Question

Find the area of a triangle whose base is 2√3 in. and height is 3√3 in.

Answer 1

area of a triangle is \(\bf \cfrac{1}{2} \times base \times height\)

Answer 2

Ok but I do not understand how to solve for the square root stuff

Answer 3

you multiply outside with outsides, and insides with insides, for the same roots, in this case both have the same root, root 2

Answer 4

so 2x3 and 3x3 divided by 2

Answer 5

ahemm, yes, but recall the 3's are inside the root, so you keep the insides, inside the root

Answer 6

\(\bf a\sqrt{x} \times b\sqrt{y} \implies ab\sqrt{xy}\)

Answer 7

I'm lost

Answer 8

well, just follow as => \(a\sqrt{x} \times b\sqrt{y} \implies ab\sqrt{xy}\)

Answer 9

what would that give you?

Answer 10

Forget it this is pointless.

Answer 11

jeez
what is \(\sqrt{3}\times \sqrt{3}\) ??

Answer 12

\(\sqrt3\) is the number whose square is 3, so what do you get when you square it?
kind of like "who is buried in grant's tomb?"
since \(\sqrt3\) is the number you square to get 3, when you square it, you get \(3\)

Answer 13

in other words
\[2\sqrt3\times 3\sqrt3=2\times 3\times \sqrt3\times \sqrt3=6\times 3\]

Answer 14

18