Question

The height of a cone is equal To it's base diameter then slant hight of the cone is

Answer 1

$\huge\boxed{\fcolorbox{blue}{orange}{HELLO\:MATE}}$

For figure refer to attachment:

GIVEN:

♠ Height of a cone is equal to it' s base diameter.

TO FIND:

♠Slant height of the cone.

ANSWER:

Let us take the diameter of the base be 'x'.

So,

Radius of the base of the cone =$\dfrac{x}{2}$

$\large\purple{\boxed{Radius=\dfrac{Diameter}{2}}}$

Clearly from the figure slant height of the cone is the hypotenuse of the ∆ABC

(REFER TO FIGURE IN ATTACHMENT)

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Now, in ∆ABC,

$\large\green{\boxed{(AC) ^{2}=(AB) ^{2}+(BC) ^{2}}}$

(By Pythagoras Theorem)

=>$(AC) ^{2}=(AB) ^{2}+(BC) ^{2}$

Let AC be l.

=>$l^{2}=(x)^{2}+(\dfrac{x}{2}) ^{2}$

=>$l^{2} = x^{2}+\dfrac{x^{2}}{4}$

=>$l^{2}= \dfrac{x^{2}+4x^{2}}{4}$

=>$l^{2}=\dfrac{5x^{2}}{4}$

=>$l =\sqrt{\dfrac{5x^{2}}{4}}$

=>$l =\dfrac{\sqrt{5}x}{2}$

Therefore slant height of the cone is $\dfrac{\sqrt{5}x}{2}$

$\huge\orange{\boxed{Answer:\dfrac{\sqrt{5}x}{2}}}$