3(2)=7​​​​​​​
\(\Rightarrow\) \(a+6=7\Rightarrow a=1\)
Now, sum of n terms \({ S }_{ n }=\frac { n }{ 2 } \left[ 2a+\left( n-1 \right) d \right] \)
\(=\frac { n }{ 2 } \left[ 2\times 1+\left( n-1 \right) 2 \right] \)
\(=\frac { n }{ 2 } \left( 2+2n-2 \right) \)
​​​​​​​\(=\frac { n }{ 2 } \left( 2+2n-2 \right) \)

Let a be the first term and d be the common difference of an AP.
We know that, the nth term of an AP is 
\({ a } _ { n } = a + ( n - 1 ) d\)
Then, 11th term, \({ a } _ { 11 } = a + 10d =38\)...(i)
[\(\because a_{11} = 38,\) given]
and 16th term,   \({ a } _ { 16 } = a 15d = 73\) ...(iii)
[\(\because a_{16} = 73\) given]
On subtracting Eq. (i) from Eq. (ii), we get
\(5d=35\quad \quad \Rightarrow \quad \quad d=\frac { 35 }{ 5 } =7\)
On putting the value of d in Eq. (i), we get
\(a + 10 \times 7 =38\)
\(\Rightarrow\) a = 38 -70 = -32
Now, 31st term, \({ a } _ { 31 } = a + 30d\)
\(= -32 + 30 \times 7\)
\(= -32 + 210 = 178\) 

\((i){9\over19}(ii){1\over19}(iii){6\over19}(iv){3\over19}\)

Yes

Given: \(\triangle\)ABC, right angled at A. BL and CM are medians.
To prove :
4(BL² + CM²) = 5BC².
Proof: In \(\triangle\)ABL,
BL² = AB² + AL²
= AB² + \({ \left( \frac { AC }{ 2 } \right) }^{ 2 }\)(BL is median)
In \(\triangle\)ACM, CM² = AC² + AM²
= AC² +\({ \left( \frac { AB }{ 2 } \right) }^{ 2 }\) (CM is median) 
BL² + CM² = AB² + AC² + \({ \left( \frac { AC }{ 2 } \right) }^{ 2 }\)+\({ \left( \frac { AB }{ 2 } \right) }^{ 2 }\)
\(\Rightarrow\) 4(BL² + CM²) = 5AB²+5AC²
= 5(AB² + AC²)
= 5BC².
Hence proved.

We have, \(\triangle\)PQR~\(\triangle\)PAB  \(\left( \because \angle P is common\frac { PA }{ AQ } =\frac { PB }{ PR } \quad \right) \)

\(\Rightarrow \quad \frac { ar(\triangle PQR) }{ ar(\triangle PAB) } ={ \left( \frac { PQ }{ PA } \right) }^{ 2 }\)
\(\Rightarrow \quad \frac { 32 }{ ar(\triangle PAB) } ={ \left( \frac { 4k }{ 3k } \right) }^{ 2 }\)
are \(\triangle\)PAB = 18 cm²
area of quadrilateral AQRB
= area of \(\triangle\)PQR-area of \(\triangle\)PAB
= 32-18
= 14 cm²

It is given that diameter of a hemisphere = 21
Let radius of hemisphere be r.
\(\therefore\) \(r=\frac { 21 }{ 2 } cm\)
and total height of tropy = 24.5 cm
\(\therefore\) Height of cylinder = Height of trophy - Radius of hemisphere
\(=24.5-\frac { 21 }{ 2 } =24.5-10.5=14\quad cm\)
\(\therefore\) Volume of the metal used in the trophy
= Volume of cylinder + Volume of hemisphere
\(=\pi { r }^{ 2 }h+\frac { 2 }{ 3 } \pi { r }^{ 3 }=\pi { r }^{ 2 }\left( h+\frac { 2 }{ 3 } r \right) \)
\(=\frac { 22 }{ 7 } \times \frac { 21 }{ 2 } \times \frac { 21 }{ 2 } \left( 14+\frac { 2 }{ 3 } \times \frac { 21 }{ 2 } \right) \)
\(=\frac { 22 }{ 7 } \times \frac { 21 }{ 2 } \times \frac { 21 }{ 2 } (14+7)\)
\(=\frac { 22 }{ 7 } \times \frac { 21 }{ 2 } \times \frac { 21 }{ 2 } \times 21\)
\(=\frac { 11\times 3\times 21\times 21 }{ 2 } { cm }^{ 3 }\)
\(\therefore\) Weight of metal=\(\frac { 11\times 3\times 21\times 21 }{ 2 } \times 1.2\quad g\)
\(=11\times 3\times 21\times 21\times 0.6\)
= 8731.8g = 8.7318 kg

