Since the sum of the first n terms of an A.P. is given as 3n² + 5n
\(\therefore\)  S_n = 3n² + 5n                ....(i)
\(\therefore\)   S_n-1 = 3( n - 1)2 + 5 ( n - 1 ) 
= 3( n² + 1 - 2n ) + 5( n - 1 )
= 3n² + 3 - 6n + 5n - 5 
= 3n² - n - 2                  ...(ii)
Now,  an = S_n - S_n-1
= 3n² + 5n - 3n² + n + 2
an =  6n + 2
Put n = 1 and n = 2, we have
a₁ = 8 and a₂ = 14
Hence, the common difference is 6.

Given vertices of a triangle are A(6,4), B(4,-3) and C(8,-3).

We have, AB = \(\sqrt { { \left( 6-4 \right) }^{ 2 }+{ \left( 4+3 \right) }^{ 2 } } \) [\(\because \) distance=\(\sqrt { { \left( { x }_{ 2 }-{ x }_{ 1 } \right) }^{ 2 }-{ \left( { y }_{ 2 }-{ y }_{ 1 } \right) }^{ 2 } } \)
 =\(\sqrt { { \left( 2 \right) }^{ 2 }+{ \left( 7 \right) }^{ 2 } } \) = \(\sqrt { 4+49 } \)=\(\sqrt { 53 } \)units
and BC = \(\sqrt { { \left( 4-8 \right) }^{ 2 }+{ \left( -3+3 \right) }^{ 2 } } \) = \(\sqrt { { \left( -4 \right) }^{ 2 }+0 } \)= 4 units
\(\because \ \ \ \ AB=AC\)
So, \(\Delta \)ABC is an isosceles triangle.
Let D be the midpoint of BC, then the coordinates of D are \(\left( \frac { 4+8 }{ 2 } ,\frac { -3-3 }{ 2 } \right) \) i.e. (6, -3)        \(\left[ \because \ coordinates\ of\ mid-point=\left( \frac { { x }_{ 1 }+{ x }_{ 2 } }{2 } ,\frac { { y }_{ 1 }+{ y }_{ 2 } }{ 2 } \right) \right] \)
Now, AD =  \(\sqrt { { \left( 6-6 \right) }^{ 2 }+{ \left( 4+3 \right) }^{ 2 } } =\sqrt { 0+{ \left( 7 \right) }^{ 2 } } \)= 7 units.
Hence, length of the median AD is 7 units.

66.67m

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

20 cm

Steps of Construction : 
1. Draw a line segment AB = 13.5 cm.
2. Through A, construct an acute angle ㄥBAX ( < 90^o).
3. Mark nine points (2 + 3 + 4 = 9) at equal distances on AX Such that AA₁= A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅ =  A₅A₆ = A₆A₇ = A₇A₈ = A₈A₉.
4. Join A₉B.
5. Through A₂ and A₅ draw A₅R || A₉B and A₂Q || A₉B, intersecting AB in Q and R.
6. With Q as centre draw an arc of radius AQ.
7. With R as centre draw another arc of radius RB, intersecting previous arc in P.
8. Join PQ and PR

Thus, ΔPQR is the required triangle.

9.72 cm²

30.38 cm²

Given, dimensions of cuboidal lead = 9 cm x 11 cm x 12 cm
ஃ Volume of cuboidal= 9 x 11 x 12=1188 cm³ and diameter of shot=3 cm        [\(\because \) volume of cuboid=l x b x h]
ஃ Radius of shot, \(r=\frac { 3 }{ 2 } =1.5\quad cm\)
Now, volume of shot
\(=\frac { 4 }{ 3 } { \pi r }^{ 3 }=\frac { 4 }{ 3 } \times \frac { 22 }{ 7 } \times { (1.5) }^{ 3 }\\ =\frac { 297 }{ 21 } =14.143{ \quad cm }^{ 3 }\)
ஃ  Required number of shots\(=\frac { Volume \ of \ cuboidal \ lead }{ Volume \ of \ shot } \) 
\(=\frac { 1188 }{ 14.143 } =84(approx)\)

Let G₁, G₂, G₃ and B₁, B₂ be three girls and two boys respectively.
Since two children are selected at random, therefore the following are the possible groups:
B₁B₂, B₁G₁, B₁G₂, B₁G₃, B₂G₁, B₂G₂, B₂G₃, G₁G₂, G₁G₃, G₂G₃
Total number of cases = 10
Now, at least one boy is selected in the following ways:
B₁B₂, B₁G₁, B₁G₂, B₁G₃, B₂G₁, B₂G₂, B₂G₃
Total number of favourable cases = 7  
\(\therefore\) Required probability = \(​​\frac {7} {10}\)