Question

7. The volume of a right circular cone is 9856 cm. If the diameter of the
base is 28cm, find
(i)height of the cone.

(ii)slant height of the cone

(iii) curved surface area of the cone

(iv) volume of the cone.​

Answer 1

Answer:

Step-by-step explanation:

ANSWER:-

Given that:-

• Volume of the cone is 9856 cm^3
• Base is 28 cm

To find:-

• slant height of the cone
• curved surface area of the cone
• volume of the cone.

Let's Do!

We know that:-

$\boxed{\sf{Volume \ of \ Cone = \dfrac{1}{3}\times \pi \times r^2 \times h}}$

So, let us place the values!

$\sf{9856 = \dfrac{1}{3}\times \dfrac{22}{7} \times (14)^2 \times h}$

$\sf{9856 = \dfrac{1}{3}\times 22 \times 14 \times h}$

$\sf{\dfrac{9856 \times 3}{22 \times 14 \times 2 } = h}$

$\sf{h = 48 \ cm}$

Now, to find slant height (l)

$\sf{l^2 = h^2+b^2}$

$\sf{l^2 = (48)^2+(14)^2}$

$\sf{l = 50 \ cm}$

Now, CSA of the cone

$\rm{C.S.A = \pi \times r \times l}$

$\rm{C.S.A = \dfrac{22}{7} \times 14 \times 50}$

$\rm{= 2200 \ cm^2}$

Volume is already given, so no need.

Let us find TSA instead.

$\sf{TSA = \pi \times r \times l + \pi \times r^2}$

$\sf{TSA = \dfrac{22}{7} \times 14 \times 48 + \dfrac{22}{7} \times (14)^2}$

= 2816 cm^2 is the T.S.A

Answer 2

Answer :-

$\: \\ \large{\underline{\underline{\sf{Firstly,\; let's\; understand\;the\; concept\; used\; :-}}}}$

Here the concept CSA, TSA and Slant Height of the cone has been used. We know that if we are given volume and radius, then we can apply formula and find out the height of cone. After finding height, we can apply Pythagoras theorem and find out the slant height of cone. And then we can apply the values and find out the required things.

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★ Formula Used :-

$\: \\ \large{\boxed{\boxed{\sf{\odot\;\; Volume\; of \; Cone\; = \; \bf{\dfrac{1}{3}\:\times\: \pi r^{2}h}}}}}$

$\: \\ \large{\boxed{\boxed{\sf{\odot\;\; (Slant\;Height)^{2},\; L^{2} \; \; = \; \; \bf{(Height)^{2},\; H^{2} \;\; + \;\; (Base)^{2}, \; B^{2}}}}}}$

$\: \\ \large{\boxed{\boxed{\sf{\odot\;\; CSA\; of \; Cone \; = \; \bf{\pi rL}}}}}$

$\: \\ \large{\boxed{\boxed{\sf{\odot\;\; TSA\; of \; Cone \; = \; \bf{\pi rL\; \; + \; \; \pi r^{2}}}}}}$

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★ Question :-

The volume of a right circular cone is 9856 cm³. If the diameter of the base is 28cm, find

(i)height of the cone.

(ii)slant height of the cone

(iii) curved surface area of the cone

(iv) total surface area of cone

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★ Solution :-

Given,

» Volume of Cone = 9856 cm³

» Diameter of base of cone = d = 28 cm

» Radius of base or cone = r = ½ × d = ½ × 28

= 14 cm

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~ i) For the Height of the Cone :-

Let the Height of the cone be 'h' m.

Then,

$\: \\ \qquad \large{\sf{:\longrightarrow \: \;\; Volume\; of \; Cone\; = \; \bf{\dfrac{1}{3}\:\times\: \pi r^{2}h}}}$

$\: \\ \qquad \large{\sf{:\longrightarrow \: \;\; 9856\;\; = \; \bf{\dfrac{1}{3}\:\times\: \dfrac{22}{\cancel{7}}\:\times\: \cancel{14}\; \:\times\: h}}}$

$\: \\ \qquad \large{\sf{:\longrightarrow \; \; \: h \; \; = \; \; \bf{\dfrac{9856 \;\: \times\: 3}{22\:\times\:2\:\times\:14\;}} \: \: = \: \: \underline{\underline{48 \; cm}}}}$

$\: \\ \large{\boxed{\rm{Height\; of \; the \; cone \; = \; \bf{48 \; cm}}}}$

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~ ii) For the Slant Height of the Cone :-

$\: \\ \qquad \large{\sf{\rightarrow \;\;\; (Slant\;Height)^{2},\; L^{2} \; \; = \; \; \bf{(Height)^{2},\; H^{2} \;\; + \;\; (Base)^{2}, \; B^{2}}}}$

$\: \\ \qquad \large{\sf{\rightarrow \;\;\; L^{2} \; \; = \; \; \bf{(48)^{2} \;\; + \;\; (14)^{2}}}}$

$\: \\ \qquad \large{\sf{\rightarrow \;\;\; L^{2} \; \; = \; \; \bf{2304 \; \;\; + \;\; 196\;}}}$

$\: \\ \qquad \large{\sf{\rightarrow \;\;\; L^{2} \; \; = \; \; \bf{\sqrt{2500}\: \: = \: \: \underline{\underline{50\; cm}}}}}$

$\: \\ \large{\boxed{\rm{Slant\; Height\; of \; the \; cone \; = \; \bf{50 \; cm}}}}$

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~ iii) For the CSA of Cone :-

$\: \\ \qquad \large{\sf{:\Longrightarrow \: \: \: \;\; CSA\; of \; Cone \; = \; \bf{\pi rL}}}$

$\: \\ \qquad \large{\sf{:\Longrightarrow \: \: \: \;\; CSA\; of \; Cone \; = \; \bf{\dfrac{22}{7}\:\times\: (14)\:\times\:(50) \; \; = \; \; \underline{\underline{2200}}}}}$

$\: \\ \large{\boxed{\rm{CSA\; of \; the \; cone \; = \; \bf{2200 \; cm^{2}}}}}$

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~ iv) For the TSA of Cone :-

$\: \\ \qquad \large{\sf{:\Longrightarrow \;\;\; TSA\; of \; Cone \; = \; \bf{\pi rL\; \; + \; \; \pi r^{2}}}}$

$\: \\ \qquad \large{\sf{:\Longrightarrow \;\;\; TSA\; of \; Cone \; = \; \bf{\dfrac{22}{7}\:\times\:(14)\:\times\:(50)\; \; + \; \; \dfrac{22}{7}\:\times\:(14)^{2}}}}$

$\: \\ \qquad \large{\sf{:\Longrightarrow \;\;\; TSA\; of \; Cone \; = \; \bf{2200\; \;\; + \;\; 616\; \: \: = \; \: \underline{\underline{2816 \; cm^{2}}}}}}$

$\: \\ \large{\boxed{\rm{TSA\; of \; the \; cone \; = \; \bf{2816 \; cm^{2}}}}}$

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$\: \\ \large{\underbrace{\underbrace{\leadsto \;\;\; More \; formulas \; to \; know \; :-}}}$

• Volume of Cylinder = πr²h

• Volume of Cube = (Side)³

• Volume of Cuboid = Length × Breadth × Height

• Volume of Hemisphere = ⅔ × πr²h

• TSA of Cylinder = 2πr² + 2πrh

• CSA of Cylinder = 2πrh