Core Connections Integrated II, 2015
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Core Connections Integrated II, 2015 View details
2. Section 6.2
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Exercise 76 Page 348

Practice makes perfect
a The perimeter of a polygon is the sum of its sides. Examining the diagram, we notice that all sides are known. With this information we can calculate the perimeter.
Perimeter: 5+11.4+7=23.4 m

To find the triangle's area, we have to multiply its base and height and divide the product by 2. We have also been given this information.

Having identified the height and base, we can determine the area.
A=1/2bh
A=1/2( 7)( 3)
â–Ľ
Evaluate right-hand side
A=1/2* 21
A=21/2
A=10.5
The area is 10.5 m^2.
b To enlarge the triangle by a scale factor of 3, we have to multiply each side of â–ł ALT by 3.
rrcl Height: & 3* 3 & -15pt= -15pt& 9 [0.3em] Side 1: & 5* 3 & -15pt= -15pt& 15 Side 2: & 7* 3 & -15pt= -15pt& 21 Side 3: & 11.4* 3 & -15pt= -15pt& 34.2

Let's draw the new triangle.

Now we can determine the perimeter and area using the same process as in Part A. Perimeter:& 15+34.2+21=70.2 m Area:& 1/2(21)(9)=94.5 m^2

c If the perimeter increases by a scale factor of 3, the ratio of the image perimeter to the preimage perimeter should equal 3. Let's test this.

70.2/23.4=3 The ratio between the perimeter of the image and the preimage does equal 3. When comparing perimeters of similar figures, this ratio is called the linear scale factor.

d Like in Part C, we will examine the ratio between the image area to the preimage area. This is called the area scale factor.

94.5/10.5=9 The scale factor between the areas is 9 and not 3. In fact, when comparing the areas of two similar figures, it will always be the square of the linear scale factor. From Part C, we know that the linear scale factor is 3. This means the area scale factor should be 3^2=9, which is the case.