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Geometry - Similarity and congruence

Grade 11A Level

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

🔑Concepts

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Congruence exists when two shapes are identical in size and shape. There are four main criteria for triangle congruence: SSS (Side-Side-Side), SAS (Side-Angle-Side), ASA (Angle-Side-Angle), and RHS (Right-angle, Hypotenuse, Side).

Two congruent triangles ABC and DEF showing identical dimensions.
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Similarity occurs when one shape is an enlargement of another. All corresponding angles are equal, and all corresponding sides are in the same ratio kk. For triangles, the AA (Angle-Angle) condition is sufficient to prove similarity.

Two similar right-angled triangles with sides in ratio 1:2.
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The relationship between length scale factor kk, area scale factor, and volume scale factor is power-based. If lengths are multiplied by kk, areas are multiplied by k2k^2 and volumes are multiplied by k3k^3.

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Corresponding sides in similar triangles are located opposite to the same angles. It is essential to identify the correct orientation (e.g., in overlapping triangles with parallel lines) to set up the correct ratios.

📐Formulae

Linear Scale Factor: k=Length2Length1k = \frac{\text{Length}_2}{\text{Length}_1}

Area Ratio: Area2Area1=k2=(l2l1)2\frac{\text{Area}_2}{\text{Area}_1} = k^2 = \left(\frac{l_2}{l_1}\right)^2

Volume Ratio: Volume2Volume1=k3=(l2l1)3\frac{\text{Volume}_2}{\text{Volume}_1} = k^3 = \left(\frac{l_2}{l_1}\right)^3

Side Proportionality: aA=bB=cC\frac{a}{A} = \frac{b}{B} = \frac{c}{C}

💡Examples

Problem 1:

Two mathematically similar cylinders have heights of 4 cm and 12 cm. If the smaller cylinder has a surface area of 50 cm250\text{ cm}^2, find the surface area of the larger cylinder.

Solution:

450 cm2450\text{ cm}^2

Explanation:

First, find the linear scale factor k=124=3k = \frac{12}{4} = 3. Since we are dealing with area, use the area scale factor k2=32=9k^2 = 3^2 = 9. The area of the larger cylinder is 50×9=450 cm250 \times 9 = 450\text{ cm}^2.

Problem 2:

Triangle ABC is similar to Triangle DEF. AB = 5 cm and DE = 15 cm. If the volume of a prism with cross-section ABC is 20 cm320\text{ cm}^3, what is the volume of a similar prism with cross-section DEF?

Solution:

540 cm3540\text{ cm}^3

Explanation:

The linear scale factor k=155=3k = \frac{15}{5} = 3. For volume, the scale factor is k3=33=27k^3 = 3^3 = 27. The volume of the larger prism is 20×27=540 cm320 \times 27 = 540\text{ cm}^3.

Problem 3:

In triangle ABC, a line XY is drawn parallel to BC such that X is on AB and Y is on AC. If AX = 3 cm, XB = 6 cm, and XY = 4 cm, calculate the length of BC.

Solution:

12 cm12\text{ cm}

Explanation:

Triangles AXY and ABC are similar because XY is parallel to BC (corresponding angles are equal). The length of AB is AX+XB=3+6=9 cmAX + XB = 3 + 6 = 9\text{ cm}. The linear scale factor k=ABAX=93=3k = \frac{AB}{AX} = \frac{9}{3} = 3. Therefore, BC=XY×3=4×3=12 cmBC = XY \times 3 = 4 \times 3 = 12\text{ cm}.

Problem 4:

Two similar cones have base radii of 5 cm5\text{ cm} and 10 cm10\text{ cm} respectively. If the smaller cone has a volume of 120 cm3120\text{ cm}^3, calculate the volume of the larger cone.

Two similar cones with radii 5cm and 10cm.

Solution:

  1. Find the linear scale factor kk: k=r2r1=105=2k = \frac{r_2}{r_1} = \frac{10}{5} = 2
  2. Determine the volume scale factor: Volume ratio=k3=23=8\text{Volume ratio} = k^3 = 2^3 = 8
  3. Calculate the volume of the larger cone: Volume2=8×120=960 cm3\text{Volume}_2 = 8 \times 120 = 960\text{ cm}^3

Explanation:

Since the cones are similar, the ratio of their volumes is the cube of the ratio of their corresponding linear dimensions (radii).

Problem 5:

In the diagram, DEDE is parallel to BCBC. If AD=4 cmAD = 4\text{ cm}, DB=6 cmDB = 6\text{ cm}, and the area of triangle ADEADE is 16 cm216\text{ cm}^2, find the area of the trapezium DBCEDBCE.

Triangle ABC with parallel line DE creating similar triangle ADE.

Solution:

  1. Identify similar triangles: △ADE\triangle ADE is similar to △ABC\triangle ABC.
  2. Find the linear scale factor kk between △ADE\triangle ADE and △ABC\triangle ABC: AB=AD+DB=4+6=10 cmAB = AD + DB = 4 + 6 = 10\text{ cm} k=ABAD=104=2.5k = \frac{AB}{AD} = \frac{10}{4} = 2.5
  3. Find the area of △ABC\triangle ABC: AreaABC=k2×AreaADE=2.52×16=6.25×16=100 cm2\text{Area}_{ABC} = k^2 \times \text{Area}_{ADE} = 2.5^2 \times 16 = 6.25 \times 16 = 100\text{ cm}^2
  4. Calculate the area of trapezium DBCEDBCE: AreaDBCE=AreaABC−AreaADE=100−16=84 cm2\text{Area}_{DBCE} = \text{Area}_{ABC} - \text{Area}_{ADE} = 100 - 16 = 84\text{ cm}^2

Explanation:

Because DE∥BCDE \parallel BC, △ADE\triangle ADE and △ABC\triangle ABC are similar. The area of the larger triangle is k2k^2 times the smaller. The trapezium is the difference between the two triangles.