Question

6 a scale factor of 2 maps figure a onto figure b. (figure a and b: triangles) enter the value of x. 10

7 on a scale drawing of oak grove, two landmarks are shown 5 inches apart. these two places are actually separated by 25 miles. what is the map’s scale? 1 inch = miles.

8 the figure on the left represents a scale drawing of the figure on the right. (left: triangle 2 in, right: 5 yd) what is the scale? 1 inch : yards.

9 polygon pqrs is a scaled copy of polygon abcd. (diagrams: abcd and pqrs) name the segment in the scaled copy that corresponds to segment ad. options: a) sr, b) rq, c) op, d) ps.

10 micah has a garden. he constructs a scale model of the garden using the scale of 0 feet. (typo likely, e.g., \scale of 1 inch : 6 feet\) what is the length of the garden in micah’s model? options: a) 3 inches, b) 4 inches, c) 6 inches, d) 12 inches.

11 read all of the statements about scale drawings below. mark all of the statements below that are true. a) scale drawings are used because they are easier to work with than an actual object or piece of furniture. b) scale drawings have the same size and dimensions as the actual object. c) scale drawings should have the same proportions as the actual object. d) scale drawings are half the size of the actual object.

Explanation:

Problem 7 (Scale Drawing - Map Scale)

Step1: Define the scale relationship

Let the scale be \( 1 \) inch \( = x \) miles. The distance on the map is \( 5 \) inches, and the actual distance is \( 25 \) miles. So we can set up the proportion: \( \frac{1}{x}=\frac{5}{25} \).

Step2: Solve for \( x \)

Cross - multiply: \( 5x = 25\times1 \). Then \( x=\frac{25}{5}=5 \).

Step1: Define the scale relationship

Let the scale be \( 1 \) inch \( = y \) yards. The length on the scale drawing (left triangle) is \( 2 \) inches, and the actual length (right triangle) is \( 5 \) yards? Wait, no, looking at the figure: the left triangle has a base of \( 2 \) in, and the right triangle has a base of \( 5 \) yd? Wait, no, maybe the left is the scale drawing and the right is the actual. Wait, the problem says "the figure on the left represents a scale drawing of the figure on the right". So the scale drawing length (left) is \( 2 \) in, actual length (right) is \( 5 \) yd? Wait, no, maybe I misread. Wait, the left triangle has a base of \( 2 \) in, and the right triangle has a base of \( 5 \) yd? Wait, no, the scale is \( 1 \) inch to how many yards. Let's set up the proportion. Let \( 1 \) inch \( = k \) yards. Then \( 2 \) inches (scale) corresponds to \( 5 \) yards (actual)? Wait, no, maybe the left is the scale drawing with length \( 2 \) in, and the actual is \( 5 \) yd? Wait, no, that can't be. Wait, maybe the left triangle has a base of \( 2 \) in, and the right triangle has a base of \( 5 \) yd? Wait, no, the correct way: if the scale drawing (left) has a length of \( 2 \) inches, and the actual (right) has a length of \( 5 \) yards, then the scale is \( 1 \) inch \(=\frac{5}{2}= 2.5\) yards? Wait, no, maybe the left is the actual and the right is the scale? No, the problem says "the figure on the left represents a scale drawing of the figure on the right". So scale drawing (left) length \( l_s = 2 \) in, actual length \( l_a=5 \) yd. Then the scale is \( 1 \) inch \(=\frac{l_a}{l_s}=\frac{5}{2} = 2.5\) yards? Wait, but maybe I made a mistake. Wait, let's re - read: "the figure on the left represents a scale drawing of the figure on the right". So left is scale, right is actual. So scale factor: \( \text{scale}=\frac{\text{actual length}}{\text{scale length}} \). So if scale length is \( 2 \) in, actual length is \( 5 \) yd, then \( 1 \) inch \(=\frac{5}{2}=2.5\) yards. But maybe the numbers are different. Wait, maybe the left triangle has a base of \( 2 \) in, and the right triangle has a base of \( 5 \) yd? Wait, perhaps the correct calculation is: Let the scale be \( 1 \) inch \( = y \) yards. Then \( 2 \) inches (scale) \( = 5 \) yards (actual). So \( y=\frac{5}{2}=2.5 \).

In a scaled copy (similar figures), corresponding sides are in proportion and have the same relative position. Polygon \( PQRS \) is a scaled copy of polygon \( ABCD \). So we need to find the segment in \( PQRS \) that corresponds to \( AD \) in \( ABCD \). By looking at the order of the vertices, \( A \) corresponds to \( P \), \( B \) to \( Q \), \( C \) to \( R \), and \( D \) to \( S \). So segment \( AD \) (connecting \( A \) and \( D \)) corresponds to segment \( PS \) (connecting \( P \) and \( S \)).

Answer:

\( 5 \)

Problem 8 (Scale Drawing - Triangle Scale)