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Triangles - Criteria for Similarity of Triangles

Grade 10CBSE

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

πŸ”‘Concepts

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Definition of Similarity: Two triangles are said to be similar if their corresponding angles are equal and their corresponding sides are in the same ratio (proportional). Visually, if you imagine Ξ”ABC\Delta ABC and Ξ”PQR\Delta PQR, they have the same shape but different sizes, like a photograph and its enlargement.

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The AA (Angle-Angle) Similarity Criterion: If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar. Visually, even without knowing the side lengths, if the 'slant' of two sides is identical relative to the base, the third angle must be equal, making the triangles similar by shape.

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The SSS (Side-Side-Side) Similarity Criterion: If the corresponding sides of two triangles are proportional, then their corresponding angles are equal, and hence the triangles are similar. For example, if one triangle has sides 3,4,53, 4, 5 and another has 6,8,106, 8, 10, the ratio of all corresponding sides is 1:21:2, confirming they are similar.

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The SAS (Side-Angle-Side) Similarity Criterion: If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are proportional, the triangles are similar. Visually, this looks like a 'hinge' where the angle is fixed, and the two 'arms' are stretched or shrunk by the same scale factor.

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Basic Proportionality Theorem (BPT) or Thales Theorem: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. Visually, in Ξ”ABC\Delta ABC, if a line DEDE is drawn parallel to BCBC, it creates a smaller 'nested' triangle Ξ”ADE\Delta ADE at the top that is similar to the whole triangle.

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Converse of Basic Proportionality Theorem: If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side. This is used to prove that a segment is 'level' or parallel to the base.

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Properties of Similar Triangles: When two triangles are similar, the ratio of any two corresponding segments (like altitudes, medians, or angle bisectors) is equal to the ratio of their corresponding sides. Furthermore, the ratio of their perimeters is also equal to the ratio of their corresponding sides.

πŸ“Formulae

If Ξ”ABCβˆΌΞ”PQR\Delta ABC \sim \Delta PQR, then ∠A=∠P,∠B=∠Q,∠C=∠R\angle A = \angle P, \angle B = \angle Q, \angle C = \angle R

Proportionality Ratio: fracABPQ=fracBCQR=fracACPR\\frac{AB}{PQ} = \\frac{BC}{QR} = \\frac{AC}{PR}

Basic Proportionality Theorem: If DEparallelBCDE \\parallel BC, then fracADDB=fracAEEC\\frac{AD}{DB} = \\frac{AE}{EC}

Corollary of BPT: fracADAB=fracAEAC=fracDEBC\\frac{AD}{AB} = \\frac{AE}{AC} = \\frac{DE}{BC}

Ratio of Perimeters: fractextPerimeterofDeltaABCtextPerimeterofDeltaPQR=fracABPQ\\frac{\\text{Perimeter of } \\Delta ABC}{\\text{Perimeter of } \\Delta PQR} = \\frac{AB}{PQ}

πŸ’‘Examples

Problem 1:

In DeltaABC\\Delta ABC, DD and EE are points on sides ABAB and ACAC respectively such that DEparallelBCDE \\parallel BC. If AD=xAD = x, DB=xβˆ’2DB = x - 2, AE=x+2AE = x + 2, and EC=xβˆ’1EC = x - 1, find the value of xx.

Solution:

  1. By the Basic Proportionality Theorem, since DEparallelBCDE \\parallel BC, we have: \nfracADDB=fracAEEC\\frac{AD}{DB} = \\frac{AE}{EC}\n2. Substitute the given values: \nfracxxβˆ’2=fracx+2xβˆ’1\\frac{x}{x - 2} = \\frac{x + 2}{x - 1}\n3. Cross-multiply to solve for xx: \nx(xβˆ’1)=(x+2)(xβˆ’2)x(x - 1) = (x + 2)(x - 2)\nx2βˆ’x=x2βˆ’4x^2 - x = x^2 - 4\n4. Simplify the equation: \nβˆ’x=βˆ’4impliesx=4-x = -4 \\implies x = 4\nThus, the value of xx is 44.

Explanation:

This problem applies the Basic Proportionality Theorem (BPT). When a line is parallel to one side of a triangle, it divides the other two sides proportionally. We set up a ratio, cross-multiplied, and solved the resulting quadratic-style equation which simplified to a linear one.

Problem 2:

A vertical pole of length 6m6 m casts a shadow 4m4 m long on the ground and at the same time a tower casts a shadow 28m28 m long. Find the height of the tower.

Solution:

  1. Let ABAB be the pole and BCBC be its shadow. Let PQPQ be the tower and QRQR be its shadow. \n2. At the same time, the angle of elevation of the sun is the same for both. Therefore, angleACB=anglePRQ\\angle ACB = \\angle PRQ. \n3. Also, both the pole and the tower are vertical, so \\angle ABC = \\angle PQR = 90^\\circ. \n4. By AA similarity criterion, DeltaABCsimDeltaPQR\\Delta ABC \\sim \\Delta PQR. \n5. Therefore, the ratios of corresponding sides are equal: \nfracABPQ=fracBCQR\\frac{AB}{PQ} = \\frac{BC}{QR}\n6. Substitute the known values (AB=6,BC=4,QR=28AB=6, BC=4, QR=28): \nfrac6PQ=frac428\\frac{6}{PQ} = \\frac{4}{28}\n7. Solve for PQPQ: \n4timesPQ=6times284 \\times PQ = 6 \\times 28\nPQ=frac1684=42mPQ = \\frac{168}{4} = 42 m

Explanation:

This is a real-world application of AA Similarity. Since the sun's rays hit the earth at the same angle for both objects at the same time, the triangles formed by the objects and their shadows are similar. This allows us to use side proportions to find the unknown height.