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5. Yes. If you're familiar with construction using compass and straight edge, one of the easiest ways to construct an equilateral triangle is to draw two circles where each circle's centre lies on the other circle's edge. Drawing a line between the two intersection points and then from each intersection point to the point on one circle farthest ...
14. You can use Heron's formula for the area of a triangle to help here. A = √s(s − a)(s − b)(s − c) where s is the semi-perimeter s = a + b + c 2 To find the conditions for the maximum area, we want to compute derivatives w.r.t. a, b and c, but since they are not independent variables, we first substitute the semi-perimeter equation ...
0. The area of a triangle made by the medians taken as sides is 75% of the triangle of which the medians are given. Now you can find the area by heron formula and the area thus you get will be 75% of the area of the triangle of which the medians are given. Share. Cite.
Using simple Pythogyras' theorum, you can prove the above result. So, the area of the largest (equilateral) triangle that can be inscribed in a circle would be: Area = √3 4. a2 = √3 4. (r√3)2 = 3√3 4. r2. Where 'r' is the radius of the circle and 'a' is the side of the triangle. For a unit circle r=1 obviously.
Konstantinos Michailidis. Nov 22, 2015. Let ABC equatorial triangle inscribed in the circle with radius r. Applying law of sine to the triangle OBC, we get. a sin60 = r sin30 ⇒ a = r ⋅ sin60 sin30 ⇒ a = √3 ⋅ r. Now the area of the inscribed triangle is. A = 1 2 ⋅ AM ⋅ BC. Now AM = AO+ OM = r +r ⋅ sin30 = 3 2 ⋅ r. and BC = a ...
The square of maximum area occurs when upper corners of square touches the sides of the equilateral triangle and the bottom side of the square is on one side of the triangle. Then you can find the relation between the area of square and the equilateral triangle.
An equilateral triangle has three sides of equal length and three internal angles which are each $60^\circ$. Imagine cutting the equilateral triangle in half by an altitude. This way, there are two right triangles which have the angle pattern $30^\circ−60^\circ−90^\circ$ .
The problem merely seems to ask for the cross-sectional area A(x) A (x). Note that, for an equilateral triangle of side s s, the area is 3–√ s2/4 3 s 2 / 4. In this case, s = ex + 2 s = e x + 2, so that. A(x) = 3–√ 4 (ex + 2)2 A (x) = 3 4 (e x + 2) 2. The volume calculation is simply an integral of A(x) A (x) over x ∈ [1, 3] x ∈ [1 ...
$\begingroup$ There is a choice of which side of AB on which to build the equilateral triangle. If D is on the opposite side from C, then PC and PD are in the right position to do the geometric mean construction (if you use a semicircle).
Start with an equilateral triangle with unit area. Trisect each of the sides and then cut-off the corners. In this case, we get a regular hexagon - see the picture below. Next, trisect each of the sides of the hexagon and cut-off the corners. This will give a dodecagon, but not a regular one. Continue this process ad infinitum.