Question

A cylindrical pillar is 50cm in diameter and 3.5m in height .find the cost of painting the curved surface of the pillar at the rate of the of Rs 12.50perm2​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

Diameter of cylindrical pillar, d = 50 cm

So, Radius of cylindrical pillar, r = 25 cm = 0.25 m

Height of cylindrical pillar, h = 3.5 m

We know, Curved Surface Area of cylinder of radius r and height h is given by

$\boxed{\rm{  \:CSA_{(Cylinder)} \: = \: 2 \: \pi \: r \: h \: }}$

So, on substituting the values, we get

$\rm \: \:CSA_{(Cylindrical \: pillar)} \: = \: 2 \times \dfrac{22}{7} \times \dfrac{25}{100} \times \dfrac{35}{10}$

$\rm\implies \: \:CSA_{(Cylindrical \: pillar)} \: = \: 5.5 \: {m}^{2}$

Now, further given that

$\rm \: Cost \: of \: painting \: {1 \: m}^{2} \: =  \:Rs \: 12.50$

So,

$\rm \: Cost \: of \: painting \: {5.5 \: m}^{2} \: =5.5 \times 12.5  = \:Rs \: 68.75$

Hence,

$\rm \: Cost \: of \: painting \: CSA_{(Cylindrical \: pillar)}= \:Rs \: 68.75$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{CSA_{(cylinder)} = 2\pi \: rh}\\ \\ \bigstar \: \bf{Volume_{(cylinder)} = \pi {r}^{2} h}\\ \\ \bigstar \: \bf{TSA_{(cylinder)} = 2\pi \: r(r + h)}\\ \\ \bigstar \: \bf{CSA_{(cone)} = \pi \: r \: l}\\ \\ \bigstar \: \bf{TSA_{(cone)} = \pi \: r \: (l + r)}\\ \\ \bigstar \: \bf{Volume_{(sphere)} = \dfrac{4}{3}\pi {r}^{3} }\\ \\ \bigstar \: \bf{Volume_{(cube)} = {(side)}^{3} }\\ \\ \bigstar \: \bf{CSA_{(cube)} = 4 {(side)}^{2} }\\ \\ \bigstar \: \bf{TSA_{(cube)} = 6 {(side)}^{2} }\\ \\ \bigstar \: \bf{Volume_{(cuboid)} = lbh}\\ \\ \bigstar \: \bf{CSA_{(cuboid)} = 2(l + b)h}\\ \\ \bigstar \: \bf{TSA_{(cuboid)} = 2(lb +bh+hl )}\\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$

Answer 2

Answer:

\large\underline{\sf{Solution-}}

Solution−

Given that,

Diameter of cylindrical pillar, d = 50 cm

So, Radius of cylindrical pillar, r = 25 cm = 0.25 m

Height of cylindrical pillar, h = 3.5 m

We know, Curved Surface Area of cylinder of radius r and height h is given by

\begin{gathered}\boxed{\rm{ \:CSA_{(Cylinder)} \: = \: 2 \: \pi \: r \: h \: }} \\ \end{gathered}

CSA

(Cylinder)

=2πrh

So, on substituting the values, we get

\begin{gathered}\rm \: \:CSA_{(Cylindrical \: pillar)} \: = \: 2 \times \dfrac{22}{7} \times \dfrac{25}{100} \times \dfrac{35}{10} \\ \end{gathered}

CSA

(Cylindricalpillar)

=2×

7

22

×

100

25

×

10

35

\begin{gathered}\rm\implies \: \:CSA_{(Cylindrical \: pillar)} \: = \: 5.5 \: {m}^{2} \\ \end{gathered}

⟹CSA

(Cylindricalpillar)

=5.5m

2

Now, further given that

\begin{gathered}\rm \: Cost \: of \: painting \: {1 \: m}^{2} \: = \:Rs \: 12.50 \\ \end{gathered}

Costofpainting1m

2

= Rs12.50

So,

\begin{gathered}\rm \: Cost \: of \: painting \: {5.5 \: m}^{2} \: =5.5 \times 12.5 = \:Rs \: 68.75 \\ \end{gathered}

Costofpainting5.5m

2

=5.5×12.5 =Rs68.75

Hence,

\begin{gathered}\rm \: Cost \: of \: painting \: CSA_{(Cylindrical \: pillar)}= \:Rs \: 68.75 \\ \end{gathered}

CostofpaintingCSA

(Cylindricalpillar)

=Rs68.75

\rule{190pt}{2pt}

Additional Information :-

\begin{gathered}\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{CSA_{(cylinder)} = 2\pi \: rh}\\ \\ \bigstar \: \bf{Volume_{(cylinder)} = \pi {r}^{2} h}\\ \\ \bigstar \: \bf{TSA_{(cylinder)} = 2\pi \: r(r + h)}\\ \\ \bigstar \: \bf{CSA_{(cone)} = \pi \: r \: l}\\ \\ \bigstar \: \bf{TSA_{(cone)} = \pi \: r \: (l + r)}\\ \\ \bigstar \: \bf{Volume_{(sphere)} = \dfrac{4}{3}\pi {r}^{3} }\\ \\ \bigstar \: \bf{Volume_{(cube)} = {(side)}^{3} }\\ \\ \bigstar \: \bf{CSA_{(cube)} = 4 {(side)}^{2} }\\ \\ \bigstar \: \bf{TSA_{(cube)} = 6 {(side)}^{2} }\\ \\ \bigstar \: \bf{Volume_{(cuboid)} = lbh}\\ \\ \bigstar \: \bf{CSA_{(cuboid)} = 2(l + b)h}\\ \\ \bigstar \: \bf{TSA_{(cuboid)} = 2(lb +bh+hl )}\\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}\end{gathered}

MoreFormulae

MoreFormulae

★CSA

(cylinder)

=2πrh

★Volume

(cylinder)

=πr

2

h

★TSA

(cylinder)

=2πr(r+h)

★CSA

(cone)

=πrl

★TSA

(cone)

=πr(l+r)

★Volume

(sphere)

=

3

4

πr

3

★Volume

(cube)

=(side)

3

★CSA

(cube)

=4(side)

2

★TSA

(cube)

=6(side)

2

★Volume

(cuboid)

=lbh

★CSA

(cuboid)

=2(l+b)h

★TSA

(cuboid)

=2(lb+bh+hl)

hope it helpful To u