Verified answer

EDIT: Fixed solution below.

Since the height of an equilateral triangle in terms of its side s is s√3/2, the height of the triangle is 6√3/2 = 3√3 and so the area is (1/2)(6)(3√3) = 9√3.

If we draw a horizontal line a height of h from the base of the triangle, the region is split into two regions: the lower region consisting of a trapezoid of height h and the upper region consisting of a triangle of height 3√3 - h.

Since the upper triangle and the triangle itself are similar triangles, the base and height are proportional. If we let x denote the base of the length of the upper triangle, we have:

(S. of small triangle)/(S. of big triangle) = (Ht. of small triangle)/(Ht. of big triangle)

==> x/6 = (3√3 - h)/(3√3)

==> x = (6√3 - 2h)/√3

Thus, the area of the upper triangle is:

A = (1/2)[(6√3 - 2h)/√3](3√3 - h) = [(6√3 - 2h)(3√3 - h)]/(2√3).

(Made a dumb mistake about the height here for some reason)

Since we require that the area of this triangle is to be half of the total area (9√3/2), we need to solve:

[(6√3 - 2h)(3√3 - h)]/(2√3) = 9√3/2

==> (6√3 - 2h)(3√3 - h) = 27

==> 54 - 6h√3 - 6h√3 + 2h^2 = 27

==> 2h^2 - 12h√3 + 27 = 0.

Solving with the Quadratic Formula gives:

h = (6√3 + 3√6)/2 ≈ 8.87 units and h = (6√3 - 3√6)/2 ≈ 1.52 units.

Since h = (6√3 + 3√6)/2 would place the line outside of the triangle, we pick h = (6√3 - 3√6)/2.

Therefore, the line should be (6√3 - 3√6)/2 units from the base.

I hope this helps!

If the equilateral triangle has side length 6, then its height is 3√3 and its area is:

A = (6)²√3/4 = 9√3 sq. units

Let h be the height at which the horizontal line is placed.

The height of the upper section (Triangle) will be: 3√3 - h(not necessary since it's an equilateral triangle which is of length L let's say).

The height of the lower section (Trapezoid) will be : h

Area of the upper section of the triangle which is of length L.

L²√3 = 18√3

L = √18 = 3√2

(h)(6 + L) = 9√3

h = 9√3/[6 + 3√2] = 1.5219

The line will be 5 high. Putting the line down the middle makes a right angled triangle. we know the longest side is 6, therefore using pythagoras, 6^2=a^2 +b^2. We know one of the sides is 3, because the line drawn down the middle divides one of the original triangle's sides in two. So, 6^2=3^2+b^2. When you do the calculation, it is easy to see that b is 5. 36=9 + ?. It is now clear that ? is 25 (Square root of 25 is 5).

Hope this helps.

I can only guess the line you ******.

However using pythag it should be 6^2 - 3^2 then square routed.

Which is the route of 27. Or in surd form 3route3

DON'T GIVE UP THY DAY JOB HAR HAR HAR **snort**