Similarity — Exercise 15(D)

A model of a vehicle is made to a scale of 1 : 25. If length of the vehicle is 1.6 m; the length of its model is :

1. 6.4 cm

2. 0.64 m

3. 40 cm

4. 15.625 m

Answer

Given,

Scale factor (k) = 1 : 25 or $\dfrac{1}{25}$

By formula,

⇒ Length of model = k × Length of vehicle

⇒ Length of model = $\dfrac{1}{25} \times 1.6$ = 0.064 m = 0.064 × 100 = 6.4 cm.

Hence, Option 1 is the correct option.

The length and the breadth of the model of a rectangular object are 12 cm and 7 cm respectively. If the scale factor is 1 : 20, the actual length and breadth of object are :

1. 1 m & 20 m

2. 2.4 m & 1.4 m

3. 12 cm & 140 cm

4. 1.2 m & 0.7 m

Answer

Given,

Scale factor (k) = 1 : 20 or $\dfrac{1}{20}$

By formula,

⇒ Length of model = k × Length of object

⇒ Length of object = $\dfrac{1}{k}$ × Length of model

⇒ Length of object = $\dfrac{1}{\dfrac{1}{20}} \times 12$ = 20 × 12 = 240 cm = $\dfrac{240}{100}$ = 2.4 m.

By formula,

⇒ Breadth of model = k × Breadth of object

⇒ Breadth of object = $\dfrac{1}{k}$ × Breadth of model

⇒ Breadth of object = $\dfrac{1}{\dfrac{1}{20}} \times 7$ = 20 × 7 = 140 cm = $\dfrac{140}{100}$ = 1.4 m.

Hence, Option 2 is the correct option.

The length of a boat is 2.4 m and the length of its model is 16 cm. The scale factor is :

1. 15 : 1

2. 2 : 3

3. 1 : 15

4. 3 : 2

Answer

Let scale factor be k.

Given,

Length of model = 16 cm

Length of boat = 2.4 m = 2.4 × 100 = 240 cm.

By formula,

⇒ Length of model = Scale factor × Length of boat

Substituting values we get :

⇒ 16 = k × 240

⇒ k = $\dfrac{16}{240} = \dfrac{1}{15}$ = 1 : 15.

Hence, Option 3 is the correct option.

The floor area of a room is 96 m² and area of the floor in the model of the room is 96 cm². The scale factor is :

1. 10 : 1

2. 100 : 1

3. 1 : 10

4. 1 : 100

Answer

Let scale factor be k.

Given,

Floor area of room = 96 m²

Floor area of model = 96 cm² = $\dfrac{96}{100^2} = \dfrac{96}{10000}$ = 96 × 10^-4 m²

By formula,

Floor area of model = (Scale factor)² × Floor area of room

Substituting values we get :

⇒ 96 × 10^-4 = k² × 96

⇒ k² = $\dfrac{96 \times 10^{-4}}{96}$

⇒ k² = 10^-4

⇒ k = $\sqrt{10^{-4}}$

⇒ k = 10^-2 = $\dfrac{1}{10^2} = \dfrac{1}{100}$ = 1 : 100.

Hence, Option 4 is the correct option.

A and B are two similar containers with capacities 216 m³ and 1000 m³ respectively. The scale factor of B with respect to A is :

1. 125 : 216

2. 5 : 3

3. 3 : 5

4. 216 : 125

Answer

Let scale factor of B with respect to A be k.

Given,

A and B are two similar containers with capacities 216 m³ and 1000 m³ respectively.

Volume of container B = k³ × Volume of container A

⇒ 1000 = k³ × 216

⇒ k³ = $\dfrac{1000}{216}$

⇒ k³ = $\Big(\dfrac{10}{6}\Big)^3$

⇒ k = $\dfrac{10}{6} = \dfrac{5}{3}$ = 5 : 3.

Hence, Option 2 is the correct option.

A triangle ABC has been enlarged by scale factor m = 2.5 to the triangle A' B' C'. Calculate :

(i) the length of AB, if A'B' = 6 cm.

(ii) the length of C'A' if CA = 4 cm.

Answer

Given that, ΔABC has been enlarged by scale factor m of 2.5 to ΔA'B'C'.

(i)

∴ A'B' = m × AB

⇒ 6 = 2.5 × AB

⇒ AB = $\dfrac{6}{2.5} = 2.4$ cm.

Hence, AB = 2.4 cm

(ii)

∴ C'A' = m × CA

⇒ C'A' = 2.5 × 4

⇒ C'A' = 10 cm.

Hence, C'A' = 10 cm.

A triangle LMN has been reduced by scale factor 0.8 to the triangle L'M'N'. Calculate:

(i) the length of M'N', if MN = 8 cm.

(ii) the length of LM, if L'M' = 5.4 cm.

Answer

Given, ΔLMN has been reduced by a scale factor (k) = 0.8 to Δ L'M'N'.

(i)

∴ M'N' = k × MN

⇒ M'N' = 0.8 × 8

⇒ M'N' = 6.4 cm

Hence, M'N' = 6.4 cm.

(ii)

∴ L'M' = k × LM

⇒ 5.4 = 0.8 × LM

⇒ LM = $\dfrac{5.4}{0.8} = 6.75$ cm.

Hence, LM = 6.75 cm.

