When constructing an inscribed equilateral triangle, how many arcs will be drawn on the circle?
2 answers:
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Answer:
8 +(3/8)√53 in² ≈ 10.73 in²
Step-by-step explanation:
Given the net of a triangular pyramid with some of the dimensions filled in, you want to find the total surface area.
<h3>Triangle base</h3>The triangle bases identified by dashed lines will have a length equal to the hypotenuse of the right triangles with legs shown as solid lines. The legs of each of those right triangles are ...
a = (3 in)/2 = 1.5 in
b = 2 in . . . . . . shown as the altitude of the triangle
Then the hypotenuse is found using the Pythagorean theorem:
c² = a² +b²
c² = 1.5² +2² = 2.25 +4 = 6.25
c = √6.25 = 2.5
The dashed lines are 2.5 inches long.
<h3>Triangle altitude</h3>The altitude from the solid horizontal line to the vertex at the bottom of the figure can be found using the fact that all of the outside edge lengths of the net are the same length. That edge length is found as the length of the hypotenuse of the right triangles in the left- and right-sides of the upper portion of the net. Each of those has a leg that is (2.5 in)/2 = 1.25 in and a leg marked as 2 in.
c² = a² +b²
c² = 1.25² +2² = 1.5625 +4 = 5.5625
c = (√89)/4 ≈ 2.358 . . . in
The unmarked altitude of the bottom triangle is then ...
b² = c² -a²
b² = 89/16 -1.5² = 53/16
b = (√53)/4 ≈ 1.820 . . . in
<h3>Surface area</h3>The surface area of the figure is the sum of the areas of the four triangles that make up the net. Each triangle has an area given by the formula ...
A = 1/2bh
The left and right triangles have b=2.5, h=2, so they each have an area of ...
A = 1/2(2.5)(2) = 2.5 . . . . in²
The center triangle has dimensions of b=3, h=2, so an area of ...
A = 1/2(3)(2) = 3 . . . . in²
The bottom triangle has dimensions of b=3, h=(√53)/4, so an area of ...
A = 1/2(3)(√53/4) = (3/8)√53 ≈ 2.730 . . . . in²
The total surface area is the sum of the areas of these triangles, so is ...
A = 2.5 in² +2.5 in² +3 in² +2.73 in² = 10.73 in²
The surface area of the triangular pyramid is (64+3√53)/8 ≈ 10.73 in².
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<em>Additional comment</em>
Often we work with pyramids that are rotationally symmetrical about a vertical line through the peak. This one is not. The altitude of the bottom triangle in the net is less than the altitude of the other triangles. This short face of the pyramid will tend to be more vertical than the other two lateral faces.
