Practice Form K 7-2 Similar Polygons Determine whether the polygons are similar. If so, write a similarity statement and give the scale factor. Algebra The polygons are similar. Find the value of each variable. You want to enlarge a 5 in. by 5 in. photo. The paper you will print on is 8.5 in. by 14 in. What is the largest size the photo can be? Ella is reading a map with a scale of 20 mi. On the map, the distance Ella must drive is 4.25 in. How many miles is this?
The Correct Answer and Explanation is:
Let’s break down the problems one by one:
Problem 1: Similarity Statement and Scale Factor
For problems like this, you need to compare the corresponding sides and angles of the polygons. If the corresponding angles are equal and the corresponding sides are proportional, the polygons are similar.
For example, if in two triangles, the angles are equal and the sides are proportional, the similarity statement could be:
Triangle ABC ∼ Triangle DEF (denoting similarity between the triangles)
The scale factor is the ratio of the corresponding side lengths.
Problem 7: Enlarging the Photo
You want to enlarge a 3 in. by 5 in. photo, and the paper you will print it on is 8.5 in. by 14 in. We are asked to find the largest size the photo can be.
To keep the aspect ratio of the photo intact, you must determine how much you can enlarge the photo based on the dimensions of the paper.
- The original photo has dimensions 3 in. by 5 in.
- The paper has dimensions 8.5 in. by 14 in.
The scale factor for each dimension is:
- Width: 8.53=2.83\frac{8.5}{3} = 2.8338.5=2.83
- Height: 145=2.8\frac{14}{5} = 2.8514=2.8
Since both dimensions are proportional, the largest photo you can print without distortion is 8.4 in. by 14 in.
Problem 8: Map Scale Conversion
Ella is reading a map with a scale of 1 inch = 20 miles. The distance on the map that Ella must drive is 4.25 inches.
To convert this to the actual driving distance, multiply the map distance by the scale factor: Actual Distance=4.25×20=85 miles\text{Actual Distance} = 4.25 \times 20 = 85 \text{ miles}Actual Distance=4.25×20=85 miles
Thus, Ella must drive 85 miles.