Let the given points be A=(1,-1), B=(5,2) and C=(9,5),
Then, AB = \(\sqrt { { (5-1) }^{ 2 }+{ (2+1) }^{ 2 } } \)
\(\left[ \because \ distance=\sqrt { { \left( { x }_{ 2 }-{ x }_{ 1 } \right) }^{ 2 }{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 } } \right] \)
\(=\sqrt { { \left( 4 \right) }^{ 2 }{ (3) }^{ 2 } } =\sqrt { 16+9 } =\sqrt { 25 } =\quad 5\quad units\)
\(BC=\sqrt { { \left( 9-5 \right) }^{ 2 }+{ (5-2) }^{ 2 } } \)
\(=\sqrt { { \left( 4 \right) }^{ 2 }+{ (3) }^{ 2 } } =\sqrt { 25 } =5\quad units\)
and \(AC=\sqrt { { (9-1) }^{ 2 }+{ (5+1) }^{ 2 } } =\sqrt { { (8) }^{ 2 }+{ (6) }^{ 2 } } \)
\(=\sqrt { 100 } =10 \ units\)
Clearly, AC=AB + BC
Hence, A, B and C are collinear points.
Hence proved.

Given, diameter of cylinder = Diameter of conical cavity
= 1.4 cm

\(\therefore\) Radius of cylinder = Radius of conical cavity
\(=\frac{\text { Diameter }}{2}=\frac{1.4}{2}=0.7 \mathrm{~cm}\)
Height of the cylinder = height of the conical cavity
= 2.4 cm
\(\therefore\) Slant height of the conical cavity,
\(\begin{aligned} l & =\sqrt{b^2+r^2}=\sqrt{(2.4)^2+(0.7)^2} \\ \end{aligned}\)
\(\begin{aligned} & =\sqrt{5.76+0.49} \\ \end{aligned}\)
\(\begin{aligned} & =\sqrt{6.25}=2.5 \mathrm{~cm} \end{aligned}\)
Now, TSA of remaining solid = CSA of conical cavity + CSA of cylinder + Area of the base of the cylinder
\(\begin{aligned} & =\pi r l+2 \pi r h+\pi r^2 \\ \end{aligned}\)
\(\begin{aligned} & =\pi r(l+2 h+r) \\ \end{aligned}\)
\(\begin{aligned} & =\frac{22}{7} \times 0.7 \times(2.5+2 \times 2.4+0.7) \\ \end{aligned}\)
\(\begin{aligned} & =22 \times 0.1 \times(2.5+4.8+0.7) \end{aligned}\)
\(\begin{aligned} & =2.2 \times 8 \\ \end{aligned}\)
\(\begin{aligned} & =17.6 \approx 18 \mathrm{~cm}^2 \end{aligned}\)

Number of total outcomes, n(S)=36
(i) When product of the numbers on the top of the dice is 6.
So, the possible ways are {(1,6), (2,3), (3,2), (6,1).}
Number of possible ways, n(E₁)=4
∴ Required probability\(=\frac { n(E_{ 1 }) }{ n(S) } =\frac { 4 }{ 36 } =\frac { 1 }{ 9 } \)
(ii) When the product of the numbers on the top of the dice is 12.
So, the possible ways are {(2,6), (3,4), (4,3), (6,2)}
Number of possible ways, n(E₂)=4
∴ Required probability\(=\frac { n(E_{ 2 }) }{ n(S) }\)
\( =\frac { 4 }{ 36 } =\frac { 1 }{ 9 } \)
(iii) Product of the numbers on the top of the dice cannot be7. its probability is zero

We know that the diagonals of a parallelogram bisect each other, therefore, mid-points of two diagonals coincide each other. 

Let P (1, 2), Q (4, y), R(x, 6) and S(3, 5) are the vertices of the parallelogram PQRS. Now, using mid-point formula, we have
 For, diagonal PR 
M\((\frac{x+1}{2},\frac{6+2}{2})\) i.e., M\((\frac{x+1}{2},{4})\)
For, diagonal QS
M\((\frac{4+3}{2},\frac{y+5}{2})\) i.e., M\((\frac{7}{2},\frac{y+5}{2})\)
Now, comparing the coordinates of M, we have
\(\frac{x+1}{2}=\frac{7}{2}\) and \(\frac{y+5}{2}=4\)
⇒ x=6 and y=3
Hence, x=6 and y=3. 

