Question

Someone pls help with this been stuck on it for time now ‍♂️l​

Answer 1

Check the picture below.

so the height we're looking for is "h" as you see in the picture on the far right.

first off, let's take a look at the left side of it, we have a circle with a radius of 15, now plugging that into the area of a circle to get its area, πr² gives us π15² or just 225π, that's the whole area of the circle, let's now get the area of the sector with 144° and subtract it from it, and what we'll be left with, will just be the blue part, the 216° sector area.

`\textit{area of a sector of a circle}\\\\ A=\cfrac{r^2\theta \pi }{360}~~ \begin{cases} r=radius\\ \theta =\stackrel{in~degrees}{angle}\\[-0.5em] \hrulefill\\ r=15\\ \theta =144 \end{cases}\implies A=\cfrac{15^2(144)\pi }{360}\implies A=90\pi \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{area of the whole circle}}{225\pi }-\stackrel{\textit{area of the }144^o~sector}{90\pi }\implies \stackrel{\textit{area of the cone uses}}{135\pi }`

now, let's take a peek a the picture in the center, as we get the blue edges together from the material to form the cone, the slant-height becomes the 15 value from the radius of the circle making the cone, so our cone's slant-height is 15 then.

now, the cone is open at the top, so area wise, the only area used by the cone is the blue section, namely what we already know is 135π, oddly enough, that's the lateral area of the cone, so let's use our slant-height and lateral area of it to get the radius of the cone, let me use a "R" for this radius, to separate it from the one from the circle.

`\textit{lateral area of a cone}\\\\ LA=\pi R√(R^2+h^2)~~ \begin{cases} LA&=135\pi \\ √(R^2+h^2)&=15 \end{cases}\implies 135\pi =\pi R15 \\\\\\ \cfrac{135\pi }{15\pi }=R\implies 9=R`

now, we know the cone has a radius of 9 and we also know its's slant-height, as you can see in the far right of the picture, we can just use pythagorean theorem to get "h".

`15^2=h^2+9^2\implies 15^2-9^2=h^2\implies √(225-81)=h\implies \blacktriangleright 12=h \blacktriangleleft`