Exercise 74 Page 456

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a An equilateral triangle has three congruent sides. Therefore, if we divide the perimeter of the loop by 3 we obtain the length of the triangle's sides.

24/3=8 cm The triangle has three congruent sides of 8 centimeters. Since it is equilateral, it also has three internal angles of 180^(∘)3=60^(∘). To find the area of this triangle we need to know its height.

From the diagram, we see two congruent right triangles. Since one of the non-right angles is 30^(∘), these must be a 30^(∘)-60^(∘)-90^(∘) triangles. In such a triangle, the legs and hypotenuse have the following relationship.

Since the smaller of the triangle's legs is a= 4 cm, we can determine the height. h=asqrt(3) ⇒ h= 4sqrt(3)cm Now we can determine the area of the triangle. Area: 1/2(8)(4sqrt(3))=16sqrt(3) cm^2

b A square has four congruent sides. If we divide the perimeter of the loop by 4, we obtain the length of the square's sides.

Side: 24/4=6 cm To calculate the area of the square we have to square its side. Area: 6^2=36 cm^2 From Part A, we know that the area of the equilateral triangle is 16sqrt(3)≈ 27.7 cm^2. Therefore the area of the square is greater.

c A regular hexagon has 6 congruent sides, which means each side is 246=4 cm. The diagonals between opposite sides creates 6 isosceles congruent triangles, each with a vertex angle of 360^(∘)6=60^(∘). Let's illustrate this.

If we draw the height from the vertex angle of one of the triangles, it will bisect the vertex angle and triangle's base.

Like in Part A, we have a 30^(∘)-60^(∘)-90^(∘) triangle. Since the smaller of the triangle's legs is a= 2 cm, we can determine the height. h=asqrt(3) ⇒ h= 2sqrt(3)cm Now we can calculate the area of the triangle and finally the hexagon by multiplying this number by 6. Area: (1/2(4)(2sqrt(3)))6=24sqrt(3) cm^2 The area of the hexagon is 24sqrt(3)≈ 41.6 cm^2, which is greater than both the square's and triangle's area.

d It would seem like as the number of sides increases the area also increases. Therefore, when the number of sides goes towards infinity, the area is maximized. This kind of a polygon is a circle.

Subchapter links

8.3.1 Area Ratios of Similar Figures p.451-452

8.3.2 Ratios of Similarity p.455-457

Exercise 74 Page 456

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