Question

Help Please!!
T.A. =
(___+__sqrt__)

Answer 1

so you have a prism, notice how many sides are on it

Answer 2

you have 2 triangles, atop and below, one rectangle on the side, and other 2 rectangles upfront and behind

Answer 3

can you get the Area for the rectangles?

Answer 4

Would'nt you just times 8 by 8? assuming 8 it the length. it would end up being 64.

Answer 5

yes, length times width

Answer 6

what would you do after that? i have an idea on how to get the answer but having trouble with the square root.

Answer 7

well, right, the rectangles is just length times width
then you just get the Area of the 2 triangles, and add all Areas up to get Total Area

Answer 8

bearing in mind that a triangle's Area is \(\cfrac{1}{2}\times base \times height\)

Answer 9

so, if you take a peek at the picture above, you just need to get the "height" of the triangle to get its Area

Answer 10

and the "height" of it, you'd get with the Pythagorean theorem
\(c^2 = a^2 + b^2 \implies b = \sqrt{c^2 - a^2}\)

Answer 11

for the area of the triangle i got 12.
when i did the "height" for some reason i got -10. c=6^2 and a=4^2 right?

Answer 12

wait i think i get it! thanks!

Answer 13

so the Area end up being
the 3 rectangles
32 + 48 + 48
and the 2 triangles
\(4\sqrt{2}+4\sqrt{2}\)

Answer 14

\(4\sqrt{2}+4\sqrt{2} = 8\sqrt{2}\)

Answer 15

haemm one sec

Answer 16

how did you come up with 4sqrt of 2, okay

Answer 17

actually 1/bh for a triangle will be
\(\cfrac{1}{2}4\times 4\sqrt{2} \implies 8\sqrt{2} \)

Answer 18

I missed one thing, the base is "4" not "2" :/

Answer 19

so the Area end up being
the 3 rectangles
32 + 48 + 48
and the 2 triangles
\(8\sqrt{2} + 8\sqrt{2} \implies 16\sqrt{2}\)

Answer 20

so would it end up being=
(128+16sqrt2)

Answer 21

yes

Answer 22

Thank you very much!

Answer 23

yw