Question

perimeter of base of a cylinder is 5cm and its curved surface area is 10cm2, then the height of the cylinder is​

Answer 1

${\large{\underline{\pmb{\frak{Given...}}}}}$

→ The perimeter of base of a cylinder is 5cm

→ The curved surface area of the cylinder is 10cm²

${\large{\underline{\pmb{\frak{To \; Find...}}}}}$

→ The height of the cylinder

${\large{\underline{\pmb{\frak{Understanding \; the \; concept...}}}}}$

❍ Concept : As we have been given the perimeter ( Circumference ) of the cylinder which is 5cm and its curved surface area which is 10cm² So, now let's find the radius of the cylinder with the help of its circumference.

★ Later substitute the value of its radius in the formula to find the curved surface area and find the height of the cylinder.

${\large{\underline{\pmb{\frak{Using \; Concepts...}}}}}$

✪ Formula to find the circumference of the base :

$\tt Circumference = 2 \pi r$

✪ Formula to find the Curved Surface Area :

$\tt C.S.A = 2\pi rh$

${\large{\underline{\pmb{\sf{RequirEd \; Solution...}}}}}$

✰ The height of the cylinder is 2cm

${\large{\underline{\pmb{\frak{Full \; solution...}}}}}$

→ We know that,

⠀⠀⠀» Circumference of the base = 5cm

⠀⠀⠀» C.S.A of the cylinder is 10cm²

~ Now let's find the radius of the cylinder with the help of the formula mention below...

⋆ Formula,

→ Circumference = 2πr

~ Now let's find the radius of the cylinder with the help of the circumference of the base given taking the value of π as 22/7...

$\longrightarrow \tt Circumference = 2\pi r$

$\longrightarrow \tt 5cm = 2 \times \dfrac{22}{7} \times r$

$\longrightarrow \tt r = \dfrac{5\times 7}{2 \times 22}$

$\longrightarrow \tt r = \dfrac{35}{44}$

• Henceforth the radius of the cylinder is 35/44cm

~ Now let's find the height of the cylinder with the help of the formula mention below...

⋆ Formula,

→ C.S.A of the cylinder = 2πrh

~ Now let's find the height of the cylinder with the help of the C.S.A and the radius which we found taking the value of π as 22/7...

$\longrightarrow \tt C.S.A = 2\pi rh$

$\longrightarrow \tt 10 = 2 \times \dfrac{22}{7} \times \dfrac{35}{44} \times h$

$\longrightarrow \tt h = \dfrac{\cancel{10}\times \cancel{44} \times \cancel{ 7}}{\cancel{35}\times \cancel{ 22 }\times \cancel{ 2}}$

$\longrightarrow \tt h = 2$

• Henceforth the height of the cylinder is 2cm

${\large{\underline{\pmb{\frak{Additional \; Information...}}}}}$

⋆ Related Formulas,

$\;\;\;\;\;\;\; \leadsto \tt volume\; of \; a \; cylinder = \pi r^{2} h$

$\;\;\;\;\;\;\;{ \leadsto} \tt T.S.A \; of \; a \; cylinder= 2\pi rh + 2\pi r^{2}$

$\;\;\;\;\;\;\; \leadsto \tt Volume \; a \; cube = Side ^3$

$\;\;\;\;\;\;\; \leadsto \tt L.S.A \; of \; a \; cube = 4(L)^{2}$

$\;\;\;\;\;\;\; \leadsto \tt T.S.A \; of \; a \; cube = 6(L)^{2}$

⋆ Diagram,

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\thicklines\multiput(-0.5,-1)(26,0){2}{\line(0,1){40}}\multiput(12.5,-1)(0,3.2){13}{\line(0,1){1.6}}\multiput(12.5,-1)(0,40){2}{\multiput(0,0)(2,0){7}{\line(1,0){1}}}\multiput(0,0)(0,40){2}{\qbezier(1,0)(12,3)(24,0)\qbezier(1,0)(-2,-1)(1,-2)\qbezier(24,0)(27,-1)(24,-2)\qbezier(1,-2)(12,-5)(24,-2)}\multiput(18,2)(0,32){2}{\sf{35/44cm}}\put(9,17.5){\sf{2cm}}\end{picture}$

* Note : View the diagram from the web if difficulty in seeing from the app

See this question at : https://brainly.in/question/40343207

Answer 2

Answer:

The perimeter of base of a cylinder is 5cm

→ The curved surface area of the cylinder is 10cm²

{\large{\underline{\pmb{\frak{To \; Find...}}}}}

ToFind...

ToFind...

→ The height of the cylinder

{\large{\underline{\pmb{\frak{Understanding \; the \; concept...}}}}}

Understandingtheconcept...

