What is the area of an equilateral triangle in terms of the length of a side?

Question

I have worked it out for a Calculus problem, but I am unsure about whether or not my answer is correct. I found that, for an equilateral triangle with each side a length "b", the height of the triangle is h = (sqrt(3)/2)b; thus, the area is A = .5 b h = (sqrt(3)/4)b^2 (or .25 *sqrt(3) * b^2). Is this correct?

What is the area of an equilateral triangle in terms of the length of a side?
I have worked it out for a Calculus problem, but I am unsure about whether or not my answer is correct. I found that, for an equilateral triangle with each side a length "b", the height of the triangle is h = (sqrt(3)/2)b; thus, the area is A = .5 b h = (sqrt(3)/4)b^2 (or .25 *sqrt(3) * b^2). Is this correct?

Edit: Thank you for your help!

Answer 1

Yes this is correct.
You could also get this result by trigonometry, but getting it by calculus is a cool tihng to be able to do!
here's how you can use trig to verify your result:
A = (1/2)(bh)
Since it's equilateral, each angle is 60 degrees. Dissect the triangle into two 30 60 90 right triangles and you'll see that the altitude is sin 60 = (sqrt 3)/2 and the base is equal to the side.
So (1/2)(bh) = (1/2)((sqrt 3)/2)b = (1/4)sqrt3 b

Answer 2

I'm guessing that you're familiar with the formula for the area of a triangle - namely A = bh/2.....(1) where A is the area b is the base h is the height Now, since the triangle is equilateral, the base and the two adjacent sides are all equal Furthermore the angle between any two sides is 180/3 = 60 degrees (or PI/3 radians) By the trigonometric identity with the height of the triangle going from midway of the base to the apex h = b*sin(60)...........(b is also the length of the hypotenuse formed from one of the adjacent sides) = b*sqrt(2)/2 Substituting this into Eq(1) gives h = b^2*sqrt(3)/4 or h = (b/2)^2 * sqrt(3) So, if the length of each side is 1 unit, the area is A = sqrt(3)/4 units^2

Answer 3

Yes. You divided the triangle into two right triangle. You know two of the sides, b/2 and b (the hypotenuse). The other side must be b* square root(3) / 2. The area of each of the two right triangles is then 1/2 * (b/2 * sqrt(3)*b/2), or b^2 * sqrt(3)/8.

Add the two halves together, total is b^2 * sqrt(3)/4.

Answer 4

The alttude forms 30, 60, 90 right triangles & the altitude is opposite the 60° angle. it is equal to the hypotenuse*√3 /2.
A=(1/2)bh=(1/2)s*s*√3 /2=s^2*√3 /4

Yes. You are correct.

Answer 5

Yes, but..
Area of an equilateral triangle = Side² *sqrt of 3 over 4

a) For right triangle =>
A = bh over 2
b) For isosceles triangle (Hero's formula)=> sqrt of [p(p-a)(p-b)(p-c)]
<::>

Answer 6

Absolutely!

How I did it:

Divide the equilateral triangle into 2 triangles:

Area of big triangle =A1
Area of small triangle = A2

(A1 = 2* A2)

A2 = 0.5*(b*0.5)*h

h = sin 60 degrees * b = 0.5*sqrt(3)*b

Hence, A2 = 0.5*(b*0.5)*0.5*sqrt(3)*b

Now,
A1 = 2*A2
A1 = 2*(0.5*(b*0.5)*0.5*sqrt(3)*b) = 0.25*sqrt(3)*b^2 = your answer

Answer 7

The area of an equalateral triangle is (S = Side)
(S^2)( radical 3)
_______________
4

(S squared radical three all divided by 4)

Answer 8

Yes, correct and logically derived. Good job!

Answer 9

b^2=h^2+0.25b^2
h^2=0.75b^2
h=sqrt(0.75)b
A = h*0.5b
sqrt(0.75)b * 0.5b
0.5sqrt(0.75)b^2

Answer 10

yes, you are correct.