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Mathematics 41 Online
OpenStudy (anonymous):

Note that the apothem shown is equal to 2 square root of three.

OpenStudy (anonymous):

OpenStudy (anonymous):

All of the triangles in the figure below are congruent. What is the area of the figure? Note that all measurements are in centimeters.

OpenStudy (anonymous):

@jim_thompson5910 help meh

jimthompson5910 (jim_thompson5910):

All triangles are congruent. So that 4 that is the base of that upper right triangle is also the base of the top most triangle. Do you see this?

OpenStudy (anonymous):

yes i so but my possible answers are weird :(

OpenStudy (anonymous):

do*

jimthompson5910 (jim_thompson5910):

What are your possible answers?

OpenStudy (anonymous):

48 + 24 square root of three. 24 square root of three. 48 square root of three. 24 + 24 square root of three.

jimthompson5910 (jim_thompson5910):

so it looks like you want the area of the hexagon as well?

OpenStudy (anonymous):

those are my anwsers???

jimthompson5910 (jim_thompson5910):

Let's find the area of each triangle and ignore the hexagon for now

jimthompson5910 (jim_thompson5910):

So what is the area of that top most triangle?

OpenStudy (anonymous):

geez idk 4?

jimthompson5910 (jim_thompson5910):

|dw:1341943885344:dw| Area: A = (bh)/2 A = (4*2)/2 A = 8/2 A = 4 So you are correct

jimthompson5910 (jim_thompson5910):

There are 6 of these (identical) triangles. So the total area of the pink area is 6*4 = 24

OpenStudy (anonymous):

so which anwser is it?

jimthompson5910 (jim_thompson5910):

The hexagon breaks up like so |dw:1341944010049:dw|

jimthompson5910 (jim_thompson5910):

It's given that the value of h is 2*sqrt(3) So the area of one triangle in that hexagon is A = (bh)/2 A = (4*2*sqrt(3))/2 A = 4*sqrt(3) There are 6 of these triangles, so the area of the hexagon is 6*4*sqrt(3) = 24*sqrt(3)

jimthompson5910 (jim_thompson5910):

Finally, add up the two areas we found to get \[\Large 24+24\sqrt{3}\]

OpenStudy (anonymous):

so its the 3rd one?

jimthompson5910 (jim_thompson5910):

no

jimthompson5910 (jim_thompson5910):

looks like the 4th

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