krit.club logo

Triangles - Similar Figures

Grade 10CBSE

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

🔑Concepts

•

Similar figures have the same shape but not necessarily the same size. For polygons, they are similar if: (i) all corresponding angles are equal and (ii) all corresponding sides are in the same ratio (proportionality).

Two triangles ABC and PQR of different sizes but similar shapes.
•

Basic Proportionality Theorem (Thales's 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.

Triangle ABC with line DE parallel to BC
•

AAA (Angle-Angle-Angle) Similarity Criterion: If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio and hence the two triangles are similar.

•

SSS (Side-Side-Side) Similarity Criterion: If in two triangles, sides of one triangle are proportional to the sides of the other triangle, then their corresponding angles are equal and the triangles are similar.

•

SAS (Side-Angle-Side) Similarity Criterion: If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.

📐Formulae

If DE∥BC in △ABC, then ADDB=AEEC\text{If } DE \parallel BC \text{ in } \triangle ABC, \text{ then } \frac{AD}{DB} = \frac{AE}{EC}

If △ABC∼△PQR, then ∠A=∠P,∠B=∠Q,∠C=∠R\text{If } \triangle ABC \sim \triangle PQR, \text{ then } \angle A = \angle P, \angle B = \angle Q, \angle C = \angle R

If △ABC∼△PQR, then ABPQ=BCQR=ACPR\text{If } \triangle ABC \sim \triangle PQR, \text{ then } \frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR}

If ADDB=AEEC, then DE∥BC\text{If } \frac{AD}{DB} = \frac{AE}{EC}, \text{ then } DE \parallel BC

💡Examples

Problem 1:

In △ABC\triangle ABC, 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:

By Basic Proportionality Theorem, since DE∥BCDE \parallel BC: ADDB=AEEC\frac{AD}{DB} = \frac{AE}{EC} Substituting the given values: xx−2=x+2x−1\frac{x}{x-2} = \frac{x+2}{x-1} x(x−1)=(x+2)(x−2)x(x-1) = (x+2)(x-2) x2−x=x2−4x^2 - x = x^2 - 4 −x=−4-x = -4 x=4x = 4

Explanation:

We applied the Thales Theorem (BPT) which states that a line parallel to one side of a triangle divides the other two sides proportionally. Cross-multiplying the ratios leads to a quadratic equation where the x2x^2 terms cancel out, leaving a simple linear equation for xx.

Problem 2:

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

Solution:

Let the height of the tower be hh. The triangles formed by the pole and its shadow, and the tower and its shadow, are similar because the angle of elevation of the sun is the same for both. Height of poleHeight of tower=Shadow of poleShadow of tower\frac{\text{Height of pole}}{\text{Height of tower}} = \frac{\text{Shadow of pole}}{\text{Shadow of tower}} 6h=428\frac{6}{h} = \frac{4}{28} 4h=6×284h = 6 \times 28 h=1684h = \frac{168}{4} h=42 mh = 42\text{ m}

Explanation:

This uses the AA similarity criterion. Both the pole and the tower are vertical (90 degrees to the ground) and the sun's rays hit them at the same angle at the same time, making the triangles similar. Thus, the ratio of their heights is equal to the ratio of their shadows.

Problem 3:

In the given figure, QA⊥ABQA \perp AB and PB⊥ABPB \perp AB. If AO=20 cmAO = 20\text{ cm}, BO=12 cmBO = 12\text{ cm} and PB=18 cmPB = 18\text{ cm}, find the length of QAQA.

A figure showing two triangles AOQ and BOP connected at vertex O with right angles at A and B.

Solution:

In △AOQ\triangle AOQ and △BOP\triangle BOP:

  1. ∠OAQ=∠OBP=90∘\angle OAQ = \angle OBP = 90^{\circ} (Given)
  2. ∠AOQ=∠BOP\angle AOQ = \angle BOP (Vertically opposite angles) By AA similarity criterion, △AOQ∼△BOP\triangle AOQ \sim \triangle BOP. Therefore, the corresponding sides are proportional: QAPB=AOBO\frac{QA}{PB} = \frac{AO}{BO} QA18=2012\frac{QA}{18} = \frac{20}{12} QA=20×1812QA = \frac{20 \times 18}{12} QA=5×6=30 cmQA = 5 \times 6 = 30\text{ cm}

Explanation:

We first prove the two triangles are similar using the AA criterion by identifying equal right angles and vertically opposite angles. Then, we set up a proportion using the corresponding sides to solve for the unknown length.

Problem 4:

In △ABC\triangle ABC, DD and EE are points on sides ABAB and ACAC such that DE∥BCDE \parallel BC. If AD=2.4 cmAD = 2.4\text{ cm}, AE=3.2 cmAE = 3.2\text{ cm} and EC=4.8 cmEC = 4.8\text{ cm}, find ABAB.

Triangle ABC with side lengths labeled and line DE parallel to BC.

Solution:

According to the Basic Proportionality Theorem (BPT), since DE∥BCDE \parallel BC: ADDB=AEEC\frac{AD}{DB} = \frac{AE}{EC} 2.4DB=3.24.8\frac{2.4}{DB} = \frac{3.2}{4.8} DB=2.4×4.83.2DB = \frac{2.4 \times 4.8}{3.2} DB=2.4×32=3.6 cmDB = \frac{2.4 \times 3}{2} = 3.6\text{ cm} Now, AB=AD+DBAB = AD + DB: AB=2.4+3.6=6.0 cmAB = 2.4 + 3.6 = 6.0\text{ cm}

Explanation:

Applying the BPT allowed us to find the segment DB. Adding the given segment AD and the calculated segment DB gives the total length of side AB.