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# help. PLZ I wish I could figure out how to do this.

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The altitude of an equilateral triangle is the square root of 6 units. What is the area of the triangle, in square units? Express your answer in simplest radical form.

Nov 28, 2017
edited by Guest  Nov 28, 2017

#1
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Let  x  be the side length of the triangle. Here's a quick drawing:

From the Pythagorean theorem...

(x/2)2 + (√6)2  =  x2

x2/4 + 6  =  x2                Multiply through by  4 .

x2 + 24  =  4x2               Subtract  x2  from both sides.

24  =  3x2                       Divide both sides by  3  then take the positive square root of both sides.

x  =  √8

And...

area of triangle, in square units  =  (1/2)(x)(√6)   =   (1/2)(√8)(√6)   =   (1/2)(4√3)   =   2√3

Nov 28, 2017

#1
+7348
+2

Let  x  be the side length of the triangle. Here's a quick drawing:

From the Pythagorean theorem...

(x/2)2 + (√6)2  =  x2

x2/4 + 6  =  x2                Multiply through by  4 .

x2 + 24  =  4x2               Subtract  x2  from both sides.

24  =  3x2                       Divide both sides by  3  then take the positive square root of both sides.

x  =  √8

And...

area of triangle, in square units  =  (1/2)(x)(√6)   =   (1/2)(√8)(√6)   =   (1/2)(4√3)   =   2√3

hectictar Nov 28, 2017