The area of the considered equilateral triangle is given by: Option D: [tex]24.2 \: \rm cm^2[/tex]
For a regular polygon (a polygon with all sides of same measure), the line segment drawn from the center of the polygon to the mid point of a side of the polygon is called apothem.
Referring to the attached figure below, for the considered case, we're given that:
Since all sides of an equilateral triangle are same, let they be of 'x' cm length for this case, then:
[tex]Perimeter = x + x + x = 22.45\\\\x = \dfrac{22.45}{3} \approx 7.48 \: \rm cm[/tex]
Since the equilateral triangle has all same sides, all three triangles AOC, AOB, and BOC are congruent and have same area.
The area of ABC = Area of AOC + Area of AOB + Area of BOC = 3 × Area of AOC
The area of AOC = half of base times height. The height is length of the apothem since because of symmetry in right and left of the apothem, it is standing with equal angle on both sides of AC, thus, half of straight line angle, therefore of 180/2 = 90 degrees, thus, perpendicular.
Area AOC = [tex]\dfrac{1}{2} \times |AC| \times |OD| \approx \dfrac{1}{2}{ \times 7.48 \times 2.16 \approx 8.078\rm \: cm^2[/tex]
Thus, area of ABC = 3 × Area of AOC = [tex]3 \times 8.078 \approx 24.2 \rm \: cm^2[/tex]
Thus, the area of the considered equilateral triangle is given by: Option D: [tex]24.2 \: \rm cm^2[/tex]
Learn more about apothem here:
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