what is the area of the triangle?

Question

what is the area of a triangle which leg a= x leg b= sq.rt.(2)x and leg c= (4)sq.rt.(15)

Answer 1

It is deceiving... it looks easy, but it gets bad really quick.

 

Heron's formula says:

 

  let S = (a+b+c)/2

 

Then the area is sqrt ( S(S-A)(S-B)(S-C))

 

 

So S = (x + sqrt(2)*X + 4*sqrt(15))/2

 

S-A = ( sqrt(2)*X + 4*sqrt(15) - x)/2

 

S-B = ( X + 4*sqrt(15) - sqrt(2)*X)/2

 

S-c = ( x + sqrt(2)*X - 4*sqrt(15))/2

 

Plugging these into the formula, the denominator is 16.

 

THe numerator is the PRODUCT of the following trinomials:

         (x + sqrt(2)*x + 4*Sqrt(15)

 

         (sqrt(2)*x + 4*sqrt(15) - x)

 

         (x  - (sqrt(2)*x - 4*Sqrt(15))

 

         (x + sqrt(2)*x - 4*Sqrt(15)) 

 

Notice the first two are difference of squares pattern (A^2 - B^2) where A = sqrt(2)+4*Sqrt(15) and B=x.

Likewise the last two are also difference of squares pattern with A = x and B = sqrt(2)-4*Sqrt(15)

 

 

FOILING them out using the difference of squares pattern:

 

  [( sqrt(2)*x + 4*Sqrt(15))^2 - x^2][ x^2 - ( sqrt(2)*x - 4*Sqrt(15))^2 ] / 16

 

=  (2x^2 + 8*sqrt(30)*x + 240 - x^2 )( x^2 - 2x^2 + 8*sqrt(30)*x + 240) / 16 

 

= (8*sqrt(30)*x + 240 + x^2)(8*sqrt(30)*x + 240 - x^2)/16

 

=  [(8*Sqrt(30)*x + 240) ^2 - x^4 ]/16  <--- difference of squares pattern with A=8*sqrt(30)*x and B  = x^2

 

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Finally, I solved the problem under the assumption that B = sqrt(2)*x;

that is, the X is NOT under the square root sign.

 

I attempted to solve it WITH the x under the square root sign;

that is sqrt(2*x)

and the result is not any better, in fact, worse.

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