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Consider a right angled $\triangle PQR$ right angled at $P$ i.e ($\angle QPR=90°$) with side $PR=4$ and area$=6$. If $\triangle PQR$ is rotated through $360°$ about the side $PR$ , what is the $TSA$ of the resulting solid?

My Attempt: $$Ar.(\triangle PQR)=6$$ $$\dfrac {1}{2} {PR}\times {PQ}=6$$ $$PQ=3$$.

Again, by Pythagoras Theorem $$QR=\sqrt {PR^{2} + PQ^{2}}$$ $$QR=5$$.

How do I proceed further?

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    $\begingroup$ Surface area of a cone is $\pi r^2 + \pi rl$ where $r$ is the radius and $l$ is the slant height. $\endgroup$ – John Lou May 3 '17 at 1:15
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    $\begingroup$ Did you want to solve this using calculus, or just geometry? $\endgroup$ – MasterYoda May 3 '17 at 1:15
  • $\begingroup$ @Cocomos, Whichever is short and easier. $\endgroup$ – pi-π May 3 '17 at 1:16
  • $\begingroup$ @AlbertEinstein Use John Lou's formula. $PQ$ will be your radius since you are rotating about $PR$ and $QR$ will be your slant height. $\endgroup$ – MasterYoda May 3 '17 at 1:22
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The solid will be a cone with radius $PQ=3$ and generator $QR=5$.

Then TSA is given by

$$TSA=\pi \cdot PQ\cdot QR+\pi PQ^2=15\pi+9\pi=24\pi$$.

enter image description here

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  • $\begingroup$ How do you find that " The solid will be a cone"? $\endgroup$ – pi-π May 3 '17 at 1:24
  • $\begingroup$ the axis of rotation is $PR$, right? then you get a cone $\endgroup$ – Arnaldo May 3 '17 at 1:26
  • $\begingroup$ How? Is there any process of calculations?? $\endgroup$ – pi-π May 3 '17 at 1:28
  • $\begingroup$ @AlbertEinstein: Please, check the pícture. Can you see now? $\endgroup$ – Arnaldo May 3 '17 at 1:40
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    $\begingroup$ @Albert Einstein..just take a piece of triangular cardboard and rotate it very fastly taking any side as its axis of rotation( you can use a stick glued with one of its sides for this purpose). Now, see the surface generated which is a cone. $\endgroup$ – Nitin Uniyal May 3 '17 at 13:07
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According to the description, the resultant shape is a cone. In which, specific parts have their standard names:-

enter image description here

PR = vertical height = h; PQ = base radius = r; QR = slant height = L.


TSA = (base area) + [curved surface area] = $(\pi r^2) + [\pi \times r \times L]$.

Here is the way to find out why [curved surface area] = $\pi \times r \times L$:-

1) Cut a cone-shaped paper drinking cup along QR.

2) Lay the cut-up object flat.

3) Convince yourself that the flattened object is in the form of a sector of a new circle with the following specs:- radius = L and arc length $= 2 \pi r$.

4) Use the above info to find the central angle of that sector.

5) Find the area of that sector (whose area is exactly that of the curved surface).

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  • $\begingroup$ How do we know that we get cone after rotation? Is there any methods/calculation? $\endgroup$ – pi-π May 3 '17 at 9:53
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    $\begingroup$ @AlbertEinstein It is a mental visualization process. $\endgroup$ – Mick May 3 '17 at 10:58

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