Sydney drew a scale diagram of a circular fire pit in the centre of a circular patio with an actual circumference of 15m. The circle containing the fire put is a reduction of the circular patio by a scale factor of1/3.

Question

A.) Determine the diameter of the actual fire pit to the nearest hundredth of a metre.
B.) What is the diameter of the fire pit in the scale diagram if the scale factor used was 2/15?
C.) After construction, the raised fire pit will have a surface area of 6.47m^3. What will be the surface area of a similar fire pit that is enlarged by a scale factor of 3?
D.) After construction, the raised fire put will have a volume of0.99m^3. If the volume of the enlargement is 26.73m^3, what scale factor was used?

Answer 1

Since the actual diameter of the fire pit is 15m and the formula for the circumference is C = πd, substitute the circumference of 15m into the formula and solve for d.
15 = πd and then divide both sides by π getting d = 15/π = 5 rounded to the nearest meter. The scale drawing is smaller so (2/15) times 5 is 10/15 or 2/3 m
If the raised fire pit has a surface area of 6.47 and a similar one larger by a factor of 3 would have a surface area of
6.47 times 3 = 19.41 or 20m2 rounded to the nearest square meter.

For the last part, if the volume of the fire pit is .99m3 and the enlargement is 26.73m3, the scale factor used is

.99x = 26.73 ror x = 27