Clearly, the fez is in the shape of a frustum of a cone with radii of one base as {tex}r_1{/tex} = 10 cm and radii of another base as {tex}r_2{/tex}  = 4 cm and slant height l = 15 cm. Let A be the area of the material used. Then,
A = Curved surface area + Area of the closed base

{tex}\Rightarrow \quad A = \pi \left( r _ { 1 } + r _ { 2 } \right) l + \pi r _ { 2 } ^ { 2 }{/tex}
{tex}\Rightarrow \quad A = \left\{ \frac { 22 } { 7 } \times ( 10 + 4 ) \times 15 + \frac { 22 } { 7 } \times 4 ^ { 2 } \right\} \mathrm { cm } ^ { 2 }{/tex}
{tex}\Rightarrow \quad A = \left( 352 \times 2 \times 15 + \frac { 352 } { 7 } \right) \mathrm { cm } ^ { 2 } = \left( 660 + \frac { 352 } { 7 } \right) \mathrm { cm } ^ { 2 } = 660 + 50.28 \mathrm { cm } ^ { 2 }{/tex}= 710.28cm².

Side of square = 10cm
For incircle
diameter of circle = side of square
{tex}\Rightarrow r = \frac{{10}}{2} = 5cm{/tex}
{tex} \therefore {/tex} Area of incircle {tex}= \pi {r^2}{/tex}
{tex}= 3.14 \times 5 \times 5{/tex}
78.50 cm²
For circumcircle
AC will be the diameter because {tex} \angle B = 90^\circ {/tex}
Now, In {tex} \triangle ABC{/tex}, by pythagoras theorem
AC² = AB² + BC²
{tex}\Rightarrow {/tex} AC² = 10² + 10²
{tex}\Rightarrow {/tex} AC² = 200
{tex}\Rightarrow AC = \sqrt {200} = 10\sqrt 2 cm{/tex}
{tex} \therefore {/tex} Area of circumcircle {tex}= \pi {r^2}{/tex}
{tex}= 3.14 \times {\left( {5\sqrt 2 } \right)^2}{/tex}
{tex}= 3.14 \times 50{/tex}
= 157cm²

Often in statistics, we tend to represent a set of data by a representative value which would approximately define the entire collection. This representative value is called as the measures of central tendency. The name itself suggests that it is a value around which the data is centered.

The measures of central tendency are given by various parameters but the most commonly used are mean, median and mode

Given: ABC is a triangle in which DE {tex}\parallel{/tex} BC.
To prove: {tex}\frac { A D } { B D } = \frac { A E } { C E }{/tex}
Construction: Draw  {tex}D N \perp A E{/tex} and  {tex}E M \perp A D{/tex}., Join BE and CD.
Proof :

In {tex}\triangle ADE,{/tex}
Area of {tex} \Delta A D E = \frac { 1 } { 2 } \times A E \times D N{/tex} ...(i)
In {tex}\Delta D E C{/tex},
Area of {tex}\operatorname \Delta D C E = \frac { 1 } { 2 } \times C E \times D N{/tex} ...(ii)
Dividing equation (l) by equation (ii),
{tex}\Rightarrow\frac { \text { area } ( \Delta A D E ) } { \text { area } ( \Delta D E C ) } = \frac { \frac { 1 } { 2 } \times A E \times D N } { \frac { 1 } { 2 } \times C E \times D N }{/tex} 
{tex}\Rightarrow{/tex} {tex}\frac { \text { area } ( \Delta A D E ) } { \text { area } ( \Delta D E C ) } = \frac { A E } { C E }{/tex} ...(iii)
Similarly, In {tex}\Delta A D E{/tex},
Area of {tex}\operatorname \Delta A D E = \frac { 1 } { 2 } \times A D \times E M{/tex} ...(iv)
In {tex}\Delta D E B,{/tex}
Area of {tex}\operatorname \Delta D E B = \frac { 1 } { 2 } \times E M \times B D{/tex} ...(v)
Dividing equation (iv) by equation (v),
{tex}\Rightarrow{/tex}{tex}\frac { \text { area } ( \Delta A D E ) } { \text { area } ( \Delta D E B ) }{/tex} {tex}= \frac { \frac { 1 } { 2 } \times A D \times E M } { \frac { 1 } { 2 } \times B D \times E M }{/tex} 
{tex}\Rightarrow{/tex} {tex}\frac { \text { area } ( \Delta A D E ) } { \text { area } ( \Delta D E B ) } = \frac { A D } { B D }{/tex} ...(vi)
{tex}\Delta D E B \text { and } \Delta D E C{/tex} lie on the same base DE and between two parallel lines DE and BC.
{tex}\therefore{/tex} Area ({tex} \Delta D E B {/tex}) = Area ( {tex} \Delta D E C {/tex})
From equation (iii),
{tex}\Rightarrow\frac { \text { area } ( \Delta A D E ) } { \text { area } ( \Delta D E B ) } = \frac { A E } { C E }.{/tex}......(vii)
From equation (vi) and equation (vii),
{tex}\frac { A E } { C E } = \frac { A D } { B D }{/tex} 
{tex}\therefore{/tex} If a line is drawn parallel to one side of a triangle to intersect the other two sides in two points, then the other two sides are divided in the same ratio. 

