Similarity (As a Size Transformation)

Longest side in Δ ABC (CA) = 6 cm

Longest side in Δ A'B'C' (C'A') = 10 cm

Let scale factor be k.

C'A' = k × CA

k = $\frac{C'A'}{CA} = \frac{10}{6} = \frac{5}{3}$.

∴ The scale factor is, k = $\frac{5}{3}$

Now,

A'B' = k × (AB) = $\frac{5}{3} \times 3.6$ = 6 cm

B'C' = k × (BC) = $\frac{5}{3} \times 4.5$ = 7.5 cm

Hence, scale factor = $\frac{5}{3}$, A'B' = 6 cm and B'C' = 7.5 cm.

Smallest side in Δ ABC (BC) = 12 cm

Smallest side in Δ A'B'C' (B'C') = 4.8 cm

Let scale factor be k.

B'C' = k × BC

k = $\frac{B'C'}{BC} = \frac{4.8}{12} = \frac{2}{5}$

∴ The scale factor is, k = $\frac{2}{5}$.

Now,

⇒ A'B' = k × (AB) = $\frac{2}{5} \times 16$ = 6.4 cm

⇒ A'C' = k × (AC) = $\frac{2}{5} \times 18$ = 7.2 cm

Hence, scale factor = $\frac{2}{5}$, A'B' = 6.4 cm and A'C' = 7.2 cm.

We know that,

Area of image = k² × Area of the actual triangle Δ PQR

= (0.72)² × 62.5

= 0.5184 × 62.5

= 32.4 cm².

Hence, area of image of Δ PQR = 32.4 cm².

Let scale factor be k.

We know that,

Area of the image = k² × (Area of the rectangle)

$$\begin{aligned}\Rightarrow 135 = k^2 \times 60 \\[1em] \Rightarrow k^2 = \frac{135}{60} \\[1em] \Rightarrow k^2 = \frac{9}{4} \\[1em] \Rightarrow k = \sqrt{\frac{9}{4}} \\[1em] \Rightarrow k = \frac{3}{2} = 1.5\end{aligned}$$

Hence, scale factor of enlargement = 1.5

(i) Since, the model of the triangular plot is made to the scale of 1 : 25000.

k = $\frac{1}{25000}$.

Length of MN in model = k × (Actual length of MN)

8 = $\frac{1}{25000}$ × Actual length of MN

Actual length of MN = 8 × 25000

= 200000 cm.

= $\frac{200000}{100000}$ km.

= 2 km.

Length of LM in model = k × (Actual length of LM)

6 = $\frac{1}{25000}$ × Actual length of LM

Actual length of LM = 6 × 25000

= 150000 cm.

= $\frac{150000}{100000}$ km.

= 1.5 km.

Using pythagoras theorem:

LN² = LM² + MN²

LN² = (1.5)² + 2²

LN² = 2.25 + 4

LN² = 6.25

LN = $\sqrt{6.25}$

LN = 2.5 km

Hence, MN = 2 km and LN = 2.5 km.

(ii) We know that,

Area of triangle = $\frac{1}{2}$ × Base × Height

= $\frac{1}{2}$ × LM × MN

= $\frac{1}{2}$ × 1.5 × 2

= 1.5 km².

Hence, actual area of the plot = 1.5 sq.km.

(i) Since, the scale of map is 1 : 200000.

1 cm on map = 200000 cm on ground

200000 cm = $\frac{200000}{100000}$ = 2km

Hence, 1 cm on map represents 2 km on ground.

(ii) Given,

Area of plot on map = 3 cm²

Area of plot on map = k² × Area of actual plot

3 = $\Big(\frac{1}{200000}\Big)^2$ × Area of actual plot

Area of actual plot = 3 × 200000 × 200000 cm²

= $\frac{3 \times 200000 \times 200000}{100000 \times 100000}$ km²

= 12 km^2.

Hence, required area = 12 km².

(iii) Area of plot on map = k² × Area of actual plot

Area of plot on map = $\Big(\frac{1}{200000}\Big)^2 \times 10$ km²

Area of plot on map = $\frac{10 \times 100000 \times 100000}{200000 \times 200000}$ cm²

= 2.5 cm².

Hence, area on the map representing the plot of land is 2.5 cm².

(i) By pythagoras theorem,

⇒ AC² = AB² + BC²

⇒ AC² = 32² + 24²

⇒ AC² = 1024 + 576

⇒ AC² = 1600

⇒ AC = $\sqrt{1600}$ = 40 cm.

Given,

Scale of map = $\frac{1}{20000}$

Length of diagonal distance in map = k × Actual length of diagonal of plot

40 = $\frac{1}{20000}$ × Actual length of diagonal of plot

Actual length of diagonal of plot = 40 × 20000 = 800000 cm

= $\frac{800000}{100000}$ = 8 km.

Hence, diagonal distance of plot = 8 km.

