Ask your own question, for FREE!
Mathematics 14 Online
OpenStudy (anonymous):

How would I solve this question? All of the triangles in the figure below are congruent. What is the area of the figure? Note that all measurements are in centimeters. Note that the apothem shown is equal to 2 (sqrt)3

OpenStudy (anonymous):

OpenStudy (kinggeorge):

A nice easy trick to this problem, is to notice that the area of the equilateral triangles is the same as the area of the hexagon. So you just have to find one to find the total area. The area of the hexagon is \[A={1 \over 2}ap\]where a is the apothem and p is the perimeter. We know the apothem to be \(2\sqrt3\) and the perimeter to be \(6\cdot4=24\) So the area of the hexagon is\[{1 \over 2}48\sqrt3 ={24\sqrt3}\]Therefore, the total area is\[2\cdot(24\sqrt3)=48\sqrt3\]

OpenStudy (kinggeorge):

Wait, is the height of the triangle 3? Was I misreading the diagram?

OpenStudy (anonymous):

But, that's not an answer. D: However, 24 (sqrt)3 ^2 is. WOuld that be it?

OpenStudy (anonymous):

The height of the triangle is 3. Yes.

OpenStudy (kinggeorge):

If the height of the triangle is 3, then the area of each triangle is \[{1\over2}\cdot3\cdot4=6\]So then the answer should be \[24\sqrt3+6\cdot6=24\sqrt3+36\]

OpenStudy (kinggeorge):

There are 6 triangles, so that's why I multiplied the 6 by 6.

OpenStudy (anonymous):

That is a possible answer. :D

OpenStudy (anonymous):

Is that answer squared?

OpenStudy (kinggeorge):

No. That's the entire answer. \[24\sqrt3+36 \;\;\;\text{cm}^2\]

OpenStudy (anonymous):

Oh, sweet. :D Thanks man!

OpenStudy (kinggeorge):

You're welcome.

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!