Question

-●▬▬▬▬๑۩۩๑▬▬▬▬▬● I have made a right circular cone whose diameter of the base is 12 cm and the slant height is 10 cm. Let me write by calculating the height of the cone .



$\huge\underline\mathtt{Solve \: this \: problem...}$




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Answer 1

$\huge\underline\mathtt{♡Be Brainly♡}$

Answer 2

Let,

the radius of the base of the right circular cone is =$\sf{\frac{12}{6} = 6 cm}$

Here, OB=6 cm

⠀⠀⠀BC=10cm

Again, let

Height of the cone =OC

⠀OBC is a right-angled triangle.

$\rule{300}{2}$

Now,

From Pythagoras Theorem, we get,

$\sf{{Hypotenuse}^{2} = {Perpendicular}^{2} + {Base}^{2}}$

$\sf{ \implies{BC}^{2} = {OC}^{2} + {OB}^{2}}$

$\sf{\implies{OC}^{2} = {BC}^{2} - {OB}^{2}}$

$\sf{\implies{OC}^{2}={10}^{2} - {6}^{2}}$

$\sf{\implies OC = \sqrt{ {10}^{2} - {6}^{2} } }$

$\sf{\implies OC = \sqrt{100 - 36}}$

$\sf{\implies OC = \sqrt{64}}$

$\sf{\implies OC=8cm}$

Therefore, the height of the right circular cone is 8 cm

$\rule{300}{2}$

$\huge\underline\mathtt\color{gold}Thankuhh$

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