An equilateral triangle, a square and a regular hexagon are connected as shown. If the sides of all figures are 16 cm and the apothem of the hexagon is 13.9 cm, find the area of the entire figure. Show your work for each of the three shapes. Round your answer to the nearest tenth.

Question

01-03-2017
Mathematics

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An equilateral triangle, a square and a regular hexagon are connected as shown. If the sides of all figures are 16 cm and the apothem of the hexagon is 13.9 cm, find the area of the entire figure. Show your work for each of the three shapes. Round your answer to the nearest tenth.

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saucierbaby saucierbaby · Answer 1 · 2017-03-01T15:07:39+01:00

As each side is 10, then 1/2x base = 5 but at this stage we do not know the height and must calculate it.
Split the triangle down the middle bisecting one angle. You have now formed two right angled triangles where the hypotenuse is the length of a side = 10cm, one leg is 5cm and the other is the height.
Therefore using Pythagoras's Theorem which states that the square of the hypotenuse is equal to the sum of the squares of the other two sides
10^2 = 5^2 + height^2
100 = 25 + height squared
75 = height squared
height = sqrt 75 = 8.66cm
Therefore we can now calculate the area of the triangle
1/2* base*height = 0.5 x 10 x 8.66 = 43.3 (to the nearest tenth)
Triangle Area = 43.3 square centimeters

SQUARE = length x width = 10 x 10 = 100

HEXAGON = 1/2 apotherm x perimeter
The perimeter is calculated by multiplying the length of one side by the number of sides
Perimeter = 10cm x 6 = 60 cm
The apotherm is 8.7
Therefore the area of the hexagon = 60 x 8.7 = 522 sq cms

Therefore the area of the entire shape is the area of the triangle + the area of the square + the area of the hexagon

= 43.3 + 100 + 522 = 665.3 square centimeter