Class 10 Exam  >  Class 10 Notes  >  Mathematics (Maths) Class 10  >  Short Notes: Triangles

Triangles Class 10 Notes Maths Chapter 6

Similar Figures

  1. Two figures having the same shape but not necessary the same size are called similar figures.
  2. All congruent figures are similar but all similar figures are not congruent.

For Example: 

(i) Any two line segments are similar since length are proportional
Triangles Class 10 Notes Maths Chapter 6
(ii) Any two circles are similar since radii are proportional

Triangles Class 10 Notes Maths Chapter 6
(iii) Any two squares are similar since corresponding angles are equal and lengths are proportional.

Triangles Class 10 Notes Maths Chapter 6

Note: Similar figures are congruent if there is one to one correspondence between the figures.

Similar Polygons

Two polygons are said to be similar to each other, if:

  1. Their corresponding angles are equal, and
  2. The lengths of their corresponding sides are proportional

For Example:

From above we deduce:

Any two triangles are similar, if their
Triangles Class 10 Notes Maths Chapter 6
(i) Corresponding angles are equal

  • ∠A = ∠P
  • ∠B = ∠Q
  • ∠C = ∠R

(ii) Corresponding sides are proportional
Triangles Class 10 Notes Maths Chapter 6

Thales Theorem or Basic Proportionality Theory

Theorem 1:  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.
Triangles Class 10 Notes Maths Chapter 6

Proof of Thales Theorem

Given: In ∆ABC, DE || BC.

To prove: AD/DB = AE/EC

Const.: Draw EM ⊥ AD and DN ⊥ AE. Join B to E and C to D.

Proof: In ∆ADE and ∆BDE, 

Triangles Class 10 Notes Maths Chapter 6 ……..(i) [Area of ∆ = 1/2 x base x corresponding altitude]

In ∆ADE and ∆CDE,

Triangles Class 10 Notes Maths Chapter 6

∵ DE || BC …[Given ]

∴ ar(∆BDE) = ar(∆CDE) 

…[∵ As on the same base and between the same parallel sides are equal in area]

From (i), (ii) and (iii),
Triangles Class 10 Notes Maths Chapter 6

Question for Short Notes: Triangles
Try yourself:Two triangles have the same shape, and their corresponding angles are equal. If one triangle has a side length of 6 centimeters and the other triangle has a side length of 9 centimeters, can we conclude that these triangles are similar? Why or why not?
View Solution

Criterion for Similarity of Triangles

Two triangles are similar if either of the following three criterion’s are satisfied:

Triangles Class 10 Notes Maths Chapter 6

  • AAA similarity Criterion. If two triangles are equiangular, then they are similar.
  • Corollary(AA similarity). If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.
  • SSS Similarity Criterion. If the corresponding sides of two triangles are proportional, then they are similar.
  • SAS Similarity Criterion. If in two triangles, one pair of corresponding sides are proportional and the included angles are equal, then the two triangles are similar.

Results in Similar Triangles based on Similarity Criterion:

  • Ratio of corresponding sides = Ratio of corresponding perimeters
  • Ratio of corresponding sides = Ratio of corresponding medians
  • Ratio of corresponding sides = Ratio of corresponding altitudes
  • Ratio of corresponding sides = Ratio of corresponding angle bisector segments.

Area of Similar Triangles

Theorem 2. The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Given: ∆ABC ~ ∆DEF 
To prove:  Triangles Class 10 Notes Maths Chapter 6
Const.: Draw AM ⊥ BC and DN ⊥ EF. 
Proof: In ∆ABC and ∆DEF 
Triangles Class 10 Notes Maths Chapter 6
Triangles Class 10 Notes Maths Chapter 6…(i) ……[Area of ∆ = 1/2 x base x corresponding altitude]
∵ ∆ABC ~ ∆DEF 
Triangles Class 10 Notes Maths Chapter 6 …..(ii) …[Sides are proportional]
∠B = ∠E ……..[∵ ∆ABC ~ ∆DEF ]
∠M = ∠N …..[each 90° 
∴ ∆ABM ~ ∆DEN …………[AA similarity] 
Triangles Class 10 Notes Maths Chapter 6 …..(iii) …[Sides are proportional]
From (ii) and (iii), we have:  Triangles Class 10 Notes Maths Chapter 6
From (i) and (iv), we have:  Triangles Class 10 Notes Maths Chapter 6
Similarly, we can prove that 
Triangles Class 10 Notes Maths Chapter 6

