Question

Find the area of a triangle whose base is 2√3 in. and height is 3√3 in.
I dont understand square roots can you plz explain the basic part...

Answer 1

\[(\sqrt{2})^2 =2\]
\[(\sqrt{3})^2 =3\]
Area = (1/2)x base x height

Answer 2

\[\sqrt {a} \times \sqrt b = \sqrt{a \times b}\] \[\sqrt a \times \sqrt a = \sqrt{a \times a} = a\]

Answer 3

i.e \[\sqrt 6 \times \sqrt 2 = \sqrt{6 \times 2} = \sqrt{12}\]

Answer 4

do you get it @Lilly95 ?

Answer 5

im confused how did you get 6 and 2

Answer 6

those were just examples

Answer 7

i.e. means something like for example

Answer 8

Oh

Answer 9

in your problem area = bh/2 therefore \[A = \frac{2\sqrt 3 \times 3 \sqrt 3}{2}\] \[2\sqrt 3 \times 3 \sqrt 3 = (2\times 3)(\sqrt{3 \times 3}) = 6(3) = 18\]

Answer 10

you ahve to divide it by 2 so \[A = \frac{18}{2} = 9\]

Answer 11

you understand right @Lilly95

Answer 12

Know i do

Answer 13

Thnx