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Geometry and Measurement - Enlargements and Scale Factors

Grade 8IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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An enlargement is a transformation that changes the size of an object while preserving its shape. It is defined by a center of enlargement and a scale factor kk. If k>1k > 1, the object expands; if 0<k<10 < k < 1, the object shrinks (reduction).

A diagram showing a small triangle being enlarged from a center point C by a scale factor of 3.
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Negative scale factors (k<0k < 0) result in an image that is inverted and positioned on the opposite side of the center of enlargement. The distance from the center is still multiplied by ∣k∣|k|.

A diagram showing a rectangle reflected through a center point C representing a scale factor of -1.
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When an object is enlarged by scale factor kk, its linear dimensions change by kk, its area changes by k2k^2, and its volume changes by k3k^3.

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Center of enlargement from coordinates: If the center is (0,0)(0,0), the image of (x,y)(x, y) is (kx,ky)(kx, ky). If the center is (a,b)(a, b), the image is (a+k(x−a),b+k(y−b))(a + k(x-a), b + k(y-b)).

📐Formulae

k=Length of image sideLength of original sidek = \frac{\text{Length of image side}}{\text{Length of original side}}

Distance from center to image point=k×Distance from center to object point\text{Distance from center to image point} = k \times \text{Distance from center to object point}

Area of Image=k2×Area of Object\text{Area of Image} = k^2 \times \text{Area of Object}

Volume of Image=k3×Volume of Object\text{Volume of Image} = k^3 \times \text{Volume of Object}

New Coordinates (x′,y′)=(k×x,k×y) (for enlargement from the origin (0,0))\text{New Coordinates } (x', y') = (k \times x, k \times y) \text{ (for enlargement from the origin } (0,0))

💡Examples

Problem 1:

A triangle with a base of 5 cm5\text{ cm} is enlarged to a similar triangle with a base of 15 cm15\text{ cm}. Calculate the scale factor kk.

Solution:

k=155=3k = \frac{15}{5} = 3

Explanation:

The scale factor is found by dividing the length of the image side by the corresponding length of the original object side.

Problem 2:

A square has an area of 10 cm210\text{ cm}^2. It is enlarged by a scale factor of k=4k = 4. What is the area of the new square?

Solution:

Area of Image=42×10=16×10=160 cm2\text{Area of Image} = 4^2 \times 10 = 16 \times 10 = 160\text{ cm}^2

Explanation:

When a shape is enlarged, its area increases by the square of the scale factor (k2k^2).

Problem 3:

A point A(2,5)A(2, 5) is enlarged from the center of enlargement at the origin (0,0)(0, 0) with a scale factor of k=−2k = -2. Find the coordinates of the image A′A'.

Solution:

x′=−2×2=−4x' = -2 \times 2 = -4 y′=−2×5=−10y' = -2 \times 5 = -10 A′=(−4,−10)A' = (-4, -10)

Explanation:

To find the new coordinates when enlarging from the origin, multiply both the xx and yy coordinates of the original point by the scale factor.

Problem 4:

A model car has a volume of 50 cm350\text{ cm}^3. The actual car is a 1:201:20 enlargement of the model. Calculate the volume of the actual car in cubic meters.

Solution:

k=20k = 20 Volume=203×50=8000×50=400,000 cm3\text{Volume} = 20^3 \times 50 = 8000 \times 50 = 400,000\text{ cm}^3 In m3:400,0001,000,000=0.4 m3\text{In } m^3: \frac{400,000}{1,000,000} = 0.4\text{ m}^3

Explanation:

The volume changes by the cube of the scale factor (k3k^3). To convert cm3cm^3 to m3m^3, we divide by 1003100^3 (1,000,0001,000,000).

Problem 5:

A rectangle with vertices P(2,2)P(2, 2), Q(4,2)Q(4, 2), R(4,3)R(4, 3), and S(2,3)S(2, 3) is enlarged with a scale factor k=2.5k = 2.5 from the origin (0,0)(0,0). Determine the coordinates of the enlarged rectangle P′Q′R′S′P'Q'R'S' and calculate the ratio of the area of the image to the area of the object.

A coordinate plane showing a small rectangle PQRS and its enlargement P'Q'R'S' by a factor of 2.5.

Solution:

  1. Multiply each coordinate by k=2.5k = 2.5: P′=(2×2.5,2×2.5)=(5,5)P' = (2 \times 2.5, 2 \times 2.5) = (5, 5) Q′=(4×2.5,2×2.5)=(10,5)Q' = (4 \times 2.5, 2 \times 2.5) = (10, 5) R′=(4×2.5,3×2.5)=(10,7.5)R' = (4 \times 2.5, 3 \times 2.5) = (10, 7.5) S′=(2×2.5,3×2.5)=(5,7.5)S' = (2 \times 2.5, 3 \times 2.5) = (5, 7.5)

  2. The ratio of areas is k2k^2: Ratio=2.52=6.25\text{Ratio} = 2.5^2 = 6.25

Explanation:

Since the enlargement is from the origin, we apply the scale factor directly to the coordinates. The area scale factor is always the square of the linear scale factor.

Problem 6:

A cylindrical water tank has a radius of 2 m2\text{ m} and a height of 5 m5\text{ m}. A geometrically similar larger tank is built using a scale factor of k=1.5k = 1.5. Calculate the volume of the larger tank.

Two cylinders side-by-side, the second being 1.5 times larger in every dimension than the first.

Solution:

  1. Calculate the volume of the original tank (V1V_1): V1=πr2h=π×22×5=20π≈62.83 m3V_1 = \pi r^2 h = \pi \times 2^2 \times 5 = 20\pi \approx 62.83\text{ m}^3

  2. Use the volume scale factor k3k^3 to find the new volume (V2V_2): V2=V1×k3V_2 = V_1 \times k^3 V2=20π×(1.5)3V_2 = 20\pi \times (1.5)^3 V2=20π×3.375=67.5π≈212.06 m3V_2 = 20\pi \times 3.375 = 67.5\pi \approx 212.06\text{ m}^3

Explanation:

When all linear dimensions (radius and height) are increased by kk, the total volume increases by k3k^3. Multiplying the initial volume by 1.531.5^3 gives the volume of the enlarged tank.