πr\(\left(l+\frac{\mathrm{r}}{4}\right)\)

Let radius of a cone = 4x
and slant height of a cone = 7x
 Now, curved surface area of a cone = 792 cm2
\(\Rightarrow \frac { 22 }{ 7 } \times 4x\ \times \ 7x=792\)
[ \(\therefore \) curved surface area of a cone  = \(\pi rl]\)
\(\Rightarrow 88x^{ 2 }=792\)
\(\Rightarrow x^{ 2 }=\frac { 792 }{ 88 } =9\)
\(\therefore x=3\)
[On taking positive square root]

Rs 137280

130 m

213.52 m

1 : 4

\(\sqrt { 297 } cm\)

Curved surface area = \(\pi\)rl
\(=\frac{22}{7} \times 7 \times 10 \mathrm{~cm}^{2}\) 
= 220 cm²

Here, h = 16 cm and r = 12 cm.
So, from l² = h² + r², we have
\(l=\sqrt{16^{2}+12^{2}} \mathrm{~cm}=20 \mathrm{~cm}\)
So, curved surface area = \(\pi\)rl
= 3.14 x 12 x 20 cm²
= 753.6 cm²
Further, total surface area = \(\pi\)rl + \(\pi\)r²
= (753.6 + 3.14 x 12 x 12) cm²
= (753.6 + 452.16) cm²
= 1205.76 cm²

550 m²

62.8 m

\(\frac { 8712 }{ 7 } { cm }^{ 2 }\)

Radius of the base of the tent (r) = 7 m 
Height (h) = 24 m
Slant height (l) = \(\sqrt{\mathrm{r}^{2}+\mathrm{h}^{2}}=\sqrt{7^{2}+24^{2}}\) 
= \(\sqrt{49+576}=\sqrt{625}\) = 25 m
Now, curved surface area of the conical tent = πrl
=\(\frac{22}{7}\) x 7 x 25 m²
= 22 x 25 m² = 550 m²
Let 'l' be the length of the cloth.
∴ l x b = 550
⇒ l x 5 = 550
⇒ l = \(\frac{550}{5} \mathrm{~m}\) = 110m
∴ Cost of the cloth = 50 x 110 = ₹5500.

Here, h = 24 cm and r = 7 cm
Since, l = \(\sqrt{\mathrm{h}^{2}+\mathrm{r}^{2}}=\sqrt{24^{2}+7^{2}} \mathrm{~cm}\)
\(=\sqrt{576+49}=\sqrt{625} \mathrm{~cm}\)
= 25 cm
Slant height = 25 cm.

Here r= 5 m, h = 12 m.
Slant height l=\(\sqrt { { 5 }^{ 2 }+{ 12 }^{ 2 } } =13\quad m\)
curved surface area of tent = \(\pi\)rl
\(=\frac { 22 }{ 7 } \times 5\times 13\quad { m }^{ 2 }\)
\(=\frac { 1430 }{ 7 } { m }^{ 2 }\)
\(\therefore\) Area of cloth required = \(\frac { 1430 }{ 7 } { m }^{ 2 }\)
Width of cloth = \(1\frac { 4 }{ 7 } m=\frac { 11 }{ 7 } m\)
Length of cloth\(=\frac { 1430 }{ 7 } \div \frac { 11 }{ 7 } \)
= 130 m.

Curved surface area of the tent = 552-2
= 550 m²
Radius(r) = 7 m
\(\therefore \ \pi \times 7\times l=550\)
\(\Rightarrow\) l = 25 m
\(\therefore \ h=\sqrt { { 25 }^{ 2 }-{ 7 }^{ 2 } } \)
= 24 m
Volume of the tent = \(\frac { 1 }{ 3 } \times \frac { 22 }{ 7 } \times 7\times 7\times 24\)
= 1232 m³.

471 m

1232 cm³

Base diameter = 40 cm
\(\therefore\) Base radius (r) = \(\frac { 40 }{ 2 } cm=20cm\)
\(=\frac { 20 }{ 100 } m=0.2m\)
Height (h) = 1 m
 \(\therefore l=\sqrt { { r }^{ 2 }+{ h }^{ 2 } } \)
\(\\ =\sqrt { { \left( 0.2 \right) }^{ 2 }+{ \left( 1 \right) }^{ 2 } } =\sqrt { 0.04+1 } \)
\(\\ =\sqrt { 1.04 } =1.02m\)
\(\therefore\)  Curved surface area = \(\pi rl\)
= 3.14 \(\times\) 0.2 \(\times\) 1.02
= 0.64056 m²
\(\therefore\)  Curved surface area of 50 cones
= 0.64056 \(\times\) 50 m²
= 32.028 m²
\(\therefore\)  Cost of painting all these cones
= 32.028 \(\times\) 12
= 384.336 = Rs 384.34 (approximately).

Base radius (r) = 7 cm
Height (h)  = 24 cm
\(\therefore \)  Slant height (l) = \(\sqrt { { r }^{ 2 }+{ h }^{ 2 } } \)
 \(=\sqrt { { \left( 7 \right) }^{ 2 }+{ \left( 24 \right) }^{ 2 } } =\sqrt { 49+576 } \)
\(\\ =\sqrt { 625 } =25cm\)
\(\therefore \)  Curved surface area of a cap = \(\pi rl\)
\(=\frac { 22 }{ 7 } \times 7\times 25=550{ cm }^{ 2 }\)
\(\therefore \)  Curved surface area of 10 caps
= 550 \(\times\) 10 = 5500 cm²
 Hence, the area of the sheet required to make 10 such caps is 5500 cm² .

