Triangly-Circly Cone!!!

Let the midpoint of $BC$ be $F$ and draw in $PA$ , $PC$ , and $PF$ :

Since $\angle ADP = \angle AEP = 90°$ , and since $AP = AP$ by the reflexive property, and since we are given that $DP = PE$ , then $\triangle ADP \cong \triangle AEP$ by the HL congruence theorem , and since corresponding parts of congruent triangles are congruent, $AD = AE$ .

Since $DP$ and $EP$ are perpendicular bisectors of two chords $AC$ and $AB$ , $P$ must be at the center of the circle.

Since $PF$ goes through the center of the circle and bisects chord $BC$ , $\angle PFB$ must be a right angle.

Since $\angle PFB = \angle PEB = 90°$ , and since $PB = PB$ by the reflexive property, and since we are given that $\angle PBF = \angle PBE$ , then $\triangle PFB \cong \triangle PEB$ by the AAS congruence theorem , and since corresponding parts of congruent triangles are congruent, $BF = BE$ .

Therefore, $CD = DA = AE = EB = BF = FC = \sqrt{3}$ , which means $\triangle ABC$ is an equilateral triangle with a side length of $2\sqrt{3}$ and a radius of $r_{\text{circle}} = 2$ .

That means $\widehat{BC}$ is one third the circumference of the original circle, or $\widehat{BC} = \frac{1}{3}C_{\text{circle}} = \frac{1}{3}\cdot 2 \pi r_{\text{circle}} = \frac{1}{3}\cdot 2 \pi \cdot 2 = \frac{4}{3}\pi$ . Since this length is the same length as the whole circumference of the base circle of the cone, $C_{\text{cone}} = \frac{4}{3}\pi$ , so the radius of the base circle of the cone is $r_{\text{cone}} = \frac{C_{\text{cone}}}{2 \pi} = \frac{\frac{4}{3}\pi}{2 \pi} = \frac{2}{3}$ .

The slant height $l$ of the cone is the same as the radius of the original circle, so $l = r_{\text{circle}} = 2$ .

By the Pythagorean Theorem, the height $h$ of the cone is $h = \sqrt{l^2 - r_{\text{cone}}^2} = \sqrt{2^2 - (\frac{2}{3})^2} = \frac{4\sqrt{2}}{3}$ .

Therefore, the radius of the original circle is $2$ , and the height of the cone is $\frac{4\sqrt{2}}{3}$ .

radius is little greater than root 3 so it is 2 also height cannot be greater than slant height(radius) so this rules out 3rd option and the answer is 1st option

Triangle is equilater side 2*3^(1/2)

Base cone = long circle/3 => radius 2/3 => hight per pitagoras

$\triangle BDA \cong \triangle BDC$ by ASA

• $\angle ABD = \angle CBD$ (given)
• $BD$ is the common line
• $\angle BDA = 90\degree = \angle BDC$

Hence, $\triangle ABC$ is an isosceles triangle; $BA = BC$ .

Draw a line joining $\textcolor{#D61F06}{AP}$ ,

$\triangle APD \cong \triangle APE$ by RHS

• $\angle ADP = 90\degree = \angle AEP$
• $AP$ is the common line
• $PD = PE$ (given)

Thus, $AD = AE$ and so $AC = 2 \sqrt{3} = AB$

Therefore, $\triangle ABC$ is an equilateral triangle as three sides are equal length $2 \sqrt{3}$ .

Height of $\triangle ABC$ = $\sqrt{ (2 \sqrt{3})^2 - ( \sqrt{3})^2 } = 3$

As $P$ is the centroid of $\triangle ABC$ as well as the centre of the circle, radius $\textcolor{#D61F06}{AP} = \cfrac{2}{3} \times 3 = \boxed{2}$ Properties of centroid of a triangle

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The centroid divides the equilateral triangle into 3 equal pieces.

Since we use $\cfrac{1}{3}$ the circle to form a cone,

• the slant length of the cone $= OC = 2$
• the circumference of the base of the cone = $\cfrac{1}{3}$ circumference of the original circle,
• thus, the radius $\textcolor{#69047E}{R}$ , of the cone's base circle = $\cfrac{1}{3} \times$ radius of the original circle = $\cfrac{1}{3} \times 2 = \cfrac{2}{3}$

Thus, the height of the cone $= \sqrt{ OC^2 - R^2 } =\sqrt{ 4 - \cfrac{4}{9} } = \boxed{\cfrac{4 \sqrt{2}}{3} }$