Exercise 113 Page 575

Let's start by copying the given diagram onto paper.

Now we will dilate one of the points with our compass. Place the needle point of the compass at V and open up the compass to the length of VB.

Without changing the compass setting, move the needle point to B and mark the same distance along the same ray. Where the arc intersects the ray, we find B'.

Let's repeat the same procedure for the remaining points.

Let's connect the points we drew.

From the description of △ A'B'C', we know that it is a dilation of △ ABC by a factor of 2. This means △ A'B'C' is twice the size of △ ABC.

If we multiply the perimeter of △ ABC with the linear scale factor between the triangles, we can find the perimeter of △ A'B'C'.

P_(△ A'B'C')=15(2)=30 units

To find the area of △ A'B'C', we should first find the area scale factor between the triangles. The area scale factor is the square of the linear scale factor. From Part B, we know that the linear scale factor was 2. With this information, we can determine the area scale factor.

(Linear scale factor)^2=( 2)^2 ⇓ Area scale factor=4 If we multiply the area of △ ABC by the area scale factor between the triangles, we can determine the area of △ A'B'C'. A_(△ A'B'C')&=(19)(4)=76 units^2

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Exercise 113 Page 575

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