Math, asked by veebhortyagi856, 3 months ago

the lateral surface area of a cylinder is 1056cm² and it's height is 10cm find radius of cylinder ​

Answers

Answered by ItzArchimedes
26

Radius = 16.8 cm

Given ,

  • Lateral surface area of cylinder [LSA] = 1056cm²
  • Height of cylinder [h] = 10cm

We need to find ,

  • Radius [r] = ?

As we know that ,

  • LSA of cylinder = 2πrh

So ,

=> 1056 = 2 × (22/7) × (r) × (10)

=> 1056/20 × 7/22 = radius

=> 52.8 × 7/22 = radius

=> 369.6/22 = radius

=> Radius = 16.8 cm

Hence , radius = 16.8cm


Anonymous: Nice dear mod !. (:
Answered by Anonymous
8

Required solution :

Given that -

The lateral surface area of a cylinder is 1056 cm²

✠ Height of cylinder is 10 cm

To find -

✠ Radius of cylinder

Solution -

✠ Radius of cylinder = 16.8 cm

Using concept -

✠ Formula to find lateral surface area of cylinder.

Using formula -

✠ LSA = 2πrh

Full solution -

~ Let us use the formula to find lateral surface area of cylinder..!

➝ LSA = 2πrh

Where,

✨ LSA denotes Lateral Surface Area

✨ Value of π is 3.14 or 22/7

✨ r denotes radius

✨ h denotes height

Here,

✨ LSA = 1056 cm²

✨ Radius = ?

✨ π = 3.14

✨ Height = 10 cm

~ Let us put the values,

➝ 1056 = 2(3.14)(r)(10)

➝ 1056 = 2 × 3.14 × r × 10

➝ 1056 = 20 × 3.14 × r

➝ 1056 = 62.80 × r

➝ 1056 / 62.80 = r

➝ 16.8 = r

➝ r = 16.8 cm

Knowledge booster :

\; \; \; \; \; \; \;{\sf{\bold{\leadsto TSA \: of \: cube \: = \: 6(side)^{2}}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto LSA \: of \: cube \:= \: 4(side)^{2}}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Volume \: of \: cube \: = \: (side)^{3}}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Diagonal \: of \: cube \: = \: \sqrt(l^{2} + b^{2} + h^{2}}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: cube \: = \: 4(l+b+h)}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto CSA \: of \: sphere \: = \: 2 \pi r^{2}}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto SA \: of \: sphere \: = \: 4 \pi r^{2}}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto TSA \: of \: sphere \: = \: 3 \pi r^{2}}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Diameter \: of \: circle \: = \: 2r}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Radius \: of \: circle \: = \: \dfrac{d}{2}}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Volume \: of \: sphere \: = \: \dfrac{4}{3} \pi r^{3}}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: rectangle \: = \: Length \times Breadth}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: rectangle \: = \:2(length+breadth)}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: square \: = \: 4 \times sides}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: square \: = \: Side \times Side}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: triangle \: = \: \dfrac{1}{2} \times breadth \times height}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: paralloelogram \: = \: Breadth \times Height}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: circle \: = \: \pi b^{2}}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: triangle \: = \: (1st \: + \: 2nd \: + 3rd) \: side}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: paralloelogram \: = \: 2(a+b)}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto TSA \: of \: cuboid \: = \: 2(l \times b + b \times h + l \times h}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto LSA \: of \: cuboid \: = \: 2h(l+b)}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Volume \: of \: cuboid \: = \: L \times B \times H}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Diagonal \: of \: cuboid \: = \: \sqrt 3l}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: cuboid \: = \: 12 \times Sides}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Volume \: of \: cylinder \: = \: \pi r^{2}h}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Surface \: area \: of \: cylinder \: = \: 2 \pi rh + 2 \pi r^{2}}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Lateral \: area \: of \: cylinder \: = \: 2 \pi rh}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Base \: area \: of \: cylinder \: = \: \pi r^{2}}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Height \: of \: cylinder \: = \: \dfrac{v}{\pi r^{2}}}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Radius \: of \: cylinder \: = \:\sqrt frac{v}{\pi h}}}}


Anonymous: Amazing ❤ appreciated :claps:
Anonymous: Haye ❤ Aapka Shukriya ❤ :smile:
Anonymous: !!♡
Anonymous: Well designed!!
Anonymous: Nice
Anonymous: Shukriya ❤
Anonymous: @MrChocolatty Aapko bhi thank uh.! (:
Anonymous: Well explained v
Anonymous: Again thx :) ❤
Similar questions