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Geometry - Congruence and Similarity

Grade 9IGCSE

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 shape and size. All corresponding sides and angles are equal.

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Conditions for Triangle Congruence: SSS (Side-Side-Side), SAS (Side-Angle-Side), ASA (Angle-Side-Angle), AAS (Angle-Angle-Side), 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 proportion.

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Conditions for Triangle Similarity: AA (Two angles are equal), SSS (All three pairs of sides are in the same proportion), or SAS (Two pairs of 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 Scale Factor (k2k^2): The ratio of the areas of two similar figures is the square of the linear scale factor.

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Volume Scale Factor (k3k^3): The ratio of the volumes of two similar solids is the cube of the linear scale factor.

📐Formulae

Linear Scale Factor (k)=Length2Length1\text{Linear Scale Factor } (k) = \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

For similar triangles: ABDE=BCEF=ACDF\text{For similar triangles: } \frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF}

💡Examples

Problem 1:

Triangle ABC is similar to Triangle DEF. In △ABC\triangle ABC, AB=4AB = 4 cm and the area is 2020 cm². In △DEF\triangle DEF, the corresponding side DE=12DE = 12 cm. Find the area of △DEF\triangle DEF.

Solution:

180 cm²

Explanation:

First, find the linear scale factor k=DEAB=124=3k = \frac{DE}{AB} = \frac{12}{4} = 3. The area scale factor is k2=32=9k^2 = 3^2 = 9. Therefore, Area of △DEF=9×Area of △ABC=9×20=180\text{Area of } \triangle DEF = 9 \times \text{Area of } \triangle ABC = 9 \times 20 = 180 cm².

Problem 2:

Two similar cylinders have heights of 5 cm and 10 cm. If the volume of the smaller cylinder is 50 cm³, find the volume of the larger cylinder.

Solution:

400 cm³

Explanation:

The linear scale factor k=105=2k = \frac{10}{5} = 2. The volume scale factor is k3=23=8k^3 = 2^3 = 8. The volume of the larger cylinder is 50×8=40050 \times 8 = 400 cm³.

Problem 3:

In △PQR\triangle PQR, a line STST is drawn parallel to QRQR such that SS is on PQPQ and TT is on PRPR. If PS=2PS = 2 cm, SQ=4SQ = 4 cm, and ST=3ST = 3 cm, find the length of QRQR.

Solution:

9 cm

Explanation:

Because ST∥QRST \parallel QR, △PST\triangle PST is similar to △PQR\triangle PQR (AA criterion). The length of PQ=PS+SQ=2+4=6PQ = PS + SQ = 2 + 4 = 6 cm. The scale factor k=PQPS=62=3k = \frac{PQ}{PS} = \frac{6}{2} = 3. Therefore, QR=k×ST=3×3=9QR = k \times ST = 3 \times 3 = 9 cm.