(ii) Area of plot on map = AB × BC = 24 × 32 = 768 cm²

Area of plot on map = k² × Area of actual plot

768 = $\Big(\frac{1}{20000}\Big)^2$ × Area of actual plot

Area of actual plot = 768 × 20000 × 20000 cm²

= $\frac{768 \times 20000 \times 20000}{100000 \times 100000}$ km²

= 30.72 km²

Hence, area of plot = 30.72 km².

(i) Given,

Scale factor (k) = 1 : 60 = $\frac{1}{60}$

Dimensions of model = l × b × h = 1 m × 60 cm × 1.25 m.

= 1 m × 0.60 m × 1.25 m.

We know that,

Model dimensions = k × Actual dimension

$$\begin{aligned}\text{Actual length} = \frac{\text{ Model length}}{k} \\[1em] = \frac{1}{\frac{1}{60}} \\[1em] = 60 \text{ m}. \\[1em] \text{Actual breadth} = \frac{\text{ Model breadth}}{k} \\[1em] = \frac{0.60}{\frac{1}{60}} \\[1em] = 0.60 \times 60\\[1em] = 36 \text{ m}. \\[1em] \text{Actual height} = \frac{\text{ Model height}}{k} \\[1em] = \frac{1.25}{\frac{1}{60}} \\[1em] = 1.25 \times 60 \\[1em] = 75 \text{ m}.\end{aligned}$$

Actual dimensions (in meters) = 60 m × 36 m × 75 m.

By formula,

Floor area of model room = k² × Floor area of building room

250 = $\frac{1}{60} \times \frac{1}{60}$ × Floor area of building room

Floor area of building room = 60 × 60 × 250 cm²

= 900000 cm²

= $\frac{900000}{10000}$ m².

= 90 m².

Hence, length = 60m, breadth = 36 m, height = 75m, floor area of a room of the building = 90 m².

(ii) By formula,

Volume of a room of model = k³ × Volume of a room of building

= $\frac{1}{60} \times \frac{1}{60} \times \frac{1}{60} \times 648$

= $\frac{1}{60} \times \frac{1}{60} \times \frac{1}{60} \times 648$ m³

= 0.003 x 100 x 100 x 100 cm³

= 3000 cm³.

Hence, volume of a room in model = 3000 cm³.

(i) Since, the model of the ship is made to the scale of 1 : 250.

∴ K = $\frac{1}{250}$.

Length of model = k × (Actual length of the ship)

1.2 = $\frac{1}{250}$ × Actual length of the ship

Actual length of the ship = 1.2 × 250 = 300 m.

Hence, the length of the ship is 300 m.

(ii) Area of the deck of the model = k² × (Area of the deck of the ship)

1.6 = $\Big(\frac{1}{250}\Big)^2$ x Area of the deck of the ship

1.6 = $\frac{1}{62500}$ x Area of the deck of the ship

Area of the deck of the ship = 62500 x 1.6 = 100000 m²

Hence, the area of the deck of the ship is 100000 m².

(iii) Given,

Volume of ship = 1 km³ = (1000)³ m³

Let the volume of the model be x.

Volume of the model = k³ × (Volume of the ship)

⇒ x = $\Big(\frac{1}{250}\Big)^3$ × 1000³

⇒ x = $\frac{1000 \times 1000 \times 1000}{250 \times 250 \times 250}$

⇒ x = 4 × 4 × 4

⇒ x = 64 m³.

Hence, the volume of the model of the ship is 64 m³.

(i) Since, the model of the ship is made to the scale of 1 : 200.

∴ K = $\frac{1}{200}$.

Length of model ship = k × (Actual length of the ship)

4 = $\frac{1}{200}$ × Actual length of the ship

Actual length of the ship = 4 × 200 = 800 m.

Hence, the length of the ship is 800 m.

(ii) Area of the deck of the model = k² × (Area of the deck of the ship)

Let area of deck of model be x.

⇒ x = $\Big(\frac{1}{200}\Big)^2$ × (160000)

⇒ x = $\frac{1}{40000}$ × (160000)

⇒ x = 4 m².

Hence, the area of the deck of model is 4 m².

(iii) Given,

Volume of model = 200 litres = $\frac{200}{1000} \text{ m}^3 = 0.2 \text{ m}^3$

Volume of the model = k³ × (Volume of the ship)

0.2 = $\Big(\frac{1}{200}\Big)^3$ × Volume of the ship

0.2 = $\frac{1}{8000000}$ × Volume of the ship

Volume of the ship = 8000000 × 0.2 = 1600000.

Hence, the volume of the ship is 1600000 m³.

Figures are said to be similar if: their shapes are identical,but their sizes may differ.

Hence, option 3 is the correct option.

All regular polygons with the same number of sides are always similar, regardless of their size.

Hence, option 1 is the correct option.

Two circles are always similar because similarity means that the figures have the same shape but possibly different sizes.

