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Area of Similar Triangles

Triangles II - Concepts

Calculating Perimeter of Triangle

Area of Similar Triangles

Similarity Related To Right Triangle

Pythagoras Theorem

Results Deduced From Pythagoras Theorem

Class - 10th IMO Subjects

Concept Explanation

Area of Similar Triangles

Theorem: Ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

Given: $large Delta ABC sim Delta PQR$

To Prove:

$large frac{area(Delta ABC)}{area(Delta PQR)}=frac{AB^2}{PQ^2}=frac{BC^2}{QR^2}=frac{AC^2}{PR^2}$

Construction: Draw AD $large perp$ BC and PS $large perp$ QR

Proof: In $large Delta$ ABD and $large Delta$ PQS

$large angle B = angle Q$ $large [;Delta ABC sim Delta PQR;]$

$large angle D = angle S$ $large [ ;Each ;90^0;]$

$large Rightarrow Delta ABD sim Delta PQS$ [ By AA Similarity Criterion ]

Ratio of corresponding sides of similar triangle

$large frac{AB}{PQ}=frac{AD}{PS}$ ..........................(1)

But we are given that $large Delta ABC sim Delta PQR$

$large frac{AB}{PQ}=frac{AC}{PR}=frac{BC}{QR}$ ..........................(2)

From Eq (1) and (2)

$large frac{BC}{QR}=frac{AD}{PS}$

Now

$large frac{area(Delta ABC)}{area(Delta PQR)}=frac{frac{1}{2}X BC;X;AD}{frac{1}{2}X QR;X;PS}$ $large [;area ;of ;Delta = frac{1}{2};X;base;X;height;]$

$large frac{area(Delta ABC)}{area(Delta PQR)}=frac{BC;X;AD}{QR;X;PS}=frac{BC^2}{QR^2}$

As $large frac{AB}{PQ}=frac{AC}{PR}=frac{BC}{QR}$

Hence $large frac{area(Delta ABC)}{area(Delta PQR)}=frac{AB^2}{PQ^2}=frac{BC^2}{QR^2}=frac{AC^2}{PR^2}$

Illustration: If the area of two similar triangles are equal, prove that they are congruent.

Let $large Delta ABC sim Delta PQR$

$large Rightarrow large frac{AB}{PQ}=frac{AC}{PR}=frac{BC}{QR}$

Given

$large frac{area(Delta ABC)}{area(Delta PQR)}=1$

We know that

$large frac{area(Delta ABC)}{area(Delta PQR)}=frac{AB^2}{PQ^2}=frac{BC^2}{QR^2}=frac{AC^2}{PR^2}$

$large Rightarrow frac{AB^2}{PQ^2}=frac{BC^2}{QR^2}=frac{AC^2}{PR^2}=1$

$large Rightarrow$ AB = PQ , BC= QR and AC = PR

Hence

$large Delta ABC cong Delta PQR$ [By SSS Criterion of Congruence]

Sample Questions

(More Questions for each concept available in Login)

Question : 1

In the following figure, $bigtriangleup ABCsim bigtriangleup bigtriangleup DEF$ . If AB = 2DE and area of $bigtriangleup ABC=56;cm^{2}$ , find the area of $bigtriangleup DEF.$
A. 16	B. 18	C. 14	D. 20
Right Option : C
View Explanation

Question : 2

In the following figure, $bigtriangleup ABCsim bigtriangleup DEF,AB=1.2;cm;and;DE=1.4;cm$ . Find the ratio of areas of $bigtriangleup ABC;to; bigtriangleup DEF$ .
A. $frac{6}{7}$	B. $frac{12}{14}$	C. $frac{36}{49}$	D. none of these
Right Option : C
View Explanation

Question : 3

$large Delta AMBsim Delta CMD$ . Also 2ar ( $large Delta AMB$ ) = ar ( $large Delta CMD$ ) the length of MD is ___________________
A. $large sqrt2 ;MB$	B. $large sqrt2 ;MD$	C. $large frac{sqrt2}{MB}$	D. $large frac{sqrt2}{MD}$
Right Option : A
View Explanation

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Language - English

Chapters

Real Numbers I

Real Numbers II

Pair of Linear Equations I

Pair of Linear Equations II

Polynomial I

Polynomials II

Quadratic Equations I

Quadratic Equations II

Introduction to Trigonometry I

Introduction to Trigonometry II

Arithmetic Progression I

Arithmetic Progression II

Triangles I

Triangles II

Some Applications of Trigonometry

Coordinate Geometry I

Coordinate Geometry II

Introduction to Trigonometry

Direct and Inverse Variation

Surface Area and Volume I

Surface Area and Volume II

Areas Related to Circles

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