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Geometry - Similarity and Congruence

Grade 10IGCSE

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

🔑Concepts

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Congruence: Two shapes are congruent if they are identical in size and shape. Corresponding sides and angles are equal.

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Congruence Criteria (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 are similar if they 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 (Triangles): AA (two angles are equal), SSS (all three sides are proportional), or SAS (two sides are 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 Ratios: In similar figures, the ratio of areas is k2k^2 and the ratio of volumes is k3k^3.

📐Formulae

Linear Scale Factor: k=length2length1k = \frac{\text{length}_2}{\text{length}_1}

Area Ratio: Area2Area1=k2\frac{\text{Area}_2}{\text{Area}_1} = k^2

Volume Ratio: Volume2Volume1=k3\frac{\text{Volume}_2}{\text{Volume}_1} = k^3

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

💡Examples

Problem 1:

Triangle ABC is similar to Triangle DEF. AB = 5 cm and DE = 15 cm. If the area of Triangle ABC is 12 cm², calculate the area of Triangle DEF.

Solution:

108 cm²

Explanation:

First, find the linear scale factor: k=DEAB=155=3k = \frac{DE}{AB} = \frac{15}{5} = 3. Since the shapes are similar, the area scale factor is k2k^2. Therefore, k2=32=9k^2 = 3^2 = 9. Area of DEF = 9×Area of ABC=9×12=1089 \times \text{Area of ABC} = 9 \times 12 = 108 cm².

Problem 2:

Two mathematically similar cylinders have heights of 4 cm and 6 cm. If the volume of the smaller cylinder is 32 cm³, find the volume of the larger cylinder.

Solution:

108 cm³

Explanation:

Find the linear scale factor: k=64=1.5k = \frac{6}{4} = 1.5. The volume scale factor is k3=(1.5)3=3.375k^3 = (1.5)^3 = 3.375. Volume of larger cylinder = 3.375×32=1083.375 \times 32 = 108 cm³.

Problem 3:

In triangle PQR, a line XY is drawn parallel to QR such that X lies on PQ and Y lies on PR. If PX = 3 cm, XQ = 6 cm and XY = 4 cm, find the length of QR.

Solution:

12 cm

Explanation:

Triangles PXY and PQR are similar because they share angle P and have corresponding angles (XY || QR). The length of PQ = PX + XQ = 3 + 6 = 9 cm. The linear scale factor k=PQPX=93=3k = \frac{PQ}{PX} = \frac{9}{3} = 3. Therefore, QR = k×XY=3×4=12k \times XY = 3 \times 4 = 12 cm.