Slant height (l) = 25 m
Base diameter (d) = 14 m
\(\therefore \) Base radius (r) = \(\frac { 14 }{ 2 } m=7m\)
\(\therefore \) Curved surface area of the tomb =\(\pi rl\)
\(=\frac { 22 }{ 7 } \times 7\times 25=550{ m }^{ 2 }\)
\(\therefore \)  Cost of white-washing the curved surface of the tomb at the rate of Rs 210 per 100 m²
= Rs \(\frac { 210 }{ 100 } \times 550\) = Rs 1155.

For conical tent
h = 8 m, r = 6 m
\(\therefore l=\sqrt { { r }^{ 2 }+{ h }^{ 2 }\ }\)
\( \\ =\sqrt { { \left( 6 \right) }^{ 2 }+{ \left( 8 \right) }^{ 2 } } =\sqrt { 36+64 }\)
\( \\ =\sqrt { 100 } =10m\)
\(\therefore \) Curved surface area = \(\pi rl\)
= 3.14 \(\times\) 6 \(\times\) 10 = 188.4 m².
Width of tarpaulin = 3 m
\(\therefore \) Length of tarpaulin 
Extra length of the material required
= 20 cm = 0.2 m
\(\therefore \) Actual length of tarpaulin required
= 62.8 m + 0.2 m = 63 m.  

(i) h = 10 m
r = 24 m
\(l=\sqrt { { r }^{ 2 }+{ h }^{ 2 } } =\sqrt { { \left( 24 \right) }^{ 2 }+{ \left( 10 \right) }^{ 2 } } \)
\(\\ =\sqrt { 576+100 } =\sqrt { 676 } =26m\)
Hence, the slant height of the tent is 26 m.
(ii) Curved surface area of the tent = \(\pi rl\)
 \(=\frac { 22 }{ 7 } \times 24\times 26{ m }^{ 2 }\)
\(\therefore \) Cost of the canvas required to make the tent, if the cost of 1 m² canvas is Rs 70.
= Rs \(\frac { 22 }{ 7 } \times 24\times 26\times 70\)
= Rs 137280.
Hence, the cost of the canvas is Rs 137280.

(i) Slant height (l) = 14 cm 
Curved surface area = 308 cm²
\(\Rightarrow \pi rl=308\)
\(\\ \Rightarrow \frac { 22 }{ 7 } \times r\times 14=308\)
\(\\ \Rightarrow r=\frac { 308\times 7 }{ 22\times 14 } \)
\(\\ \Rightarrow r=7cm\)
Hence, the radius of the base is 7 cm.
(ii) Total surface area of the cone = \(\pi r\left( l+r \right) \)
\(=\frac { 22 }{ 7 } \times 7\times \left( 14+7 \right)\)
\( \\ =\frac { 22 }{ 7 } \times 7\times 21=462{ cm }^{ 2 }\)
Hence, the total surface area of the cone is 462 cm².

Slant height (l)         = 21 m
Diameter of base     = 24 m
\(\therefore \) Radius of base (r) = \(\frac { 24 }{ 2 } m=12m\)
\(\therefore \) Total curved surface area of the cone
\(=\pi r\left( l+r \right) \)
\(\\ =\frac { 22 }{ 7 } \times 12\times \left( 21+12 \right) \)
\(\\ =\frac { 22 }{ 7 } \times 12\times 33=\frac { 8712 }{ 7 } =1244.57{ m }^{ 2 }.\)

\(\because \) Diameter of the base   = 10.5 cm
\(\therefore \) Radius of the base (r) \(=\frac { 10.5 }{ 2 } cm\)
= 5.25 cm
Slant height (l) = 10 cm
\(\therefore \) Curved surface area of the cone = \(\pi rl\)
 \(=\frac { 22 }{ 7 } \times 5.25\times 10=165{ cm }^{ 2 }.\)

Let the height of the right circular cone be h cm.
r = 56 cm
Curved surface area = 12320 cm²
\(\Rightarrow \pi rl=12320\)
\(\\ \Rightarrow \pi r\sqrt { { r }^{ 2 }+{ h }^{ 2 } } =12320\)
\(\\ \Rightarrow \frac { 22 }{ 7 } \times 56\times \sqrt { { \left( 56 \right) }^{ 2 }+{ h }^{ 2 } } =12320\)
\(\\ \Rightarrow \sqrt { { \left( 56 \right) }^{ 2 }+{ h }^{ 2 } } =\frac { 12320\times 7 }{ 22\times 56 } \)
\(\\ \Rightarrow \sqrt { { \left( 56 \right) }^{ 2 }+{ h }^{ 2 } } =70\)
\(\\ \Rightarrow { \left( 56 \right) }^{ 2 }+{ h }^{ 2 }={ \left( 70 \right) }^{ 2 }\)
Squaring both sides
\(\Rightarrow { h }^{ 2 }={ \left( 70 \right) }^{ 2 }-{ \left( 56 \right) }^{ 2 }\)
\(\\ \Rightarrow { h }^{ 2 }=\left( 70-56 \right) \left( 70+56 \right) \)
\(\\ \Rightarrow { h }^{ 2 }=\left( 14 \right) \left( 126 \right) \)
\(\\ \Rightarrow { h }^{ 2 }=\left( 14 \right) \left( 14\times 9 \right)\)
\( \\ \Rightarrow h=14\times 3=42\)
Extracting square root
Hence, the height of the right circular cone is 42 cm.