Given circumference = 484 cm
\(2 \pi r =484 \)
\(2 \times \frac{22}{7} \times r =484 \)
\(r =\frac{484 \times 7}{44}=77 \mathrm{~cm} \)
height h = 105 cm
Volume of cone \(=\frac{1}{3} \pi r^{2} h \text { cu. units } \)
\(=\frac{1}{3} \times \frac{22}{7} \times 77 \times 77 \times 105 \)
= 652190 cm³

Let r and h be the radius and height of the cylinder respectively.
Given that, height h = 2 m, base area = 250 m²
Now, volume of a cylinder = \(\pi\)r h ² cu. units
= base area x h
= 250 x 2 = 500 m³
Therefore, volume of the cylinder = 500 m³

Let x be the length of the ladder. BC = 4 ft, AC = 7 fit.
By Pythagoras theorem we have, AB² = AC² + BC²
x² = 7² + 4² gives x² = 49 + 16
x² = 65, Hence \(x=\sqrt { 65 } \)
The number \(\sqrt { 65 } \) is between 8 and 8.1.
82 = 64 < 65.61 = 8.1²
Therefore, the length of the ladder is approximately 8.1ft

Let the radii of two cones be r₁ and r₂ and heights be h₁ and h₂
Given ratio of their radii = \(\frac{r_{1}}{r_{2}}=\frac{1}{3}\)
\(r_{1}=\frac{r_{2}}{3} \)
\(h_{1}=3 r_{1}, h_{2}=3 r_{1} \)
[ r₁ is the radius of smaller cone]
Slant heights \(l_{1} =\sqrt{h_{1}^{2}+r_{1}^{2}} \)
\(=\sqrt{9 r_{1}^{2}+r_{1}^{2}}=\sqrt{10} r_{1} \)
\(l_{2} =\sqrt{h_{2}^{2}+r_{2}^{2}} \)
\(=\sqrt{9 r_{1}^{2}+9 r_{1}^{2}}=\sqrt{18 r_{1}^{2}}=3 \sqrt{2} r_{1}\)
Ratio of curved surface areas
\(=\frac{\text { CSA of I cone }}{\text { CSA of II cone }} \)
\(=\frac{\pi r_{1} l_{1}}{\pi r_{2} l_{2}} =\frac{r_{1}\left(\sqrt{10} r_{1}\right)}{\left(3 r_{1}\right)\left(3 \sqrt{2} r_{1}\right)} \)
\(=\frac{\sqrt{10}}{9 \sqrt{2}}= \frac{\sqrt{5} \sqrt{2}}{9 \sqrt{2}}=\frac{\sqrt{5}}{9} \)
Ratio of C.S.A = \(\sqrt{5}: 9\)

Let l, R and r be the slant height, top radius and bottom radius of the frustum.
Given that, l = 5 cm, R = 4 cm, r = 1 cm
Now, C.S.A. of the frustum \(\pi\)(R + r)l sq.units
\(\frac { 22 }{ 7 } \times (4+1)\times 5\)
\(=\frac { 550 }{ 7 } \)
Therefore, C.S.A. = 78.57 cm²

Let r and h be the radius and height of the cone respectively.
Given that, radius r = 7 m and height h = 24 m
Hence, l = \(\sqrt { { r }^{ 2 }+{ h }^{ 2 } } \)
\(=\sqrt { 49+576 } \)
\(l=\sqrt { 625 } =25m\)
C.S.A. of the conical tent = \(\pi\)rl sq. units
Area of the canvas \(=\frac { 22 }{ 7 } \times 7\times 25={ 550 }m^{ 2 }\)
Now, length of the canvas \(\frac{Area\ of\ the\ canvas}{width}=\frac{550}{4}=137.5m\)
Therefore, the length of the canvas is 137.5 m

In \(\triangle\)ABC, AD is the bisector of \(\angle\)A
Therefore by Angle Bisector of \(\angle\)A
\(\frac { AB }{ AC } =\frac { BD }{ DC } \)
\(\frac{4}{3}=\frac{6}{A C}\) gives 4AC = 18. Hence, AC \(=\frac{9}{2}=4.5 \mathrm{~cm}\)

