a right-angled triangle ABC

I've assumed that BC is not the hypotenuse. A cone is formed:

The formula for the surface area (including the base) is pi*r² + pi*r*s where s is the slant height (= √(h² + r²).  

Here: r = BC = 10cm, h = AB = 8cm

I'll let you do the numerics.

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Thanks Alan,

I think I am suffering from an embarrasing mental block.

C(10,0)     A(0,8)

equation AC:    

$$y=-0.8x+8$$

or

$$x= \frac{y-8}{-0.8}$$

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For the SA of the curved surface why can't I just say

$$\\SA=\int_0^8\;\;2\pi x\;dy\\\\ SA=2\pi\int_0^8\;\;\frac{y-8}{-0.8}\;dy\\\\ SA=\frac{2\pi}{-0.8}\int_0^8\;\;y-8\;dy\\\\ SA=\frac{2\pi}{-0.8}\left[\;\;\frac{y^2}{2}-8y\right]_0^8\\\\\ SA=\frac{2\pi}{-0.8}\left[\frac{8^2}{2}-64\right]\\\\ SA=\frac{2\pi}{-0.8}\left[\frac{-62}{2}\right]\\\\ SA=80\pi\\\\$$

I know that this is wrong but I don't know why.

This is how the surface area of the slanted surface should be calculated:

Note that A is written as 2*pi*sqrt(1+(h/r)²)*r²/2  =  pi*sqrt(r² + h²)*r = pi*s*r

You must use ds, not just dx or dy. 

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Thanks Alan

Melody, we can also do this in this manner .

Surface area = 2pi ∫f(x) √[1 + [ f ' (x)]^2 ] dx  from a to b

If we rotate the line   f(x) = (10/8)x  = 1.25x  around the x axis  from x = 0 to x = 8 we have the same cone laid on its side..... and we have....  {Remember that f 'x  = 1.25.....and we must square this under the root }

2 pi ∫ (1.25x ) √[1 + [1.25]^2] dx   from  0 to 8 =

2pi √[1 + [1.25]^2] [ 1.25/2] x^2    from 0  to 8  =

 1.25pi √[1 + [1.25]^2] * 64  =  402.32 sq units

And we can add the area of the base to this.

Chris, I don't think you mean √[1 + [ f ' (x)]^2 ]^.5, but just √[1 + [ f ' (x)]^2 ] or [1 + [ f ' (x)]^2 ]^.5,

and this, multiplied by dx, is just ds!

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Thanks, Alan.....I forgot I already had the root in there....

Thanks Alan and Chris,  I have to take time out to think about all this.    (◐‿◑)