Question 179757
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The area of a triangle is given by:  *[tex \Large A =  \frac {bh}{2}] where <i>b</i> represents the measure of the base and <i>h</i> represents the measure of the height.


For your problem:  *[tex \Large A =  \frac {(sqrt{30})(sqrt{6})}{2}] 


Since *[tex \Large (sqrt{a})(sqrt{b}) = sqrt{ab} \text {  } \math \forall a, b \in \R, a \geq 0, b \geq 0] we can write:


*[tex \Large A =  \frac {sqrt{180}}{2}]


But:  *[tex \Large 180 = 4 \times 9 \times 5]


So:  *[tex \Large \frac {(2)(3)sqrt{5}}{2}=3 sqrt{5}]



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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