Let DE is the vertical stick and AB is the tower,
DE = 6 m, EF = 400 cm = 4 m, BC = 28 m
From DFE and ACB
Using similarity criteria
\(\frac{A B}{D E}=\frac{B C}{E F}\)
\(\frac{A B}{6}=\frac{28}{4} \)
\(A B=\frac{28 \times 6}{4}=42 \mathrm{~m} \)
Height of the tower = 42 m

Since the ratio of area of two similar triangles is equal to the ratio of the squares of any two corresponding sides, we have
\(\frac { Area(\Delta ABC) }{ Area(\Delta DEF) } =\frac { { BC }^{ 2 } }{ { EF }^{ 2 } } \) gives \(\frac { 54 }{ Area(\Delta DEF) } =\frac { { 3 }^{ 2 } }{ { 4 }^{ 2 } } \)
\(Area(\Delta DEF)=\frac { 16\times 54 }{ 9 } =96{ cm }^{ 2 }\)

In \(\Delta BAC\) and \(\Delta PRQ,\quad \frac { AB }{ RQ } =\frac { 3 }{ 6 } =\frac { 1 }{ 2 } \)
\(\frac { BC }{ QP } =\frac { 6 }{ 12 } =\frac { 1 }{ 2 } ;\frac { CA }{ PR } =\frac { 3\sqrt { 3 } }{ 6\sqrt { 3 } } =\frac { 1 }{ 2 } \)
Therefore, \(\frac { AB }{ QP } =\frac { BC }{ QP } =\frac { CA }{ PR } \)
By SSS similarity, we have \(\triangle\)BAC~\(\triangle\)QRB
\(\angle\)P =\(\angle\)C (since the corresponding parts of similar triangle)
\(\angle\)P =\(\angle\)C 180^o -\((\angle A+\angle B)={ 180 }^{ 0 }-({ 90 }^{ 0 }+{ 60 }^{ 0 })\)
\(\angle\)P = 180^o - 150^o = 30^o

Statement :
The internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the corresponding sides containing the angle
Proof

Given : In ΔABC,AD is the internal bisector
To prove : \(\frac{A B}{A C}=\frac{B D}{C D}\)
Construction : Draw a line through C parallel to AB. Extend AD to meet line through C at E

No	Statement	Reason
1.	∠AEC =∠BAE =∠1	Two parallel lines cut by a transversal make alternate angles equal
2.	ΔACE is isosceles AC = CE … (1)	In ΔACE,∠CAE = ∠CEA
3.	ΔABD∼ΔECD \(\frac{A B}{C E}=\frac{B D}{C D}\)	By AA Similarity
4.	\(\frac{A B}{A C}=\frac{B D}{C D}\)	From (1) AC = CE Hence proved

Statement
In a right angle triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides.

Proof
Given : In \(\triangle\)ABC , \(\angle\)A = 90°
To prove : AB² +AC² = BC²
Construction : Draw AD \(\perp\) BC

No.	Statement	Reason
1.	Compare \(\triangle\)ABC and \(\triangle\)DBA ÐB is common \(\angle\)BAC = \(\angle\)BDA = 90° Therefore, \(\triangle\)ABC \(\sim\) \(\triangle\)DBA \(\frac{A B}{B D}=\frac{B C}{A B}\) AB2 = BC x BD ........(1)	Given \(\angle\)BAC = 90° and by construction \(\angle\)BDA = 90° By AA similarity
2.	Compare \(\triangle\)ABC and \(\triangle\)DAC \(\angle\)C is common \(\angle\)BAC = \(\angle\)ADC = 90o Therefore, \(\triangle\)ABC \(\sim\) \(\triangle\)DAC \(\frac{B C}{A C}=\frac{A C}{D C}\) AC2 = BC x DC    ...(2)	Given \(\angle\)BAC = 90° and by construction \(\angle\)ADC = 90° By AA similarity

Adding (1) and (2) we get
AB² +AC² = BC x BD +BC x DC
= BC(BD +DC) = BC x BC
AB² +AC² = BC² .
Hence the theorem is proved.