In right triangle ABP,
{tex}\tan 30 ^ { \circ } = \frac { \mathrm { AB } } { \mathrm { BP } }{/tex}
{tex}\Rightarrow \frac { 1 } { \sqrt { 3 } } = \frac { \mathrm { AB } } { \mathrm { BP } }{/tex}
BP = AB{tex}\sqrt3{/tex} ........ (i)
In right triangle ABQ,
{tex}tan\;60^0\;={AB\over BQ}{/tex}
{tex}\Rightarrow \sqrt { 3 } = \frac { A B } { B Q }{/tex}
{tex}\Rightarrow _ { B Q } = \frac { A B } { \sqrt { 3 } }{/tex}....... (ii)
{tex}\because{/tex} PQ = BP - BQ
{tex}\therefore{/tex} PQ = AB{tex}\sqrt { 3 } - \frac { A B } { \sqrt { 3 } } = \frac { 3 A B - A B } { \sqrt { 3 } } = \frac { 2 A B } { \sqrt { 3 } }{/tex} = 2BQ [From eq. (ii)]
{tex}\Rightarrow{/tex} BQ = {tex}\frac12{/tex}PQ
{tex}\because{/tex} Time taken by the car to travel a distance PQ = 6 seconds.
{tex}\therefore{/tex} Time taken by the car to travel a distance BQ, i.e. {tex}\frac12{/tex}PQ = {tex}\frac12{/tex} {tex}\times{/tex} 6 = 3 seconds.
Hence, the further time taken by the car to reach the foot of the tower is 3 seconds.

We have,
LHS = cosec²{tex}\theta{/tex} + sec²{tex}\theta{/tex}
{tex}\Rightarrow \quad \mathrm { LHS } = \frac { 1 } { \sin ^ { 2 } \theta } + \frac { 1 } { \cos ^ { 2 } \theta }{/tex}
{tex}\Rightarrow \quad \mathrm { LHS } = \frac { \cos ^ { 2 } \theta + \sin ^ { 2 } \theta } { \cos ^ { 2 } \theta \sin ^ { 2 } \theta } = \frac { 1 } { \sin ^ { 2 } \theta \cos ^ { 2 } \theta } = \frac { 1 } { \sin ^ { 2 } \theta } \times \frac { 1 } { \cos ^ { 2 } \theta } = \operatorname { cosec } ^ { 2 } \theta \sec ^ { 2 } \theta{/tex} = RHS

Given,

{tex}sec\ \theta+ tan\ \theta  = p{/tex} ...(i)
Also, we know that,

 {tex}sec^2 \theta  - tan^2 \theta  = 1{/tex}
{tex}\Rightarrow{/tex} (sec {tex}\theta{/tex} - tan {tex}\theta{/tex}) (sec {tex}\theta{/tex} + tan {tex}\theta{/tex}) = 1 [{tex}\because a^2-b^2=(a+b)(a-b){/tex}]
{tex}\Rightarrow{/tex} (sec {tex}\theta{/tex} - tan {tex}\theta{/tex})p = 1  [using equation (i)]
{tex}\Rightarrow{/tex} sec {tex}\theta{/tex} - tan {tex}\theta{/tex} {tex}=\frac{1}{p}{/tex} ...(ii)
(i)+(ii),  we get,

{tex}sec\theta + tan\theta+ sec\theta - tan\theta = p+ \frac{1}{p}{/tex}
{tex}\Rightarrow 2sec\theta = \frac{p^2+1}{p}{/tex}

{tex}\Rightarrow sec\theta = \frac{p^2+1}{2p}{/tex}

{tex}\Rightarrow \frac{1}{cos\theta} =\frac{p^2+1}{2p}{/tex}

{tex}\Rightarrow cos\theta =\frac{2p}{p^2+1}{/tex}------(iii)

Now, we know that,

{tex}sin\theta = \sqrt( 1- cos^2\theta) {/tex}

put the value of {tex}cos\theta{/tex} from eq. (iii), we get,

{tex}sin\theta = \sqrt(1-(\frac{2p}{p^2+1})^2){/tex}

{tex}\Rightarrow sin\theta = \sqrt(1-\frac{4p^2}{(p^2+1)^2}){/tex}

{tex}\Rightarrow sin\theta = \sqrt(\frac{(p^2+1)^2-4p^2}{(p^2+1)^2}){/tex}

{tex}\Rightarrow sin\theta = \sqrt(\frac{p^4+1+2p^2-4p^2}{(p^2+1)^2}){/tex}

{tex}\Rightarrow sin\theta = \sqrt(\frac{p^4+1-2p^2}{(p^2+1)^2}){/tex}

{tex}\Rightarrow sin\theta = \sqrt(\frac{(p^2-1)^2}{(p^2+1)^2}){/tex}

{tex}\Rightarrow sin\theta = \frac{p^2-1}{p^2+1}{/tex}

{tex}cosec\theta = \frac{p^2+1}{p^2-1} [\because cosec\theta =\frac{1}{sin\theta}]{/tex}

hence,   {tex}cosec\ \theta{/tex} {tex}=\frac{1+p^{2}}{1-p^{2}}{/tex}

Given: Radius of base (r) and height (h) of the conical tank are 7 m and 24 m

 ⇒ Slant height (l) = r^{​​​​​2} + h^{​​​​​​2}          

{tex}= \sqrt { r ^ { 2 } + h ^ { 2 } } = \sqrt { 7 ^ { 2 } + 24 ^ { 2 } }{/tex}

{tex}= \sqrt { 625 } = 25 \mathrm { m }{/tex}

C.S.A. {tex}= \pi r l{/tex}

{tex}= \frac { 22 } { 7 } \times 7 \times 25 = 550 \mathrm { m } ^ { 2 }{/tex}

Let x m of cloth is required

CSA of tent = area of cloth.

or, {tex}5 x = 550 \text { or, } x = \frac { 550 } { 5 } = 110 \mathrm { m }{/tex}

{tex}\therefore{/tex} 110 m of cloth is required.

Cost of cloth {tex}= 25 \times 110 = Rs. 2750{/tex}