You can put this solution on YOUR website! Let s=length of a side of the equilateral triangle
Area (A) of a triangle equals (1/2)base(b)*height(h)=(1/2)bh
Height of an equilateral triangle with sides s units long is (using the Pythagorean theorm):
s^2=((1/2)s)^2+h^2 or h^2=s^2-(1/4)s^2=(3/4)s^2 and
h=(sqrt(3))/2)s so area would be
A=(1/2)s*(sqrt(3)/2)s=s^2(sqrt(3))/4)--------------eq1 (area of equilateral triangle with sides s units long).
Let l=length of the sides of the square
Area (A) of the square=l^2------------------------eq2
Now we are told that the perimeter of the triangle and square are equal. That means that 3s=4l or l=(3/4)s. Now we'll substitute this into eq2 and we get
Area of square=((3/4)s)^2=(9/16)s^2
Now for the ratio:
So (Area of square)/(Area of triangle)=(9/16)s^2/((sqrt(3))/4)s2)
The s^2 cancel on the right side so we end up with:
(Area of square)/(Area of triangle)=(9/16)/sqrt(3)/4) multiply numerator and denomionator by 4/(sqrt(3)) (Note: this will get rid of the complex fraction)
(Area of square)/(Area of triangle)=((9/16)*(4/(sqrt(3))/1 and this reduces to:
9/(4(sqrt(3)) now we multiply the numerator and denominator by sqrt(3) and we get:9(sqrt(3))/4*3=(9/12)sqrt(3) or (3/4)sqrt(3)
So,(Area of square)/(Area of triangle)=(3/4)sqrt(3)---shud be the ans
Hope this helps---ptaylor