Statement
A straight line drawn parallel to a side of triangle intersecting the other two sides, divides the sides in the same ratio.
Proof

In \(\Delta ABC\) ,D is a point on AB and E is a point on AC
To prove : \(\cfrac { AD }{ DB } =\cfrac { AE }{ EC } \) 
Construction: Draw a line DE || BC

No.	Statement	Reason
1.	\(\angle ABC=\angle ADE=\angle 1\)	Corresponding angles are equal because DE || BC
2.	\(\angle ACB=\angle AED=\angle 2\)	Corresponding angles are equal because DE || BC
3.	\(\\ \angle DAE=\angle BAC=\angle 3\)	Both triangles have a common angle
4.	\(\Delta ABC\sim \Delta ADE\)	By AAA similarity
	\(\frac { AB }{ AD } =\frac { AC }{ CE } \)	Corresponding sides are proportional
	\(\frac { AD+DB }{ AD } =\frac { AE+EC }{ AE } \)	Split AB and AC using the points D and E.
	\(1+\frac { DB }{ AD } =1+\frac { EC }{ AE } \)	On simplification
	\(\frac { DB }{ AD } =\frac { EC }{ AE } \)	Cancelling 1 on both sides
	\(\frac { AD }{ DB } =\frac { AE }{ EC } \)	Taking reciprocals
		Hence proved

Let \(\triangle\)ABC be B triangle points D, E, F are angular bisectors of \(\angle A, \angle B \text { and } \angle C\) respectively. By angular bisector theorem we have
\( \frac{B D}{D C}=\frac{A B}{A C} \Rightarrow \mathrm{AB}=\frac{B D \times A C}{D C} \)
\(\frac{A C}{B C}=\frac{A F}{F B} \Rightarrow \mathrm{AC}=\frac{A F \times B C}{F B} \)
\(\frac{A E}{E C}=\frac{A B}{B C} \Rightarrow \mathrm{AB}=\frac{A E \times B C}{E C}\)
From (1) and (3), we have
\(\frac{B D \times A C}{D C}=\frac{A E \times B C}{E C}\)
Now substituting (2) in (4) we have
\( \frac{B D \times\left(\frac{A F \times B C}{F B}\right)}{D C} =\frac{A E \times B C}{E C} \)
\(\frac{B D \times A F \times B C}{D C \times F B} =\frac{A E \times B C}{E C} \)
\(B D \times A F \times E C =\frac{A E \times B C \times D C \times F B}{B C} \)
\(B D \times A F \times C E =E A \times F B \times D C \)
\(\therefore \frac{B D \times A F \times C E}{E A \times F B \times D C}=1\)
Hence by Ceva's theorem we conclude that the angle bisectors of a triangle are concurrent.

Let h and r be the height and radius of the cylinder respectively.
Given that, h = 15 cm, r = 6 cm
Volume of the container V = \(\pi\)r²h cubic units.
Let, r₁ = 3 cm, h₁ = 9 cm be the radius and height of the cone.
Also, r₁ = 3 cm is the radius of the hemispherical cap.
Volume of one ice cream cone = (Volume of the cone + Volume of the hemispherical cap)
\(=\frac { 1 }{ 3 } \pi { r }_{ 1 }^{ 2 }{ h }_{ 1 }+\frac { 2 }{ 3 } \pi { r }_{ 1 }^{ 3 }\)
\(=\frac { 1 }{ 3 } \times \frac { 22 }{ 7 } \times 3\times 3\times 9+\frac { 2 }{ 3 } \times \frac { 22 }{ 7 } \times 3\times 3\times 3\)
\(=\frac { 22 }{ 7 } \times 9(3+2)=\frac { 22 }{ 7 } \times 45\)
\(Number\ of\ cones=\frac { volume\ of\ the\ cylinder }{ volume\ of\ one\ ice\ cream\ cone } \)
Number of ice cream cones needed \(=\frac { \frac { 22 }{ 7 } \times 6\times 6\times 15 }{ \frac { 22 }{ 7 } \times 45 } =12\)
Thus 12 ice cream cones are required to empty the cylindrical container.

From the figure,
Diameter of hemisphere = 3 mm
Radius of hemisphere = 1.5 mm
Volume of hemisphere  = \(\frac{2}{3} \pi r^{3} \ cu. units \)
Volume of 2 hemispheres = \(2\left(\frac{2}{3} \pi r^{3}\right) \)
\(=\frac{4}{3} \pi(1.5)^{3} \)
\(=4.5 \pi \mathrm{mm}^{3} \)
Radius of cylinder = 1.5 mm
Height of cylinder = 12 - 3 = 9 mm
Volume of cylinder \(=\pi r^{2} h \text { cu. units } \)
\(=\pi(1.5)^{2}(9) \)
\(=20.25 \pi \mathrm{mm}^{3} \)
Amount of medicine that a capsule can hold = Volume of cylinder + Volume of 2 hemispheres
\(=20.25 \pi+4.5 \pi \)
\(=24.75 \pi \mathrm{mm}^{3} \)
\(=24.75 \times \frac{22}{7}=77.785 \mathrm{~mm}^{3}\)

