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# another problem

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Side  $$CD$$of rectangle $$ABCD$$ measures 12 meters, as shown. Each of the three triangles with a side along segment $$CD$$ is an equilateral triangle. What is the total area of the shaded regions? Express your answer in simplest radical form

https://latex.artofproblemsolving.com/3/f/3/3f369421bbd6032e57b184bc4f0d665a4cb7cba8.png

tertre  Dec 28, 2017
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#1
+5888
+2

The three triangles along CD are all equilateral and they all have the same height, so they are all congruent to each other. The length of each side of these three triangles  =  12/3 m  =  4  m

And we know that an angle along line CD between two  60°  angles  =  180° - 60° - 60°  =  60°

So the two shaded triangles are also equilateral triangles with a side length of  4 m.

the area of the shaded region  =  2 * (area of one shaded triangle)

=  2 * ( 1/2 * base * height )

=  2 * ( 1/2 * 4 * 2√3 )

=  8√3     sq meters

hectictar  Dec 29, 2017
#2
+311
+2

1/2 sol: An altitude of an equilateral triangle is therefore $$\sqrt{3}$$ times the length of half the side length of the triangle. Therefore, an equilateral triangle with side length 4 has altitude length $$\sqrt{3}(4/2) = 2\sqrt{3}$$, and area $$(2\sqrt{3})(4)/2 = 4\sqrt{3}$$ square units. The shaded regions consist of two of these equilateral triangles, so their total area is $$2(4\sqrt{3}) = \boxed{8\sqrt{3}}$$ .

ant101  Dec 29, 2017

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