Here, sum of first seven terms. i.e., S₇ = 119
\({7 \over 2}[2a+(7-1)d]=119\)
14a + 42d = 238
a + 3d = 17
S₁₇ = 714
\({17 \over 2}(2a+16d)=714\)
\(\Rightarrow\)    17a + 136d = 714
a + 8d = 42            ....(ii)
Subtracting (i) from (ii), we have
5d = 25 \(\Rightarrow\) d = 5Therefore, a + 3 X 5 = 17  \(\Rightarrow\)  a = 2   [ Using (i) ]
\(\Rightarrow\)    S_n = \({n\over 2}[2\times2+(n-1)5]\)
[ Using S_n = \({n \over 2}[2a+(n-1)d]\)]
\(={n \over 2}(4+5n-5)\)
\(={n \over 2}[5n-1]\)
\(={(n)(5n-1)\over2}\)
S_n = \({5{n}^{2}-n\over 2}\)

(a) A bastman hits a boundary 9 times out of 45 balls he plays.
P(A) = P (hits a target) = \(\frac{9}{45}=\frac{1}{5}\)
Since, \(P(A)+P(\bar{A})=1\)
\(\begin{aligned} \Rightarrow \quad P(\bar{A}) & =1-P(A) \end{aligned}\)
\(\begin{aligned} =1-\frac{1}{5}=\frac{4}{5} \end{aligned}\)
\(\therefore\) Probability of not hitting the boundary is \(\frac{4}{5}\)
\(\therefore\) Both Assertion (A) and reason (R) are correct.

(d) If Assertion is incorrect but Reasonis correct.

(i) (b): (b) \(\triangle\)s PQM and RSM
(ii) (d): AA similarity
As \(\angle \mathrm{PQM}=\angle \mathrm{RSM}=90^{\circ}\)
and angle of incidence is equal to angle of reflection.
Then, \(\angle \mathrm{PMQ}=\angle \mathrm{RMS}\)
(iii) (a): Since, \(\Delta \mathrm{PMQ} \sim \Delta \mathrm{RMS}\)
Then, \(\frac{\mathrm{PQ}}{\mathrm{RS}}=\frac{\mathrm{QM}}{\mathrm{MS}} \Rightarrow \mathrm{PQ}=\frac{2.8 \times 2}{1.4}\)
= 4m
(iv) (c): Distance between Rashmi and Gulmohar tree
QS = QM + MS = 2.8 + 1.4 = 4.2 m

(i) (b): capacity of first glass \(=\pi r^{2} h-\frac{2}{3} \pi r^{3}\)
\(=\pi r^{2}\left(h-\frac{2}{3} r\right)\)
\(=\pi(3)^{2}\left(10-\frac{2}{3} \times 3\right)\)
\(=9 \pi(10-2)\)
72\(\pi\) cm³
(ii) (a): capacity of second glass \(=\pi r^{2} \mathrm{H}-\frac{1}{3} \pi r^{2} h\)
\(=\pi r^{2}\left(\mathrm{H}-\frac{1}{3} h\right)\)
\(=\pi(3)^{2}\left(10-\frac{1}{3} \times 1.5\right)\)
\(=9 \pi(10-0.5)\)
= 85.5 \(\pi\)cm³
(iii) (a): \(\text { Ratio }=\frac{\text { Capacity of first glass }}{\text { Capacity of second glass }}=\frac{72 \pi}{85.5 \pi}=\frac{16}{19}=16: 19 .\)
(iv) (d): From part (i) and (ii)
Suresh got more quantity of juice.
\(=85.5 \pi \mathrm{cm}^{3}-72 \pi \mathrm{cm}^{3}=13.5 \pi \mathrm{cm}^{3}\)
(v) (a): Total quantity of juice is purchased by Suresh = (72 \(\pi\) + 85.5 \(\pi\) ) cm³
= 157.5\(\pi\) cm³

Total number of coins = 50 + 48 + 36 + 28 + 8 = 170
(i) (c): Number of \(\begin{equation} ₹ \end{equation} \) 5 coins = 36
Required probability = \(\begin{equation} \frac{36}{70}=\frac{18}{85} \end{equation}\)
(ii) (b): Number of \(\begin{equation} ₹ \end{equation} \)20 coins = 8
Required probability = \(\begin{equation} \frac{8}{170}=\frac{4}{85} \end{equation}\)
(iii) (d): Number of \(\begin{equation} ₹ \end{equation} \)10 coins = 28
Probability (coin is of \(\begin{equation} ₹ \end{equation} \)10) = \(\begin{equation} \frac{28}{170} \end{equation}\)
Required probability = 1 - P(coin is of 10)
\(\begin{equation} =1-\frac{28}{170}=\frac{142}{170}=\frac{71}{85} \end{equation}\)
(iv) (a) : Total number of coins of \(\begin{equation} ₹ \end{equation} \)10 and \(\begin{equation} ₹ \end{equation} \) 20
= 28 + 8 = 36
Required probability = \(\begin{equation} \frac{36}{170}=\frac{18}{85} \end{equation}\)
(v) (a): Total number of coins of \(\begin{equation} ₹ \end{equation} \)1, \(\begin{equation} ₹ \end{equation} \)2 and \(\begin{equation} ₹ \end{equation} \) 5
 = 50 + 48 + 36 = 134
Required probability = \(\begin{equation} \frac{134}{170}=\frac{67}{85} \end{equation}\)