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Geometry and Measurement - Similarity and Congruence

Grade 8IB

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

🔑Concepts

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Congruent figures are identical in shape and size. Two triangles are congruent if they satisfy one of the four criteria: SSS (Side-Side-Side), SAS (Side-Angle-Side), ASA (Angle-Side-Angle), or RHS (Right angle-Hypotenuse-Side). In congruent shapes, all corresponding sides and angles are equal.

Two identical triangles labeled ABC and PQR illustrating congruence.
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Similar figures have the same shape but different sizes. One figure is an enlargement of the other by a scale factor kk. For similarity, corresponding angles must be equal, and corresponding sides must be in the same ratio. For triangles, common criteria include AA (Angle-Angle), SAS (ratio of two sides and equal included angle), or SSS (ratio of all three sides).

Two right-angled triangles where one is twice the size of the other, illustrating similarity.
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When two shapes are similar with a linear scale factor kk, the ratio of their areas is k2k^2 and the ratio of their volumes is k3k^3. If side lengths double, the area quadruples and the volume increases by eight times.

A small square and a large square demonstrating the area scale factor.
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In similar triangles formed by parallel lines (the 'Ladder' or 'A-frame' theorem), the smaller triangle is similar to the larger triangle because they share a common angle and have corresponding angles created by the parallel lines.

A triangle with a parallel line segment inside creating a smaller similar triangle.

📐Formulae

Scale Factor (k)=Corresponding Side of ImageCorresponding Side of Object\text{Scale Factor } (k) = \frac{\text{Corresponding Side of Image}}{\text{Corresponding Side of Object}}

ad=be=cf=k (for similar triangles with sides a,b,c and d,e,f)\frac{a}{d} = \frac{b}{e} = \frac{c}{f} = k \text{ (for similar triangles with sides } a,b,c \text{ and } d,e,f)

Area1Area2=(side1side2)2=k2\frac{\text{Area}_1}{\text{Area}_2} = \left( \frac{\text{side}_1}{\text{side}_2} \right)^2 = k^2

Volume1Volume2=(side1side2)3=k3\frac{\text{Volume}_1}{\text{Volume}_2} = \left( \frac{\text{side}_1}{\text{side}_2} \right)^3 = k^3

💡Examples

Problem 1:

In ΔABC\Delta ABC and ΔPQR\Delta PQR, ∠A=∠P=50∘\angle A = \angle P = 50^\circ, AB=6 cmAB = 6\text{ cm}, AC=8 cmAC = 8\text{ cm}, PQ=9 cmPQ = 9\text{ cm}, and PR=12 cmPR = 12\text{ cm}. Determine if the triangles are similar and find the scale factor.

Solution:

We check the ratio of the corresponding sides: PQAB=96=1.5\frac{PQ}{AB} = \frac{9}{6} = 1.5 and PRAC=128=1.5\frac{PR}{AC} = \frac{12}{8} = 1.5. Since two sides are in the same proportion and the included angle is equal (∠A=∠P=50∘\angle A = \angle P = 50^\circ), the triangles are similar by SASSAS similarity. The scale factor kk is 1.51.5.

Explanation:

To prove similarity using SASSAS, we must show that the ratio of two pairs of sides is equal and the angle between those sides is identical.

Problem 2:

Two similar rectangles have a scale factor of 3:43:4. If the area of the smaller rectangle is 18 cm218\text{ cm}^2, find the area of the larger rectangle.

Solution:

The scale factor k=43k = \frac{4}{3}. The ratio of the areas is k2k^2: Area ratio=(43)2=169\text{Area ratio} = \left( \frac{4}{3} \right)^2 = \frac{16}{9} Let xx be the area of the larger rectangle: x18=169\frac{x}{18} = \frac{16}{9} x=18×169=2×16=32 cm2x = 18 \times \frac{16}{9} = 2 \times 16 = 32\text{ cm}^2

Explanation:

When objects are similar, the ratio of their areas is the square of the ratio of their corresponding linear dimensions.

Problem 3:

In ΔXYZ\Delta XYZ and ΔLMN\Delta LMN, XY=5 cmXY = 5\text{ cm}, YZ=7 cmYZ = 7\text{ cm}, XZ=10 cmXZ = 10\text{ cm}, LM=5 cmLM = 5\text{ cm}, MN=7 cmMN = 7\text{ cm}, and LN=10 cmLN = 10\text{ cm}. Are the triangles congruent?

Solution:

Comparing the sides of the two triangles: XY=LM=5XY = LM = 5 YZ=MN=7YZ = MN = 7 XZ=LN=10XZ = LN = 10 All three corresponding sides are equal. Therefore, ΔXYZ≅ΔLMN\Delta XYZ \cong \Delta LMN by the SSSSSS congruence criterion.

Explanation:

Congruence requires all corresponding parts to be equal. SSSSSS is one of the standard rules to prove that two triangles are identical in size and shape.

Problem 4:

In the diagram provided, DEDE is parallel to BCBC. Given AD=4 cmAD = 4\text{ cm}, DB=6 cmDB = 6\text{ cm}, and DE=5 cmDE = 5\text{ cm}, calculate the length of BCBC.

Triangle ABC with segment DE parallel to BC. AD=4, DB=6, DE=5.

Solution:

  1. Identify the similar triangles: ΔADE∼ΔABC\Delta ADE \sim \Delta ABC because ∠A\angle A is common and ∠ADE=∠ABC\angle ADE = \angle ABC (corresponding angles).
  2. Find the total length of side ABAB: AB=AD+DB=4+6=10 cmAB = AD + DB = 4 + 6 = 10\text{ cm}
  3. Set up the ratio of corresponding sides: ADAB=DEBC\frac{AD}{AB} = \frac{DE}{BC}
  4. Substitute the known values: 410=5BC\frac{4}{10} = \frac{5}{BC}
  5. Solve for BCBC: 4×BC=504 \times BC = 50 BC=504=12.5 cmBC = \frac{50}{4} = 12.5\text{ cm}

Explanation:

Since the lines are parallel, the triangles are similar by AA criterion. The scale factor is determined by comparing the full side ABAB to the segment ADAD.

Problem 5:

Two similar cylinders have heights of 5 cm5\text{ cm} and 15 cm15\text{ cm}. If the volume of the larger cylinder is 540π cm3540\pi\text{ cm}^3, find the volume of the smaller cylinder.

Two cylinders of different sizes labeled with heights 5 and 15.

Solution:

  1. Find the linear scale factor kk (larger to smaller): k=HeightsmallHeightlarge=515=13k = \frac{\text{Height}_{small}}{\text{Height}_{large}} = \frac{5}{15} = \frac{1}{3}
  2. Find the volume scale factor: Volume ratio=k3=(13)3=127\text{Volume ratio} = k^3 = \left(\frac{1}{3}\right)^3 = \frac{1}{27}
  3. Calculate the volume of the smaller cylinder (VsV_s): Vs=Vl×127V_s = V_l \times \frac{1}{27} Vs=540π×127V_s = 540\pi \times \frac{1}{27} Vs=20π cm3V_s = 20\pi\text{ cm}^3

Explanation:

Similarity in 3D objects requires using the cube of the linear scale factor to relate volumes.