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.
<span>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. </span>
<span>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: </span>
<span>(S. of small triangle)/(S. of big triangle) = (Ht. of small triangle)/(Ht. of big triangle) </span>
<span>==> x/6 = (3√3 - h)/(3√3) </span>
<span>==> x = (6√3 - 2h)/√3 </span>
<span>Thus, the area of the upper triangle is: </span>
<span>A = (1/2)[(6√3 - 2h)/√3](3√3 - h) = [(6√3 - 2h)(3√3 - h)]/(2√3). </span>
<span>(Made a dumb mistake about the height here for some reason) </span>
<span>Since we require that the area of this triangle is to be half of the total area (9√3/2), we need to solve: </span>
<span>[(6√3 - 2h)(3√3 - h)]/(2√3) = 9√3/2 </span>
<span>==> (6√3 - 2h)(3√3 - h) = 27 </span>
<span>==> 54 - 6h√3 - 6h√3 + 2h^2 = 27 </span>
<span>==> 2h^2 - 12h√3 + 27 = 0. </span>
<span>Solving with the Quadratic Formula gives: </span>
<span>h = (6√3 + 3√6)/2 ≈ 8.87 units and h = (6√3 - 3√6)/2 ≈ 1.52 units. </span>
<span>Since h = (6√3 + 3√6)/2 would place the line outside of the triangle, we pick h = (6√3 - 3√6)/2. </span>
<span>Therefore, the line should be ==> (6√3 - 3√6)/2 units from the base. </span>
<span>I hope this helps! ^^ Brainliest Please?</span><span>
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Answer:
Step-by-step explanation:
I believe the answer is B. The y intercept is when x is 0.
Answer:
54,000
Step-by-step explanation:
The midpoint here which is = to M is (8, 1). Hope this helps