Understandingtheconcept...

❍ Concept : As we have been given the perimeter ( Circumference ) of the cylinder which is 5cm and its curved surface area which is 10cm² So, now let's find the radius of the cylinder with the help of its circumference.

★ Later substitute the value of its radius in the formula to find the curved surface area and find the height of the cylinder.

{\large{\underline{\pmb{\frak{Using \; Concepts...}}}}}

UsingConcepts...

UsingConcepts...

✪ Formula to find the circumference of the base :

\tt Circumference = 2 \pi rCircumference=2πr

✪ Formula to find the Curved Surface Area :

\tt C.S.A = 2\pi rhC.S.A=2πrh

{\large{\underline{\pmb{\sf{RequirEd \; Solution...}}}}}

RequirEdSolution...

RequirEdSolution...

✰ The height of the cylinder is 2cm

{\large{\underline{\pmb{\frak{Full \; solution...}}}}}

Fullsolution...

Fullsolution...

→ We know that,

⠀⠀⠀» Circumference of the base = 5cm

⠀⠀⠀» C.S.A of the cylinder is 10cm²

~ Now let's find the radius of the cylinder with the help of the formula mention below...

⋆ Formula,

→ Circumference = 2πr

~ Now let's find the radius of the cylinder with the help of the circumference of the base given taking the value of π as 22/7...

\longrightarrow \tt Circumference = 2\pi r⟶Circumference=2πr

\longrightarrow \tt 5cm = 2 \times \dfrac{22}{7} \times r⟶5cm=2×

7

22

×r

\longrightarrow \tt r = \dfrac{5\times 7}{2 \times 22}⟶r=

2×22

5×7

\longrightarrow \tt r = \dfrac{35}{44}⟶r=

44

35

Henceforth the radius of the cylinder is 35/44cm

~ Now let's find the height of the cylinder with the help of the formula mention below...

⋆ Formula,

→ C.S.A of the cylinder = 2πrh

~ Now let's find the height of the cylinder with the help of the C.S.A and the radius which we found taking the value of π as 22/7...

\longrightarrow \tt C.S.A = 2\pi rh⟶C.S.A=2πrh

\longrightarrow \tt 10 = 2 \times \dfrac{22}{7} \times \dfrac{35}{44} \times h⟶10=2×

7

22

×

44

35

×h

\longrightarrow \tt h = \dfrac{\cancel{10}\times \cancel{44} \times \cancel{ 7}}{\cancel{35}\times \cancel{ 22 }\times \cancel{ 2}}⟶h=

35

×

22

×

2

10

×

44

×

7

\longrightarrow \tt h = 2⟶h=2

Henceforth the height of the cylinder is 2cm

{\large{\underline{\pmb{\frak{Additional \; Information...}}}}}

AdditionalInformation...

AdditionalInformation...

⋆ Related Formulas,

\;\;\;\;\;\;\; \leadsto \tt volume\; of \; a \; cylinder = \pi r^{2} h⇝volumeofacylinder=πr

2

h

\;\;\;\;\;\;\;{ \leadsto} \tt T.S.A \; of \; a \; cylinder= 2\pi rh + 2\pi r^{2}⇝T.S.Aofacylinder=2πrh+2πr

2

\;\;\;\;\;\;\; \leadsto \tt Volume \; a \; cube = Side ^3⇝Volumeacube=Side

3

\;\;\;\;\;\;\; \leadsto \tt L.S.A \; of \; a \; cube = 4(L)^{2}⇝L.S.Aofacube=4(L)

2

\;\;\;\;\;\;\; \leadsto \tt T.S.A \; of \; a \; cube = 6(L)^{2}⇝T.S.Aofacube=6(L)

2

⋆ Diagram,

\setlength{\unitlength}{1mm}\begin{picture}(5,5)\thicklines\multiput(-0.5,-1)(26,0){2}{\line(0,1){40}}\multiput(12.5,-1)(0,3.2){13}{\line(0,1){1.6}}\multiput(12.5,-1)(0,40){2}{\multiput(0,0)(2,0){7}{\line(1,0){1}}}\multiput(0,0)(0,40){2}{\qbezier(1,0)(12,3)(24,0)\qbezier(1,0)(-2,-1)(1,-2)\qbezier(24,0)(27,-1)(24,-2)\qbezier(1,-2)(12,-5)(24,-2)}\multiput(18,2)(0,32){2}{\sf{35/44cm}}\put(9,17.5){\sf{2cm}}\end{picture}

* Note : View the diagram from the web if difficulty in seeing from the app

See this question at : https://brainly.in/question/40343207