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Triangles - Introduction

Grade 10CBSE

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

🔑Concepts

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Two figures are said to be congruent if they have the same shape and the same size. Two figures having the same shape but not necessarily the same size are called similar figures. All circles, squares, and equilateral triangles are similar.

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Two triangles are similar if their corresponding angles are equal and their corresponding sides are in the same ratio (proportional). Symbolically, △ABC∼△DEF\triangle ABC \sim \triangle DEF.

Two triangles ABC and DEF of different sizes but same shape representing similarity.
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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. Conversely, if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Triangle ABC with a line DE parallel to BC intersecting AB at D and AC at E.
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Criteria for Similarity of Triangles include: AAA (Angle-Angle-Angle), AA (Angle-Angle), SSS (Side-Side-Side), and SAS (Side-Angle-Side). If any of these conditions are met, the triangles are similar.

📐Formulae

ADDB=AEEC\frac{AD}{DB} = \frac{AE}{EC}

ABPQ=BCQR=ACPR\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR}

ADAB=AEAC\frac{AD}{AB} = \frac{AE}{AC}

△ABC∼△PQR  ⟹  ∠A=∠P,∠B=∠Q,∠C=∠R\triangle ABC \sim \triangle PQR \implies \angle A = \angle P, \angle B = \angle Q, \angle C = \angle R

💡Examples

Problem 1:

In △ABC\triangle ABC, DD and EE are points on sides ABAB and ACAC respectively such that DE∥BCDE \parallel BC. If AD=1.5 cmAD = 1.5\text{ cm}, DB=3 cmDB = 3\text{ cm}, and AE=1 cmAE = 1\text{ cm}, find ECEC.

Solution:

By the Basic Proportionality Theorem (BPT), since DE∥BCDE \parallel BC, we have: ADDB=AEEC\frac{AD}{DB} = \frac{AE}{EC} Substituting the given values: 1.53=1EC\frac{1.5}{3} = \frac{1}{EC} 12=1EC\frac{1}{2} = \frac{1}{EC} EC=2 cmEC = 2\text{ cm}

Explanation:

The BPT states that a line parallel to one side of a triangle divides the other two sides proportionally. We set up the ratio and solve for the unknown variable ECEC.

Problem 2:

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

Solution:

Using BPT: ADDB=AEEC\frac{AD}{DB} = \frac{AE}{EC} xx−2=x+2x−1\frac{x}{x - 2} = \frac{x + 2}{x - 1} Cross-multiplying gives: 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 apply the proportionality property AD/DB=AE/ECAD/DB = AE/EC and use the algebraic identity (a+b)(a−b)=a2−b2(a+b)(a-b) = a^2 - b^2 to simplify the equation.

Problem 3:

In the given figure, PS∥QTPS \parallel QT. If RS=4RS = 4 cm, ST=2ST = 2 cm and RP=6RP = 6 cm, find the length of PQPQ.

Triangle RQT with line PS parallel to QT.

Solution:

In △RQT\triangle RQT, PS∥QTPS \parallel QT. By Basic Proportionality Theorem: RPPQ=RSST\frac{RP}{PQ} = \frac{RS}{ST} Substitute the given values: 6PQ=42\frac{6}{PQ} = \frac{4}{2} 6PQ=2\frac{6}{PQ} = 2 PQ=62=3 cmPQ = \frac{6}{2} = 3\text{ cm} The length of PQPQ is 3 cm3\text{ cm}.

Explanation:

Since a line segment PSPS is parallel to side QTQT of △RQT\triangle RQT, the segments on the other two sides are proportional according to the Thales's Theorem.

Problem 4:

In the figure, ABCDABCD is a trapezium with AB∥DCAB \parallel DC. If points EE and FF lie on ADAD and BCBC respectively such that EF∥ABEF \parallel AB, prove that AEED=BFFC\frac{AE}{ED} = \frac{BF}{FC}.

Trapezium ABCD with line EF parallel to AB and diagonal AC intersecting EF at G.

Solution:

Join ACAC to intersect EFEF at point GG. In △ADC\triangle ADC, EG∥DCEG \parallel DC (since EF∥ABEF \parallel AB and AB∥DCAB \parallel DC). By BPT: AEED=AGGC\frac{AE}{ED} = \frac{AG}{GC} ... (1) In △CAB\triangle CAB, GF∥ABGF \parallel AB. By BPT: CGGA=CFFB\frac{CG}{GA} = \frac{CF}{FB}, which is AGGC=BFFC\frac{AG}{GC} = \frac{BF}{FC} ... (2) From (1) and (2): AEED=BFFC\frac{AE}{ED} = \frac{BF}{FC} Hence proved.

Explanation:

By drawing a diagonal, we create two triangles. Applying the Basic Proportionality Theorem to both triangles allows us to link the ratios of the non-parallel sides through a common ratio on the diagonal.