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Triangles - Prove similarity of triangles using AA, SSS, and SAS similarity criteria

Grade 10CBSE

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

🔑Concepts

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The AA (Angle-Angle) Similarity Criterion states that if two angles of one triangle are equal to two angles of another triangle, then the triangles are similar. This implies the third angles must also be equal due to the angle sum property.

Two triangles ABC and PQR showing equal angles at B=Q and C=R for AA similarity.
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The SSS (Side-Side-Side) Similarity Criterion states that if the corresponding sides of two triangles are in the same ratio (proportional), then their corresponding angles are equal and the triangles are similar.

Triangles with sides scaled by a constant factor illustrating SSS similarity.
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The SAS (Side-Angle-Side) Similarity Criterion states that if one angle of a triangle is equal to one angle of another triangle and the sides including these angles are proportional, then the triangles are similar.

Triangle highlighting an angle and the two adjacent sides for SAS similarity.
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Property of Similarity: If ΔABC∼ΔDEF\Delta ABC \sim \Delta DEF, then the ratio of any two corresponding sides is equal to the ratio of their corresponding altitudes, medians, or perimeters.

📐Formulae

If ΔABC∼ΔPQR\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: ABPQ=BCQR=ACPR\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR}

Basic Proportionality Theorem: If DE∥BCDE \parallel BC, then ADDB=AEEC\frac{AD}{DB} = \frac{AE}{EC}

Corollary of BPT: ADAB=AEAC=DEBC\frac{AD}{AB} = \frac{AE}{AC} = \frac{DE}{BC}

Ratio of Perimeters: Perimeter of ΔABCPerimeter of ΔPQR=ABPQ\frac{\text{Perimeter of } \Delta ABC}{\text{Perimeter of } \Delta PQR} = \frac{AB}{PQ}

💡Examples

Problem 1:

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

Solution:

  1. By the Basic Proportionality Theorem, since DE∥BCDE \parallel BC, we have:
    ADDB=AEEC\frac{AD}{DB} = \frac{AE}{EC}
  2. Substitute the given values:
    xx−2=x+2x−1\frac{x}{x - 2} = \frac{x + 2}{x - 1}
  3. Cross-multiply to solve for xx:
    x(x−1)=(x+2)(x−2)x(x - 1) = (x + 2)(x - 2)
    x2−x=x2−4x^2 - x = x^2 - 4
  4. Simplify the equation:
    −x=−4  ⟹  x=4-x = -4 \implies x = 4
    Thus, 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.
  2. At the same time, the angle of elevation of the sun is the same for both. Therefore, ∠ACB=∠PRQ\angle ACB = \angle PRQ.
  3. Also, both the pole and the tower are vertical, so ∠ABC=∠PQR=90∘\angle ABC = \angle PQR = 90^\circ.
  4. By AA similarity criterion, ΔABC∼ΔPQR\Delta ABC \sim \Delta PQR.
  5. Therefore, the ratios of corresponding sides are equal:
    ABPQ=BCQR\frac{AB}{PQ} = \frac{BC}{QR}
  6. Substitute the known values (AB=6,BC=4,QR=28AB=6, BC=4, QR=28):
    6PQ=428\frac{6}{PQ} = \frac{4}{28}\
  7. Solve for PQPQ:
    4×PQ=6×284 \times PQ = 6 \times 28
    PQ=1684=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.

Problem 3:

In the given figure, QR/QS=QT/PRQR/QS = QT/PR and ∠1=∠2\angle 1 = \angle 2. Show that ΔPQS∼ΔTQR\Delta PQS \sim \Delta TQR.

Geometry figure showing a large triangle TQR with internal points P and S to demonstrate SAS similarity.

Solution:

  1. In ΔPQR\Delta PQR, given ∠1=∠2\angle 1 = \angle 2. Thus, PQ=PRPQ = PR (sides opposite to equal angles are equal).
  2. We are given the ratio: QRQS=QTPR\frac{QR}{QS} = \frac{QT}{PR}.
  3. Substituting PR=PQPR = PQ into the ratio, we get: QRQS=QTPQ\frac{QR}{QS} = \frac{QT}{PQ}.
  4. In ΔPQS\Delta PQS and ΔTQR\Delta TQR:
    • QRQS=QTPQ\frac{QR}{QS} = \frac{QT}{PQ} (from step 3)
    • ∠Q=∠Q\angle Q = \angle Q (Common angle)
  5. Therefore, by SAS similarity criterion, ΔPQS∼ΔTQR\Delta PQS \sim \Delta TQR.

Explanation:

This problem uses the property of isosceles triangles to substitute one side in a given ratio, fulfilling the SAS criteria requirements.

Problem 4:

SS and TT are points on sides PRPR and QRQR of ΔPQR\Delta PQR such that ∠P=∠RTS\angle P = \angle RTS. Show that ΔRPQ∼ΔRTS\Delta RPQ \sim \Delta RTS.

Triangle PQR with a transversal line ST such that angle RTS equals angle P.

Solution:

  1. Consider ΔRPQ\Delta RPQ and ΔRTS\Delta RTS.
  2. ∠RPQ=∠RTS\angle RPQ = \angle RTS (Given).
  3. ∠R=∠R\angle R = \angle R (Common angle).
  4. Since two angles of ΔRPQ\Delta RPQ are equal to two angles of ΔRTS\Delta RTS, by the AA similarity criterion, ΔRPQ∼ΔRTS\Delta RPQ \sim \Delta RTS.

Explanation:

The AA criterion is the most direct way to prove similarity when two angles (one given and one common) are identified.