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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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Congruent shapes are identical in size and shape. For triangles, four criteria guarantee 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, known as the linear scale factor kk.

Two similar right-angled triangles where the second triangle is twice the size of the first.
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Parallel lines often create similar triangles through shared angles (AAA property). If a line is parallel to one side of a triangle, the smaller triangle created is similar to the original large triangle.

A triangle with a parallel line segment inside, creating two similar triangles.
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The relationship between length, area, and volume in similar figures follows the power rule: if the linear scale factor is kk, the area scale factor is k2k^2 and the volume scale factor is k3k^3.

📐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.

Problem 4:

In the diagram, XYXY is parallel to BCBC. If AX=4AX = 4 cm, XB=6XB = 6 cm, and XY=5XY = 5 cm, calculate the length of BCBC.

Triangle ABC with parallel line XY. AX=4, XB=6, XY=5.

Solution:

  1. Identify the similar triangles: △AXY∼△ABC\triangle AXY \sim \triangle ABC because XY∥BCXY \parallel BC.
  2. Determine the side lengths of the larger triangle: AB=AX+XB=4+6=10AB = AX + XB = 4 + 6 = 10 cm.
  3. Find the linear scale factor kk: k=ABAX=104=2.5k = \frac{AB}{AX} = \frac{10}{4} = 2.5.
  4. Calculate BCBC: BC=k×XY=2.5×5=12.5BC = k \times XY = 2.5 \times 5 = 12.5 cm.

Explanation:

Since the lines are parallel, corresponding angles are equal, making the triangles similar. We use the ratio of the full side ABAB to the small side AXAX to find the scale factor.

Problem 5:

Two mathematically similar solid cones have surface areas of 32π32 \pi cm2^2 and 72π72 \pi cm2^2. If the smaller cone has a volume of 6464 cm3^3, find the volume of the larger cone.

Two similar cones of different sizes with their surface areas labeled.

Solution:

  1. Find the area scale factor: Area2Area1=72π32π=94\frac{Area_2}{Area_1} = \frac{72\pi}{32\pi} = \frac{9}{4}.
  2. Find the linear scale factor kk: k=94=32=1.5k = \sqrt{\frac{9}{4}} = \frac{3}{2} = 1.5.
  3. Find the volume scale factor: k3=(1.5)3=3.375k^3 = (1.5)^3 = 3.375.
  4. Calculate the volume of the larger cone: V2=64×3.375=216V_2 = 64 \times 3.375 = 216 cm3^3.

Explanation:

To move from area to volume, you must first find the linear scale factor by taking the square root of the area ratio, then cube that linear factor to find the volume ratio.

Congruence and Similarity Grade 9 Notes & Examples