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Geometry - Similarity and Congruence

Grade 12A Level

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

🔑Concepts

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Two shapes are congruent if they are identical in shape and size. This means all corresponding sides are equal (SSSSSS, SASSAS, ASAASA) and all corresponding angles are equal. Congruence is a special case of similarity where the scale factor k=1k = 1.

Two congruent triangles ABC and PQR
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Two shapes are similar if they are the same shape but different sizes. All corresponding angles are equal, and the lengths of corresponding sides are proportional by a scale factor kk. For any two similar figures, the ratio of their lengths is kk, the ratio of their areas is k2k^2, and the ratio of their volumes is k3k^3.

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Similarity in triangles occurs if: 1. Two angles are the same (AAAA), 2. Three sides are in the same proportion (SSSSSS similarity), or 3. Two sides are in proportion and the included angle is equal (SASSAS similarity). Parallel lines often create similar triangles through corresponding or alternate angles.

Triangle ABC with a line DE parallel to BC creating a smaller similar triangle ADE
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For 3D objects, if the linear scale factor from solid A to solid B is kk, then the ratio of their surface areas is SB/SA=k2S_B/S_A = k^2 and the ratio of their volumes is VB/VA=k3V_B/V_A = k^3.

📐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.

Problem 4:

Two similar containers have capacities of 250250 ml and 22 liters. If the height of the smaller container is 1010 cm, calculate the height of the larger container.

Two similar rectangular containers of different sizes

Solution:

  1. Convert volumes to the same units: V1=250V_1 = 250 ml, V2=2000V_2 = 2000 ml.
  2. Find the volume scale factor: k3=V2V1=2000250=8k^3 = \frac{V_2}{V_1} = \frac{2000}{250} = 8.
  3. Find the linear scale factor: k=83=2k = \sqrt[3]{8} = 2.
  4. Calculate the height of the larger container: H2=k×H1=2×10=20H_2 = k \times H_1 = 2 \times 10 = 20 cm.

Explanation:

Since the containers are mathematically similar, the ratio of their volumes is the cube of the ratio of their heights. By finding the cube root of the volume ratio, we find the linear scale factor kk, which is then applied to the known height.

Problem 5:

A map has a scale of 1:500001:50000. A forest on the map has an area of 44 cm2\text{cm}^2. Calculate the actual area of the forest in km2\text{km}^2.

Diagram showing a small square on a map representing a much larger area in reality

Solution:

  1. Linear scale factor k=50000k = 50000.
  2. Area scale factor k2=(50000)2=2,500,000,000k^2 = (50000)^2 = 2,500,000,000.
  3. Actual area in cm2\text{cm}^2: 4×2,500,000,000=10,000,000,0004 \times 2,500,000,000 = 10,000,000,000 cm2\text{cm}^2.
  4. Convert to m2\text{m}^2: 10,000,000,000/10,000=1,000,00010,000,000,000 / 10,000 = 1,000,000 m2\text{m}^2.
  5. Convert to km2\text{km}^2: 1,000,000/1,000,000=11,000,000 / 1,000,000 = 1 km2\text{km}^2.

Explanation:

Map scales are linear ratios. To find the area ratio, we must square the linear ratio. After calculating the actual area in square centimeters, we convert it to square kilometers using the factors (1002)(100^2) for m2\text{m}^2 and (10002)(1000^2) for km2\text{km}^2.