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# If altitude $CD$ is centimeters, what is the number of square centimeters in the area of ​ ABC$? -1 98 6 +-291 If altitude$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
+22
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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!

Guest Mar 28, 2020