Question

Curved surface area of a cone is 251.2 cm2 and radius of its base is 8cm. Find its slant
height and perpendicular height. (n = 3.14)​

Answer 1

Answer:

Diαgram :-

$\setlength{\unitlength}{1.2mm}\begin{picture}(5,5)\thicklines\put(0,0){\qbezier(1,0)(12,3)(24,0)\qbezier(1,0)(-2,-1)(1,-2)\qbezier(24,0)(27,-1)(24,-2)\qbezier(1,-2)(12,-5)(24,-2)}\put(-0.5,-1){\line(1,2){13}}\put(25.5,-1){\line(-1,2){13}}\multiput(12.5,-1)(2,0){7}{\line(1,0){1}}\multiput(12.5,-1)(0,4){7}{\line(0,1){2}}\put(18,1.6){\sf{8\ cm}}\put(9.5,10){\sf{6\ cm}}\end{picture}$

$\begin{gathered}\end{gathered}$

Given :

• Curved surface area of a cone is 251.2 cm².
• Radius of its base is 8cm.

$\begin{gathered}\end{gathered}$

To Find :

• Height
• Perpendicular height

$\begin{gathered}\end{gathered}$

Using Formulαs :

$\longrightarrow{\underline{\boxed{\pmb{\sf{CSA = \pi rl}}}}}$

$\longrightarrow{\underline{\boxed{\pmb{\sf{h= \sqrt{{l}^{2} - {r}^{2}}}}}}}$

Where :-

• CSA = Curved surface area
• π = 3.14
• h = height
• l = slant height
• r = radius

$\begin{gathered}\end{gathered}$

Solution :

$\small\bigstar$ Finding the slant height of cone by substituting the given values in the formula:-

$\begin{gathered} \qquad\longrightarrow{\bf{CSA = \pi rl}} \\ \\ \quad\longrightarrow{\sf{251.2= 3.14 \times 8 \times l}} \\ \\ \qquad\longrightarrow{\sf{251.2= 25.12\times l}} \\ \\ \quad\longrightarrow{\sf{l = \frac{251.2}{25.12}}} \\ \\\quad\longrightarrow{\sf{l = \frac{251.2 \times 100}{25.12 \times 100}}} \\ \\ \quad\longrightarrow{\sf{l = \frac{25120}{2512}}} \\ \\ \quad\longrightarrow{\sf{l = \cancel{\frac{25120}{2512}}}} \\ \\ \quad\longrightarrow{\sf{l = 10 \: cm}} \\ \\ \qquad\bigstar{\underline{\boxed{\sf{\pink{Slant \: height = 10 \: cm}}}}}\end{gathered}$

∴ The slant height of cone is 10 cm.

$\rule{300}{1.5}$

$\small\bigstar$ Now, finding the perpendicular height of cone by substituting the given values in the formula:-

$\begin{gathered} \qquad\longrightarrow{\bf{h= \sqrt{{l}^{2} - {r}^{2}}}} \\ \\ \qquad\longrightarrow{\sf{h= \sqrt{{(10)}^{2} - {(8)}^{2}}}} \\ \\ \qquad\longrightarrow{\sf{h= \sqrt{{(10 \times 10)} - {(8 \times 8)}}}} \\ \\ \quad\longrightarrow{\sf{h= \sqrt{100 - 64}}} \\ \\ \quad\longrightarrow{\sf{h= \sqrt{36}}} \\ \\ \quad\longrightarrow{\sf{h= \sqrt{6 \times 6}}} \\ \\ \longrightarrow{\sf{h= 6 \: cm }} \\ \\ \quad\bigstar{\underline{\boxed{\sf{\pink{Height= 6 \: cm }}}}}\end{gathered}$

∴ The height of cone is 6 cm.

$\begin{gathered}\end{gathered}$

Leαrn More :

$\begin{array}{|c|c|c|}\cline{1-3}\bf Shape&\bf Volume\ formula&\bf Surface\ area formula\\\cline{1-3}\sf Cube&\tt l^3}&\tt 6l^2\\\cline{1-3}\sf Cuboid&\tt lbh&\tt 2(lb+bh+lh)\\\cline{1-3}\sf Cylinder&\tt {\pi}r^2h&\tt 2\pi{r}(r+h)\\\cline{1-3}\sf Hollow\ cylinder&\tt \pi{h}(R^2-r^2)&\tt 2\pi{rh}+2\pi{Rh}+2\pi(R^2-r^2)\\\cline{1-3}\sf Cone&\tt 1/3\ \pi{r^2}h&\tt \pi{r}(r+s)\\\cline{1-3}\sf Sphere&\tt 4/3\ \pi{r}^3&\tt 4\pi{r}^2\\\cline{1-3}\sf Hemisphere&\tt 2/3\ \pi{r^3}&\tt 3\pi{r}^2\\\cline{1-3}\end{array}$

$\rule{300}{1.5}$

Answer 2

Corrected question:

• Curved surface area of a cone is 251.2 cm² and radius of its base is 8cm. Find its slant height and perpendicular height. (π = 3.14)

Information provided with us:

• Curved surface area of a cone is 251.2 cm2 and radius of its base is 8cm

What we have to calculate:

• Slant height and perpendicular height

Formulas needed to be applied:

Curved surface area of cone:-

• C.S.A. = π × r × l

Here,

• Value of π is 3.14
• r is radius
• l is slant height

Slant height:-

• S² = r² + h²

Here,

• S is slant height
• r is radius
• h is height

Putting the values in the formula of C.S.A. of cone,

➺ 251.2 = 3.14 × 8 × l

➺ 2512 / 10 = 3.14 × 8 × l

➺ 2512 / 10 = (314 / 100) × 8 × l

➺ 2512 × 100 / 10 = 314 × 8 × l

• Cancelling the 0,

➺ 2512 × 10 / 1 = 314 × 8 × l

➺ 2512 × 10 = 314 × 8 × l

• On multiplying 2512 by 10 we gets 25120,

➺ 25120 = 314 × 8 × l

• Transposing the sides,

➺ I = 25120 / 314 × 8

➺ I = 12560 / 314 × 4

➺ I = 6280 / 314 × 2

➺ I = 3140 / 314

➺ I = 1570 / 157

➺ I = 10

• Henceforth, l is 10 cm

Substituting the values in the formula of slant height,

➺ ( 10 )² = ( 8 )² + h²

➺ ( 10 × 10 ) = ( 8 )² + h²

➺ ( 100 ) = ( 8 )² + h²

➺ ( 100 ) = ( 8 × 8 ) + h²

➺ ( 100 ) = ( 64 ) + h²

Transposing the sides,

➺ h² = 100 - 64

➺ h² = 36

➺ h = √36

➺ h = 6

Henceforth, perpendicular height is of 6 cm.

____________

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