Results based on Area Theorem: 

  • Ratio of areas of two similar triangles = Ratio of squares of corresponding altitudes
  • Ratio of areas of two similar triangles = Ratio of squares of corresponding medians
  • Ratio of areas of two similar triangles = Ratio of squares of corresponding angle bisector segments.

Note: If the areas of two similar triangles are equal, the triangles are congruent.

Question for Short Notes: Triangles
Try yourself:Which criterion is used to establish the similarity of two triangles when one pair of corresponding sides are proportional, and the included angles are equal?
View Solution

Pythagoras Theorem

Theorem 3: In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Proof of Pythagoras Theorem

Given: ∆ABC is a right triangle right-angled at B.

To prove: AB2 + BC2 = AC2

Const.: Draw BD ⊥ AC

Proof: In ∆s ABC and ADB, 

Triangles Class 10 Notes Maths Chapter 6

∠A = ∠A …[common]

∠ABC = ∠ADB …[each 90°]

∴ ∆ABC ~ ∆ADB …[AA Similarity]

Triangles Class 10 Notes Maths Chapter 6 ………[sides are proportional] 

⇒ AB2 = AC.AD 

Now in ∆ABC and ∆BDC

∠C = ∠C …..[common]

∠ABC = ∠BDC ….[each 90°]

∴ ∆ABC ~ ∆BDC …..[AA similarity] 

Triangles Class 10 Notes Maths Chapter 6  ……..[sides are proportional] 

BC² = AC.DC …(ii)

On adding (i) and (ii), we get

AB2 + BC= ACAD + AC.DC

⇒ AB2 + BC2 = AC.(AD + DC)

AB2 + BC2 = AC.AC

∴ AB2 + BC2 = AC2 

Converse of Pythagoras Theorem

Theorem 4: In a triangle, if square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.
Triangles Class 10 Notes Maths Chapter 6

Proof of

Converse of Pythagoras Theorem

Given: In ∆ABC, AB2 + BC2 = AC2

To prove: ∠ABC = 90°

Const.: Draw a right angled ∆DEF in which DE = AB and EF = BC

Proof: In ∆ABC,

AB2 + BC2 = AC2 …(i) [given]

In rt. ∆DEF

DE2 + EF2 = DF2 …[by pythagoras theorem]

AB2 + BC2 = DF2 …..(ii) …[DE = AB, EF = BC]

From (i) and (ii), we get

AC2 = DF2

⇒ AC = DF

Now, DE = AB …[by cont]

EF = BC …[by cont]

DF = AC …….[proved above]

∴ ∆DEF ≅ ∆ABC ……[SSS congruence]

∴ ∠DEF = ∠ABC …..[CPCT]

∠DEF = 90° …[by cont]

∴ ∠ABC = 90° 

Results based on Pythagoras’ Theorem:
(i) Result on obtuse Triangles.
If ∆ABC is an obtuse angled triangle, obtuse angled at B,
If AD ⊥ CB, then
AC2 = AB2 + BC+ 2 BC.BD 
Triangles Class 10 Notes Maths Chapter 6

(ii) Result on Acute Triangles.
If ∆ABC is an acute angled triangle, acute angled at B, and AD ⊥ BC, then
AC2 = AB+ BC2 – 2 BD.BC.
Triangles Class 10 Notes Maths Chapter 6

The document Triangles Class 10 Notes Maths Chapter 6 is a part of the Class 10 Course Mathematics (Maths) Class 10.
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