# Problems on Size Transformation

Here we will solve different types of problems on size transformation.

1. A map of a rectangular park is drawn to a scale of 1 : 5000.

(i) Find the actual length of the park if the length of the same in the map is 25 cm.

(ii) If the actual width of the park is 1 km, find its width in the map.

Solution:

(i) Given scale = 1 : 5000

Let the actual length of the park be p

Therefore, $$\frac{\textrm{Length in Map}}{\textrm{Actual Length of the Park}}$$ = $$\frac{1}{5000}$$

⟹ $$\frac{25 cm}{p}$$ = $$\frac{1}{5000}$$

⟹ p = 25 × 5000 cm

⟹ p = 125000 cm

⟹ p = 125000 × 10$$^{-5}$$ km

⟹ p = 1.25 km

(ii) Let the width in the map be q

Therefore, $$\frac{\textrm{Width in Map}}{\textrm{Actual Width}}$$ = $$\frac{1}{5000}$$

⟹ $$\frac{q}{1 km}$$ = $$\frac{1}{5000}$$

⟹ q = $$\frac{1 km}{5000}$$

⟹ q = $$\frac{1 × 10^{5}}{5000}$$ cm

⟹ q = 20 cm

2. The length of a building is 40 m and the length of its model is 50 cm. Find the area of the terrace of the building if the area of the terrace in the model is 1400 cm$$^{2}$$.

Solution:

Let the scale factor = k

Therefore, k = $$\frac{\textrm{Length of the Building}}{\textrm{Length of the Model}}$$

= $$\frac{40 m}{50 cm}$$

= $$\frac{40 × 100 cm}{50 cm}$$

= 80 cm

Let the area of the terrace of the building = a

$$\frac{\textrm{Area of the Terrace of the Building}}{\textrm{Area of the Terrace in the Model}}$$ = k$$^{2}$$

Therefore, $$\frac{a}{1400 cm^{2}}$$ = 80 × 80

⟹ a = 80 × 80 × 1400 cm$$^{2}$$

⟹ a = $$\frac{80 × 80 × 1400}{100 × 100}$$ m$$^{2}$$

⟹ a = 896 m$$^{2}$$