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anonymous
 one year ago
A side of an equilateral triangle is 20yd long. What is the area of the triangle? ( Picture Attached )
anonymous
 one year ago
A side of an equilateral triangle is 20yd long. What is the area of the triangle? ( Picture Attached )

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mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1An equilateral triangle is also equiangular. All sides have length 20 yd, and all angles measure 60 deg. dw:1437540579933:dw

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1To find the area, we can use the formula \(\Large A = \dfrac{bh}{2} \) The base can be any side of the triangle. The height is the altitude.

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1Let's use the bottom side as the base. Then we can draw the height as shown below. dw:1437540791741:dw

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1Now we need the length of the height. dw:1437540867820:dw

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1Each small triangle is a 306090 triangle. We recal that a 306090 triangle has sides in the ratio \(1 : \sqrt 3 : 2\) The three sides of a 306090 triangle are the two legs and the hypotenuse. The two legs form the right angle. The two legs are the short leg opposite the 30deg angle, the long leg opposite the 60deg angle. The hypotenuse is the longest side and is opposite the 90deg angle. The ratio tells us that the hypotenuse is twice the length of the short leg. The long leg is \(\sqrt 3\) times the length of the short leg.

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1In the small triangle to the right, the short leg measures 10 yd. That means the long leg measures \(10 \sqrt 3~yd\). The long leg of the small triangle to the right is also the height of the large triangle. Now we have the height we need to find the area.

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.1\(\large A = \dfrac{bh}{2} = \dfrac{20 ~yd \times 10\sqrt 3~yd}{2} = 100 \sqrt 3~yd^2\)
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