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\(=\frac { Volume \ of \ cuboidal \ lead }{ Volume \ of \ shot } \) 
\(=\frac { 1188 }{ 14.143 } =84(approx)\)

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.

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 = \(​​\frac {7} {10}\) 

30.38 cm²

9.72 cm²

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.

20 cm

66.67m

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, 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