A model of an aeroplane is made to a scale of 1 : 400. Calculate :

(i) the length, in cm of the model; if the length of the aeroplane is 40 m.

(ii) the length, in m, of the aeroplane, if length of its model is 16 cm.

Answer

(i) Given,

Scale-factor (k) = $\dfrac{1}{400}$

⇒ Length of model = k × Length of aeroplane

⇒ Length of model = $\dfrac{1}{400} \times 40 = \dfrac{1}{10}$ = 0.1 m = 10 cm.

Hence, length of model = 10 cm.

(ii) Given,

Scale-factor (k) = $\dfrac{1}{400}$

⇒ Length of model = k × Length of aeroplane

⇒ 16 = $\dfrac{1}{400}$ × Length of aeroplane

⇒ 16 × 400 = Length of aeroplane

⇒ Length of aeroplane = 6400 cm = 64 m.

Hence, length of aeroplane = 64 m.

The dimensions of the model of a multistorey building are 1.2 m × 75 cm × 2 m. If the scale factor is 1 : 30; find the actual dimensions of the building.

Answer

Dimensions of model = 1.2 m × 75 cm × 2 m = 1.2 m × 0.75 m × 2 m

So,

Length = 1.2 m

Breadth = 0.75 m

Height = 2 m.

$\phantom{\Rightarrow} \dfrac{\text{Length of model}}{\text{Length of building}} = k \\[1em] \Rightarrow \dfrac{1.2}{\text{Length of building}} = \dfrac{1}{30} \\[1em] \Rightarrow \text{Length of building} = 1.2 \times 30 = 36 \text{ m}. \\[2em] \phantom{\Rightarrow} \dfrac{\text{Breadth of model}}{\text{Breadth of building}} = k \\[1em] \Rightarrow \dfrac{0.75}{\text{Breadth of building}} = \dfrac{1}{30} \\[1em] \Rightarrow \text{Breadth of building} = 0.75 \times 30 = 22.5 \text{ m}. \\[2em] \phantom{\Rightarrow} \dfrac{\text{Height of model}}{\text{Height of building}} = k \\[1em] \Rightarrow \dfrac{\text{2}}{\text{Height of building}} = \dfrac{1}{30} \\[1em] \Rightarrow \text{Height of building} = 2 \times 30 = 60 \text{ m}.$

Hence, actual dimension of building = 36 m × 22.5 m × 60 m.

On a map drawn to scale of 1 : 2,50,000; a triangular plot of land has the following measurements : AB = 3 cm, BC = 4 cm and angle ABC = 90°.

Calculate :

(i) the actual length of AB and BC in km

(ii) the area of the plot in sq. km.

Answer

(i) Given,

Scale factor for reduction (k) = $\dfrac{1}{250000}$

⇒ Length of AB in model = k × Length of AB

⇒ 3 = $\dfrac{1}{250000}$ × Length of AB

⇒ Length of AB = 3 × 250000 = 750000 cm = $\dfrac{750000}{100 \times 1000}$ = 7.5 km.

⇒ Length of BC in model = k × Length of BC

⇒ 4 = $\dfrac{1}{250000}$ × Length of BC

⇒ Length of BC = 4 × 250000 = 1000000 cm = $\dfrac{1000000}{100 \times 1000}$ = 10 km.

Hence, actual length of AB = 7.5 km and BC = 10 km.

(ii) Area of plot = $\dfrac{1}{2} \times$ base × height

= $\dfrac{1}{2}$ × BC × AB

= $\dfrac{1}{2}$ × 10 × 7.5

= 37.5 sq. km.

Hence, area of plot = 37.5 sq. km.

A model of a ship is made to a scale 1 : 300.

(i) The length of the model of the ship is 2 m. Calculate the length of the ship.

(ii) The area of the deck of the ship is 180,000 m². Calculate the area of the deck of the model.

(iii) The volume of the model is 6.5 m³. Calculate the volume of the ship.

Answer

(i) Given,

Scale factor (k) = 1 : 300 = $\dfrac{1}{300}$

$\phantom{\Rightarrow} \dfrac{\text{Length of model of ship}}{\text{Length of ship}} = k \\[1em] \Rightarrow \dfrac{2}{\text{Length of ship}} = \dfrac{1}{300} \\[1em] \Rightarrow \text{Length of ship} = 2 \times 300 = 600 \text{ m}.$

Hence, length of ship = 600 m.

(ii)

$\phantom{\Rightarrow} \dfrac{\text{Area of deck of model}}{\text{Area of deck of ship}} = k^2 \\[1em] \Rightarrow \dfrac{\text{Area of deck of model}}{180000} = \dfrac{1}{300} \times \dfrac{1}{300} \\[1em] \Rightarrow \text{Area of deck of model} = 180000 \times \dfrac{1}{90000} = 2 \text{ m}^2.$

Hence, area of deck of model = 2 m².

(iii)

$\Rightarrow \dfrac{\text{Volume of model of ship}}{\text{Volume of ship}} = k^3 \\[1em] \Rightarrow \dfrac{6.5}{\text{Volume of ship}} = \Big(\dfrac{1}{300}\Big)^3 \\[1em] \Rightarrow \text{Volume of ship} = 6.5 \times (300)^3 = 6.5 \times 27000000 \text{ m}^3 \\[1em] = 175500000 \text{ m}^3.$

Hence, volume of ship = 175500000 m³.