In △ A B C and △ P Q R shown below. A B = Q R , B C = R P , C A = P Q Are C D and P S equal? Why?

In △ A B C and △ P Q R A B = Q R , B C = R P and C A = P Q ∴ △ A B C ≅ △ Q R P ...( By S S S test) ∴ ∠ A B C = ∠ Q R P . . . 1 Now, i n △ C D B and △ P S R , ∠ D B C = ∠ S R P . . . 2 (From (1)) ∠ C D B = ∠ P S R = 90 ° Thus, in above triangles two angles are equal. ∴ ∠ D C B = ∠ R P S . . . 3 [ ∵ By using angle sum property] According to property, If one side of a triangle and angle at its ends are equal to one side of another triangle and the angles at its ends, then the third angles are also equal and the sides opposite equal angle are equal. Here, common side are B C and R P ∠ D C B = ∠ R P S [From (3)] And ∠ D B C = ∠ S R P [From (2)] Applying it in △ C D B and △ P S R , we get C D = P S . . . 4 ... ( ∵ CD is opposite ∠ CBD , PS is opposite ∠ PRS and ∠ CBD = ∠ PRS ) Hence, they are equal.

In △ A B C and △ P Q R shown below. A B = Q R , B C = R P , C A = P Q What is the relation between the areas of △ A B C and △ P Q R .

In △ A B C and △ P Q R A B = Q R , B C = R P and C A = P Q By observing the angles ∠ A = ∠ Q , ∠ B = ∠ R and ∠ C = ∠ P ∴ △ A B C ≅ △ P Q R ...( By S S S test) Also, congruent triangles have equal areas. ⇒ a r △ A B C = a r △ P Q R

In the quadrilateral A B C D shown below, the sides A B and C D are parallel. M is the midpoint of the side B C . The lines D M and A B extended, meet at N . Are the areas of △ D C M and △ B M N equal? Why?

Given: A B ∥ C D and B M = M C In △ D C M and △ B M N , ∠ D M C = ∠ B M N ...(vertically opposite angles) B M = M C ....(Given) ∠ D C M = ∠ M B N ...(Alternate angle, D C ∥ B N with B C as transversal). Using the property, If one side of a triangle and angle at its ends are equal to one side of another triangle and the angles at its ends, then the third angles are also equal and the sides opposite equal angle are equal. ⇒ ∠ C D M = ∠ M N B ⇒ D M = M N and D C = B N As all three sides and angles are equal. △ D C M ≅ △ N B M Also, congruent triangles have equal areas. Hence, area ( △ D C M ) = area ( △ B M N )

In the quadrilateral A B C D shown below, the sides A B and C D are parallel. M is the midpoint of the side B C . The lines D M and A B extended, meet at N . What is the relation between the areas of the quadrilateral A B C D and the triangle A N D ?

Given: A B ∥ C D and B M = M C In △ D C M and △ B M N , ∠ D M C = ∠ B M N ...(vertically opposite angles) B M = M C ....(Given) ∠ D C M = ∠ M B N ...( Alternate angle , D C ∥ B N with B C as transversal ). Using the property, If one side of a triangle and angle at its ends are equal to one side of another triangle and the angles at its ends, then the third angles are also equal and the sides opposite equal angle are equal. ⇒ ∠ C D M = ∠ M N B ⇒ D M = M N and D C = B N As all three sides and angles are equal. △ D C M ≅ △ N B M Also, congruent triangles have equal areas. Hence, area ( △ D C M ) = area ( B M N ) ...(i) We know that, Area of a parallelogram = base × height And area of a triangle = 1 2 × base × height Thus, area of parallelogram = area ( A B C D ) = area ( A D M B ) + area ( △ D C M ) . . . i i And area of triangle = area ( △ A N D ) = area ( A D M B ) + area ( △ B M N ) . . . i i i = area ( A D M B ) + area ( △ D C M ) ... [From (i)] = area ( A B C D ) Hence, area ( A B C D ) = a r e a △ A N D

E M B I B E

Standard 8 Mathematics Part - 1>Chapter 1 - Equal Triangles>Exercise 3

Kerala Board Solutions for Chapter: Equal Triangles, Exercise 3: Exercise 3

Author:Kerala Board

Kerala Board Mathematics Solutions for Exercise - Kerala Board Solutions for Chapter: Equal Triangles, Exercise 3: Exercise 3

Attempt the free practice questions on Chapter 1: Equal Triangles, Exercise 3: Exercise 3 with hints and solutions to strengthen your understanding. Standard 8 Mathematics Part - 1 solutions are prepared by Experienced Embibe Experts.

Questions from Kerala Board Solutions for Chapter: Equal Triangles, Exercise 3: Exercise 3 with Hints & Solutions

EASY

8th Kerala Board

IMPORTANT

Standard 8 Mathematics Part - 1>Chapter 1 - Equal Triangles>Exercise 3>Q (2)

Draw a line of $5.5 centimetres$ along. Using a set square, locate the midpoint of this line.

MEDIUM

8th Kerala Board

IMPORTANT

Standard 8 Mathematics Part - 1>Chapter 1 - Equal Triangles>Exercise 3>Q (3)

In the figure, $A B C D E$ is pentagon with all sides of the same length and all angles of the same size. The sides $A B$ and $A E$ extended, meet the side $C D$ extended at $P$ and $Q$ .

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Are the sides of $△ B P C$ equal to the sides of $△ Q E D$ . Why?

MEDIUM

8th Kerala Board

IMPORTANT

Standard 8 Mathematics Part - 1>Chapter 1 - Equal Triangles>Exercise 3>Q (3)

In the figure, $A B C D E$ is pentagon with all sides of the same length and all angles of the same size. The sides $A B$ and $A E$ extended, meet the side CD extended at $P$ and $Q$ .

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