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Geometry question
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Not sure how I got this one right - can someone please explain. Answer is marked.
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Are of triangle = (t^2sqrt3)/4
Area of square s^2
(t^2sqrt3)/4=s^2 if you get sqrt of both side
you will get s=(t*4throot of 3)/2
which is your answer (upside down)
It is very hard to write like this let me know if you did not uderstand it
Good Luck
Area of square s^2
(t^2sqrt3)/4=s^2 if you get sqrt of both side
you will get s=(t*4throot of 3)/2
which is your answer (upside down)
It is very hard to write like this let me know if you did not uderstand it
Good Luck
rosh26 wrote:Not sure how I got this one right - can someone please explain. Answer is marked.
That is the formula for equilateral triangles but if you want it you can do it yourself
area = h*t/2
height (90 degree) will divide the t into 2 equal parts now we have 2 equal right angle triangle. lets find what is h
t^2=(t/2)^2+h^2
t^2=t^2/4+h^2
t^2-(t^2/4) = h^2
(4t^2-t^2)/4=h^2
3t^2/4=h^2
h= t*sqrt3/2
area = t*h/2= t*t*sqrt3/4= t^2*sqrt3/4
area = h*t/2
height (90 degree) will divide the t into 2 equal parts now we have 2 equal right angle triangle. lets find what is h
t^2=(t/2)^2+h^2
t^2=t^2/4+h^2
t^2-(t^2/4) = h^2
(4t^2-t^2)/4=h^2
3t^2/4=h^2
h= t*sqrt3/2
area = t*h/2= t*t*sqrt3/4= t^2*sqrt3/4
rosh26 wrote:I dont understand how you got the area of the trianhe. How did you get t^2sqrt3???