Hence, option 2 is the correct option.

The resulting figure, after the transformation is applied, is called the image.

Hence, option 2 is the correct option.

If the transformation is a reduction, the image is smaller than the object, meaning the scale factor k is between 0 and 1: 0 < k < 1.

Hence, option 3 is the correct option.

The length of any side in the resulting figure (the image) is equal to the length of the corresponding side in the given figure (the object) multiplied by the scale factor k.

Hence, option 2 is the correct option.

If k = 1, the resulting figure (image) is congruent to the original figure (object), meaning it has the same size and the same shape. This specific transformation is called an identity transformation because it maps the object onto itself.

Hence, option 1 is the correct option.

Volume of the resulting figure = k³ × (the volume of given figure)

Hence, option 3 is the correct option.

We know that,

If scale factor = k, then :

Area of model = k² × (Area of actual figure)

Hence, option 1 is the correct option.

Given,

Scale = 1 : p = $\frac{1}{p}$.

Hence, option 2 is the correct option.

Scale factor of a picture and the actual height = 20 cm : 1.6 m = 20 cm : 160 cm

$$\begin{aligned}\therefore \frac{\text{Height in picture}}{\text{Actual height}} = \frac{\text{20 cm}}{\text{160 cm}} \\[1em] \Rightarrow \frac{18}{\text{Actual height}} = \frac{\text{20 cm}}{\text{160 cm}} \\[1em] \Rightarrow \text{Actual height} = \frac{18 \times 160}{20} = \text{144 cm} = 1.44 \text{ m}.\end{aligned}$$

Hence, option 4 is the correct option.

Two circles always have the same characteristic shape, so any two circles are always similar.

So, Assertion (A) is true.

Similar shapes have the same shape but not necessarily the same size, so the Reason is incorrect.

So, Reason (R) is false.

Hence, option 3 is the correct option.

If the scale factor k is greater than 1, the resulting image is larger than the object, which is an enlargement, not a reduction.

So, Assertion (A) is false.

A size transformation is a process in which a given figure is reduced (0 < k < 1) or enlarged (k > 1) by a certain scale factor such that the resulting figure is similar to the given figure.

So, Reason (R) is true.

Hence, option 4 is the correct option.

Scale factor (k) = 2 cm to 1 km or 2 cm to 100000 cm = $\frac{2}{100000} = \frac{1}{50000}$

(a) Length of cart track on map = k × Length of cart track on ground

$7.6 = \frac{1}{50000}$ × Length of cart track on ground

Length of cart track on ground = 7.6 × 50000 = 380000 cm = $\frac{380000}{100000}$ = 3.8 km

Hence, length of actual cart track on ground = 3.8 km

(b) Area of model grid square = k² × Area of actual grid square

⇒ 4 = $\Big(\frac{1}{50000}\Big)^2$ × Area of actual grid square

⇒ 4 = $\Big(\frac{1}{25 \times 10^8}\Big)$ × Area of actual grid square

⇒ Area of actual grid square = $4 \times 25 \times 10^8$

⇒ Area of actual grid square = 100 × 10⁸ = 10² × 10⁸ = 10¹⁰ cm²

= $\frac{10^{10}}{10^{10}}$ km² = 1 km².

Hence, actual area of a grid square = 1 km².

(a) Scale factor is greater than 1 as volume of model is greater than the original eye.

(b)

(i) By formula,

$$\begin{aligned}\Rightarrow k^3 = \frac{\text{Volume of model}}{\text{Volume of original human eye}} \\[1em] \Rightarrow k^3 = \frac{1404}{6.5} \\[1em] \Rightarrow k^3 = \frac{216}{1} \\[1em] \Rightarrow k^3 = 6^3 \\[1em] \Rightarrow k = 6.\end{aligned}$$

Hence, scale factor (k) = 6.

(ii) By formula,

$$\begin{aligned}\Rightarrow k = \frac{\text{Radius of model}}{\text{Radius of human eye}} \\[1em] \Rightarrow 6 = \frac{7.2}{\text{Radius of human eye}} \\[1em] \Rightarrow \text{Radius of human eye} = \frac{7.2}{6} = 1.2 \text{ cm}\end{aligned}$$

Diameter of human eye = 1.2 × 2 = 2.4 cm

Hence, diameter of human eye = 2.4 cm.

(iii) By formula,

$$\begin{aligned}\Rightarrow k^2 = \frac{\text{Surface area of model}}{\text{Surface area of human eye}} \\[1em] \Rightarrow 6^2 = \frac{651.6}{\text{Surface area of human eye}} \\[1em] \Rightarrow \text{Surface area of human eye} = \frac{651.6}{6^2} \\[1em] \Rightarrow \text{Surface area of human eye} = \frac{651.6}{36} = 18.1 \text{ cm}^2.\end{aligned}$$

Hence, surface area of human eye = 18.1 cm².