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\(^{2}\)









9th Grade Math

From Problems on Size Transformation to HOME PAGE


New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.



Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.