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Geometry and Trigonometry - Similarity and congruence

Grade 9IB

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 size and shape. There are four criteria for triangle congruence: SSS (Side-Side-Side), SAS (Side-Angle-Side), ASA (Angle-Side-Angle), and RHS (Right angle-Hypotenuse-Side). When triangles are congruent, all corresponding sides and angles are equal.

Two congruent triangles ABC and PQR demonstrating identical shape and size.
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Similarity occurs when two shapes have the same shape but different sizes. One is an enlargement of the other. For triangles, similarity is established if corresponding angles are equal (AA) or if corresponding sides are in the same ratio (a1a2=b1b2=c1c2\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}).

Two similar right-angled triangles where the larger has sides twice the length of the smaller.
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The Linear Scale Factor kk represents the ratio of corresponding side lengths. If the linear scale factor is kk, then the Area Scale Factor is k2k^2 and the Volume Scale Factor is k3k^3. This is vital for solving problems involving similar 2D shapes or 3D solids.

Diagram showing that doubling the length results in four times the area (k=2, k squared = 4).
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In similarity, the ratio of any linear measurement (height, radius, perimeter, median) follows the scale factor kk. For example, if two circles have radii in ratio 1:31:3, their circumferences are also in ratio 1:31:3, but their areas are in ratio 1:91:9.

📐Formulae

k=fractextLengthofImagetextLengthofObjectk = \\frac{\\text{Length of Image}}{\\text{Length of Object}}

fraca1a2=fracb1b2=fracc1c2=k\\frac{a_1}{a_2} = \\frac{b_1}{b_2} = \\frac{c_1}{c_2} = k

fractextArea1textArea2=k2=left(fracl1l2right)2\\frac{\\text{Area}_1}{\\text{Area}_2} = k^2 = \\left(\\frac{l_1}{l_2}\\right)^2

fractextVolume1textVolume2=k3=left(fracl1l2right)3\\frac{\\text{Volume}_1}{\\text{Volume}_2} = k^3 = \\left(\\frac{l_1}{l_2}\\right)^3

💡Examples

Problem 1:

In triangle ABCABC, DEDE is parallel to BCBC. If AD=4AD = 4 cm, DB=6DB = 6 cm, and DE=5DE = 5 cm, find the length of BCBC.

Solution:

triangleADEsimtriangleABC\\triangle ADE \\sim \\triangle ABC because angleA\\angle A is common and angleADE=angleABC\\angle ADE = \\angle ABC (corresponding angles). The scale factor kk is: k=fracABAD=fracAD+DBAD=frac4+64=frac104=2.5k = \\frac{AB}{AD} = \\frac{AD + DB}{AD} = \\frac{4 + 6}{4} = \\frac{10}{4} = 2.5 Now, BC=ktimesDEBC = k \\times DE: BC=2.5times5=12.5textcmBC = 2.5 \\times 5 = 12.5 \\text{ cm}

Explanation:

Since DEparallelBCDE \\parallel BC, triangles ADEADE and ABCABC are similar by AAAA criterion. We find the ratio of the full side ABAB to ADAD to get the scale factor, then multiply the base DEDE by this factor.

Problem 2:

Two similar solid cones have surface areas of 50textcm250 \\text{ cm}^2 and 450textcm2450 \\text{ cm}^2. If the height of the smaller cone is 66 cm, find the height of the larger cone.

Solution:

First, find the area scale factor k2k^2: k2=frac45050=9k^2 = \\frac{450}{50} = 9 Find the linear scale factor kk: k=sqrt9=3k = \\sqrt{9} = 3 The height of the larger cone HH is: H=ktimesh=3times6=18textcmH = k \\times h = 3 \\times 6 = 18 \\text{ cm}

Explanation:

The ratio of areas is the square of the linear scale factor. By taking the square root of the area ratio, we find kk, which can then be applied to the height.

Problem 3:

A model car is built to a scale of 1:201:20. if the volume of the model's petrol tank is 10textcm310 \\text{ cm}^3, calculate the volume of the actual car's tank in liters.

Solution:

The linear scale factor is k=20k = 20. The volume scale factor is: k3=203=8000k^3 = 20^3 = 8000 Actual volume in textcm3\\text{cm}^3: V=10times8000=80000textcm3V = 10 \\times 8000 = 80000 \\text{ cm}^3 Convert to liters (since 1000textcm3=1textliter1000 \\text{ cm}^3 = 1 \\text{ liter}): V=frac800001000=80textlitersV = \\frac{80000}{1000} = 80 \\text{ liters}

Explanation:

Volume scale factor is the cube of the linear scale factor. After calculating the volume in cubic centimeters, we convert it to liters using the standard conversion.

Problem 4:

A cylindrical water tank has a height of 22 m and a capacity of 500500 liters. A similar cylindrical tank has a height of 33 m. Calculate the capacity of the larger tank.

Two cylinders representing water tanks with heights 2m and 3m.

Solution:

k=h2h1=32=1.5k = \frac{h_2}{h_1} = \frac{3}{2} = 1.5 V2V1=k3=(1.5)3\frac{V_2}{V_1} = k^3 = (1.5)^3 V2=500×3.375V_2 = 500 \times 3.375 V2=1687.5 litersV_2 = 1687.5 \text{ liters}

Explanation:

Since the tanks are similar, we first find the linear scale factor kk by dividing the heights. Because volume scales by k3k^3, we cube the scale factor and multiply it by the original volume to find the new capacity.

Problem 5:

In the figure, ABAB is parallel to DEDE. If AC=5AC = 5 cm, CD=10CD = 10 cm, and AB=4AB = 4 cm, find the length of DEDE.

An hourglass shape formed by two triangles with parallel bases AB and DE meeting at vertex C.

Solution:

∠BAC=∠EDC (Alternate angles)\angle BAC = \angle EDC \text{ (Alternate angles)} ∠ABC=∠DEC (Alternate angles)\angle ABC = \angle DEC \text{ (Alternate angles)} △ABC∼△DEC (AA Similarity)\triangle ABC \sim \triangle DEC \text{ (AA Similarity)} DEAB=CDAC\frac{DE}{AB} = \frac{CD}{AC} DE4=105\frac{DE}{4} = \frac{10}{5} DE4=2\frac{DE}{4} = 2 DE=8 cmDE = 8 \text{ cm}

Explanation:

Because AB∥DEAB \parallel DE, alternate interior angles are equal, making the triangles similar. We use the ratio of corresponding sides CDCD and ACAC to find the scale factor, then apply it to ABAB to find DEDE.