1. **State the problem:** We have a floor plan with a scale of 2 cm : 5 m. We need to find: a) Actual length and width of room 1. b) Width of the front door on the floor plan if actual width is 0.8 m. c) Actual area of the floor of the house to the nearest square metre. 2. **Understand the scale:** The scale means 2 cm on the plan corresponds to 5 m in reality. This can be written as: $$\frac{2 \text{ cm}}{5 \text{ m}} = \frac{1 \text{ cm}}{x \text{ m}}$$ Solving for $x$: $$x = \frac{5}{2} = 2.5 \text{ m/cm}$$ So, every 1 cm on the plan equals 2.5 m actual length. 3. **a) Actual length and width of room 1:** - Measure length and width of room 1 on the floor plan in cm (to nearest mm). - Suppose length on plan is $L_p$ cm and width is $W_p$ cm. - Actual length $L_a = L_p \times 2.5$ m. - Actual width $W_a = W_p \times 2.5$ m. 4. **b) Width of front door on floor plan:** - Actual width of door $D_a = 0.8$ m. - Using scale, width on plan $D_p$ cm satisfies: $$\frac{D_p \text{ cm}}{0.8 \text{ m}} = \frac{2 \text{ cm}}{5 \text{ m}}$$ - Cross multiply: $$5 D_p = 2 \times 0.8$$ $$5 D_p = 1.6$$ - Divide both sides by 5: $$D_p = \frac{1.6}{5} = \cancel{\frac{1.6}{5}} = 0.32 \text{ cm}$$ 5. **c) Actual area of the floor of the house:** - Measure the area of the house on the floor plan in square cm, call it $A_p$. - Since 1 cm corresponds to 2.5 m, area scale factor is: $$ (2.5)^2 = 6.25 \text{ m}^2/\text{cm}^2 $$ - Actual area $A_a = A_p \times 6.25$ m$^2$. - Round to nearest square metre. **Summary:** - Use ruler to measure lengths and area on plan. - Multiply lengths by 2.5 to get actual meters. - Multiply area by 6.25 to get actual square meters. - Width of front door on plan is 0.32 cm. **Note:** Actual measurements depend on your ruler measurements on the plan.