Question

if a right circular cone has raidus has 4cm and slant height 5cm then what is the its volume ​

Answer 1

GIVEN:

• Radius of right circular cone = 4 cm
• Slant height of right circular cone = 5 cm

TO FIND:

• What is the volume of cone ?

SOLUTION:

Let the height of the cone be 'h' cm

To find the height of the cone, we use the formula:-

$\large{\boxed{\bf{\star \: l^2 = r^2 + h^2 \: \star}}}$

On putting the given values in the formula, we get

$\sf{\longmapsto (5)^2 = (4)^2 + h^2 }$

$\sf{\longmapsto (5)^2 - (4)^2 = h^2 }$

$\sf{\longmapsto 25 - 16 = h^2 }$

$\sf{\longmapsto 9 = h^2}$

$\sf{\longmapsto \sqrt{9} = h }$

$\bf{\longmapsto 3 \: cm = h }$

• Height = h = 3 cm

We know that the formula for finding the volume of cone is:-

$\large{\boxed{\bf{\star \: VOLUME = \dfrac{1}{3} \pi r^2 h \: \star}}}$

According to question:-

$\sf{\longmapsto VOLUME = \dfrac{1}{\cancel{3}} \times \dfrac{22}{7} \times 4^2 \times \cancel{3}}$

$\sf{\longmapsto VOLUME = \dfrac{22}{7} \times 16 }$

$\sf{\longmapsto VOLUME = \cancel\dfrac{352}{7}}$

$\bf{\longmapsto VOLUME = 50.28 \: cm^3}$

❝ Hence, the volume of cone is 50.28 cm³ ❞

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Answer 2

QUESTION:-

✯ɪғ ᴀ ʀɪɢʜᴛ ᴄɪʀᴄᴜʟᴀʀ ᴄᴏɴᴇ ʜᴀs ʀᴀɪᴅᴜs ʜᴀs 4ᴄᴍ ᴀɴᴅ sʟᴀɴᴛ ʜᴇɪɢʜᴛ 5ᴄᴍ ᴛʜᴇɴ ᴡʜᴀᴛ ɪs ᴛʜᴇ ɪᴛs ᴠᴏʟᴜᴍᴇ

ANSWER✔

$\Large\underline\bold{GIVEN,}$

$\sf\dashrightarrow radius \:of\:of\:cone\:=4cm$

$\sf\dashrightarrow slant\:height\:of\:the\:cone=5cm$

$\Large\underline\bold{TO\:FIND}$

$\sf\dashrightarrow height\:of\:cone\:$

$\sf\large\dashrightarrow volume\:of\:the\:cone$

LET,

$\sf\large\therefore taking\:two\:cases\:$

$\sf\therefore in\:case\:1,\:we\:will\:find\:height\:of\:the\:$

$\sf\therefore in\:case\:2\:we\:will\:find\:volume\:of\:cone$

$\Large\underline\bold{SOLUTION,}$

$\large{\fbox {CASE:-1}}$

$\sf\therefore let\:the\:height\:of\:the\:cone\:be\:x\:cm$

USING FORMULA,

$\sf\large\therefore (hypo)^2=(S)^2+(S)^2$

" OR,

$\sf\large\therefore (L)^2=(r)^2+(x)^2$

$\sf\therefore radius=4cm$

$\sf\therefore slant\:height= 5cm$

NOW,

$\sf{\implies (5)^2 = (4)^2 + (x)^2 }$

$\sf{\implies (5)^2 - (4)^2 = (x)^2 }$

$\sf{\implies 25 - 16 = (x)^2 }$

$\sf{\implies (x)^2 =9}$

$\sf{\implies x= \sqrt{9} }$

$\sf{\implies x=3cm }$

$\sf{\boxed{\sf{height=3cm}}}$

$\large {\fbox {CASE:-2}}$

$\sf\large\therefore to\:find\:the\:volume\:of\:cone$

$\sf\therefore H=3CM$

$\sf\therefore R=4CM$

NOW,

$\Large\underline\bold{BY\:USING\: FORMULA}$

$\sf\large\therefore VOLUME\:OF\:CONE = \dfrac{1}{3} \pi r^2 h$

$\sf\therefore VOLUME\:OF\:CONE = \dfrac{1}{3} \times \dfrac{22}{7} \times (4)^2 \times h$

$\sf{\implies \dfrac{1}{\cancel {3}} \times \dfrac{22}{7} \times (4)^2 \times \cancel{3}}$

$\sf{\implies \dfrac{22}{7} \times 4 \times 4 }$

$\sf{\implies \dfrac{22}{7} \times 16 }$

$\sf{\implies \cancel\dfrac{352}{7}}$

$\sf{\therefore volume\:of\:cone = 50.28 \: cm^3}$

$\large{\boxed{\sf{volume\:of\:cone\:50.28cm^3}}}$

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✯ADDITIONAL INFORMATION,

✯DIAGRAM OF RIGHT TRIANGLE,

$\setlength{\unitlength}{1.6mm}\begin{picture}(30,20)\linethickness{0.1mm}\put(0,0){\line(0,1){37.5}}\put(0,0){\line(1,0){25}}\put(25,0){\line(-2,3){25}}\end{picture}\put(-32,18){3}\put(-18,-1.5){4}\put(-18.2,20){5}$

✯DIAGRAM OF RIGHT CIRcULAR CONE,

Cone

$\begin{lgathered}\setlength{\unitlength}{0.99cm}\begin{picture}(6, 4)\linethickness{0.26mm}\qbezier(5.8,2.0)(5.8,2.3728)(4.9799,2.6364)\qbezier(4.9799,2.6364)(4.1598,2.9)(3.0,2.9)\qbezier(3.0,2.9)(1.8402,2.9)(1.0201,2.6364)\qbezier(1.0201,2.6364)(0.2,2.3728)(0.2,2.0)\qbezier(0.2,2.0)(0.2,1.6272)(1.0201,1.3636)\qbezier(1.0201,1.3636)(1.8402,1.1)(3.0,1.1)\qbezier(3.0,1.1)(4.1598,1.1)(4.9799,1.3636)\qbezier(4.9799,1.3636)(5.8,1.6272)(5.8,2.0)\put(0.2,2){\line(1,0){2.8}}\put(3.2,4){\sf{3cm}}\put(3,2){\line(0,2){4.5}}\put(1.5,1.7){\sf{4cm}}\qbezier(.2,2.05)(.7,3)(3,6.5)\qbezier(5.8,2.05)(5.3,3)(3,6.5)\put(1,4){\sf l}\put(3,2.02){\circle*{0.15}}\put(2.7,2){\dashbox{0.01}(.3,.3)}\end{picture}\\\end{lgathered}$