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If altitude $CD$ is \($\sqrt3$ \)centimeters, what is the number of square centimeters in the area of \($\Delta\) ABC$?

 Mar 28, 2020
 #1
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By the definition of 30 - 60 - 90 triangles...AC=\(2\sqrt{3}\) and \(AD=\sqrt{3}\).

 

Triangle CDB is also a 30-60-90 degree triangle...angle B is sixty(60) degrees.

 

That means BD=1, and CB=2

 

..Thus, the area of the triangle is \(\frac{1}{2}*2*2\sqrt{3}=2\sqrt{3}\)

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 Mar 28, 2020
 #4
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Instead  of  AD = sqrt(3)   should be  CD = sqrt(3)  

Guest Mar 28, 2020
 #2
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Both triangles are 30-60-90 triangles. Using triangle properties, we get that AD is 3 and DB is 1. We also know that CB is 2.  AC is \(2\sqrt{3}\). The area is 0.5*base*height so it is \(2\sqrt{3}\)

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 Mar 28, 2020
edited by MathIsCool  Mar 28, 2020
 #5
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If altitude CD = √3 in centimeters, what is the number of square centimeters in the area of  ΔABC?

 

 Mar 28, 2020
 #6
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Note that in triangle  ABC  longer leg is  2√3 cm  and the area is  2√3 cm²

 

If the value of a leg of the right triangle is 2, than the numeric value of the other leg and the area of that triangle is the same!   indecision

Guest Mar 28, 2020

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