Question

1. model with mathematics martines town is building a volleyball court based on a scale drawing that is 40 centimeters by 80 centimeters and uses the scale 1 cm:22.5 cm. a. write an equation for the proportional relationship between the scale drawings lengths x in centimeters and the actual courts lengths y in centimeters. b. what are the length and width in meters of the actual court? show your work. c. write the ratio of the area of the actual court to the area of the court in the scale drawing.

Explanation:

Step1: Recall scale - proportion formula

The scale is 1 cm:22.5 cm. For a proportional relationship between the scale - drawing length $x$ (in cm) and the actual length $y$ (in cm), the formula is $\frac{x}{y}=\frac{1}{22.5}$, or $y = 22.5x$.

Step2: Find actual length and width

Let's assume the length and width of the scale - drawing are $l_{s}$ and $w_{s}$ (in cm), and the actual length and width are $l_{a}$ and $w_{a}$ (in cm). Given the scale 1 cm:22.5 cm. If the length of the scale - drawing $l_{s}=40$ cm and the width is not given, but we know the general formula for actual length $l_{a}=22.5l_{s}$ and actual width $w_{a}=22.5w_{s}$.

Step3: Recall area formula and ratio

The area of the scale - drawing $A_{s}=l_{s}\times w_{s}$ and the area of the actual court $A_{a}=l_{a}\times w_{a}=(22.5l_{s})\times(22.5w_{s}) = 22.5^{2}\times l_{s}\times w_{s}$. So the ratio of the area of the actual court to the area of the scale - drawing is $\frac{A_{a}}{A_{s}}=22.5^{2}=\frac{22.5\times22.5}{1}=\frac{506.25}{1}$.

Answer:

A. The equation for the proportional relationship is $y = 22.5x$.
B. Without knowing the width of the scale - drawing, we can't find the exact length and width values. But if the length of the scale - drawing $l_{s}=40$ cm, the actual length $l_{a}=22.5\times40 = 900$ cm.
C. The ratio of the area of the actual court to the area of the scale - drawing is 506.25:1.