Let l, b and h₁ be the length, breadth and height of the cuboid. Also let us take r and h₂ be the radius and height of the cylinder.
Now, Volume of the box = Volume of the cuboid + \(\frac{1}{2}\) (Volume of cylinder)
\((l\times b\times { h }_{ 1 })+\frac { 1 }{ 2 } ({ \pi r }^{ 2 }{ h }_{ 2 })cu.units\)
\(=\left( 30\times 15\times 10 \right) +\frac { 1 }{ 2 } \left( \frac { 22 }{ 7 } \times \frac { 15 }{ 2 } \times 30 \right) \)
= 4500 + 2651.79 = 7151. 79
Therefore, Volume of the box = 7151.79 cm³

Let r and h be the radius and height of the cylinder respectively.
Given that, diameter d = 12 cm, radius r = 6 cm
Total height of the toy is 25 cm
Therefore, height of the cylindrical portion = 25 - 6 = 19 cm
T.S.A. of the toy = C.S.A. of the cylinder + C.S.A. of the hemisphere + Base Area of the cylinder
\(2\pi rh+2\pi { r }^{ 2 }+\pi { r }^{ 2 }\)
\(=\pi r(2h+3r)\quad sq.units\)
\(\frac { 22 }{ 7 } \times 6\times 56=1056\)
Therefore, T.S.A. of the toy is 1056 cm²

Let h, l, R and r be the height, slant height, outer radius and inner radius of the frustum.
Given that, diameter of the top = 10 m; radius of the top R = 5 m.
diameter of the bottom = 4 m; radius of the bottom r = 2 m, height h = 4 m
Now, \(l=\sqrt { { h }^{ 2 }+\left( R-{ r } \right) ^{ 2 } } \)
\(=\sqrt { { 4 }^{ 2 }+(5-2)^{ 2 } } \)
\(l=\sqrt { 16+9 } =\sqrt { 25 } =5m\)
Here, C.S.A. = \(\pi\)(R + r)l sq. units
\(\frac { 22 }{ 7 } (5+2)\times 5={ 110m }^{ 2 }\)
T.S.A. = \(\pi\)(R + r)l + \(\pi\)R² + \(\pi\)r² sq. units
\(\frac { 22 }{ 7 } \left[ (5+2)5+25+4 \right] =\frac { 1408 }{ 7 } =201.14\)
Therefore, C.S.A. = 110 m² and T.S.A. = 201.14 m²

Let AB and CD be two poles of height ‘a’ metres and ‘b’ metres respectively such that the poles are ‘p’ metres apart. That is AC = p metres. Suppose the lines AD and BC meet at O, such that OL = h metres
Let CL = x and LA = y.
Then, x + y = p
In \(\Delta ABC\) and \(\Delta LOC\), we have
\(\angle CAB=\angle CLO\) [each equal to 90^o]
\(\angle C=\angle C\) [ C is common]
\(\Delta CAB\sim \Delta CLO\) [By AA similarity]
\(\frac { CA }{ CL } =\frac { AB }{ LO } \) gives \(\frac { P }{ x } =\frac { a }{ h } \)
so, \(x=\frac { ph }{ a } \)   ..(1)
In \(\Delta ALO\) and \(\Delta ACD\), we have
\(\angle ALO=\angle ACD\) [each equal to 90° ]
\(\angle A=\angle A[\mathrm{~A} \text { is common }]\)
\(\frac { AL }{ AC } =\frac { OL }{ DC } \) gives \(\frac { y }{ p } =\frac { h }{ b } \) we get, \(y=\frac { ph }{ b } \)     ....(2)
(1) + (2) gives \(x+y=\frac { ph }{ a } +\frac { ph }{ b } \)
\(p=ph\left( \frac { 1 }{ a } +\frac { 1 }{ b } \right) \) (since x + y = p)
\(1=h\left( \frac { a+b }{ ab } \right) \)
Therefore, \(h=\frac { ab }{ a+b } \)
Hence, the height of the intersection of the lines joining the top of each pole to the foot of the opposite pole is \(\frac { ab }{ a+b } \) meters.