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Geometry - Similarity and Congruence

Grade 12IGCSE

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

🔑Concepts

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Congruence: Two shapes are identical in shape and size. All corresponding sides and angles are equal.

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Congruence Criteria for Triangles: SSS (Side-Side-Side), SAS (Side-Angle-Side), ASA (Angle-Side-Angle), and RHS (Right-angle-Hypotenuse-Side).

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Similarity: Two shapes have the same shape but different sizes. Corresponding angles are equal, and corresponding sides are in the same ratio.

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Similarity Criteria for Triangles: AA (two angles are the same), SSS ratio (all sides are proportional), and SAS ratio (two sides proportional and the included angle is equal).

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Linear Scale Factor (k): The ratio of any two corresponding lengths in similar figures.

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Area and Volume Scale Factors: If the linear scale factor is k, the area ratio is k² and the volume ratio is k³.

📐Formulae

k=Length2Length1k = \frac{\text{Length}_2}{\text{Length}_1}

Area2Area1=k2=(L2L1)2\frac{\text{Area}_2}{\text{Area}_1} = k^2 = \left(\frac{L_2}{L_1}\right)^2

Volume2Volume1=k3=(L2L1)3\frac{\text{Volume}_2}{\text{Volume}_1} = k^3 = \left(\frac{L_2}{L_1}\right)^3

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

💡Examples

Problem 1:

Two mathematically similar cylinders have heights of 5 cm and 10 cm. If the surface area of the smaller cylinder is 40 cm², find the surface area of the larger cylinder.

Solution:

k=105=2k = \frac{10}{5} = 2. Area ratio=k2=22=4\text{Area ratio} = k^2 = 2^2 = 4. New Area=40×4=160 cm2\text{New Area} = 40 \times 4 = 160\text{ cm}^2.

Explanation:

First, find the linear scale factor (k) by dividing the corresponding heights. Since we are looking for area, square the scale factor to find the area scale factor, then multiply the original area by this factor.

Problem 2:

In triangle ABC, a line DE is drawn parallel to BC such that D is on AB and E is on AC. If AD = 3cm, DB = 6cm, and BC = 12cm, find the length of DE.

Solution:

△ADE∼△ABC\triangle ADE \sim \triangle ABC. AB=AD+DB=3+6=9 cmAB = AD + DB = 3 + 6 = 9\text{ cm}. k=ADAB=39=13k = \frac{AD}{AB} = \frac{3}{9} = \frac{1}{3}. DE=BC×k=12×13=4 cmDE = BC \times k = 12 \times \frac{1}{3} = 4\text{ cm}.

Explanation:

Because DE is parallel to BC, ∠ADE=∠ABC\angle ADE = \angle ABC and ∠AED=∠ACB\angle AED = \angle ACB (corresponding angles). By AA criteria, the triangles are similar. We use the ratio of the small side to the full side of the large triangle to find the scale factor.

Problem 3:

Two similar solid spheres have volumes in the ratio 27:64. If the radius of the larger sphere is 20 cm, calculate the radius of the smaller sphere.

Solution:

k3=2764  ⟹  k=27643=34k^3 = \frac{27}{64} \implies k = \sqrt[3]{\frac{27}{64}} = \frac{3}{4}. r=20×34=15 cmr = 20 \times \frac{3}{4} = 15\text{ cm}.

Explanation:

The volume ratio is the cube of the linear scale factor. Take the cube root of the volume ratio to find the linear scale factor (k). Multiply the larger radius by k to find the smaller radius.