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! ^^ Brainliest Please?

5 0

3 years ago

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Darren runs 1/4 a mile in 2.5 mins. how long will it take him to run a full mile?

rusak2 [61]

it is simple, you multiply 2.5 by 4

2.5*4=10

6 0

3 years ago