If in the given figure OR ' = 2 × OR , OS ' = 2 × OS and OT ' = 2 × OT . Prove that △ R S T ~ △ R ' S ' T ' .

Given: In ∆ O S ' R ' and ∆ O S R , ∠ S ' O R ' = ∠ S O R (Common angle) O S ' O S = O R ' O R = 2 1 . . . . ( i ) Therefore, by SAS criterion ∆ O S ' R ' ~ ∆ O S R Similarly, in ∆ O S ' T ' and ∆ O S T ∠ S ' O T ' = ∠ S O T (Common angle) O S ' O S = O T ' O T = 2 1 . . . . ( ii ) Therefore, by SAS criterion ∆ O S ' T ' ~ ∆ O S T From ( i ) and ( ii ) , we get O S ' O S = O R ' O R = O T ' O T ⇒ ∆ R ' S ' T ' ~ ∆ R S T

In triangle ACB , ∠ ACB = 90 ° and CD ⊥ AB . Prove that BC 2 AC 2 = BD AD .

Given: In ∆ C A B , ∠ A C B = 90 ° and C D ⊥ A B We know that if a perpendicular drawn from the vertex of the right angle to the hypotenuse then triangles on both sides of the perpendicular are similar to the whole triangle and to each other. ⇒ ∆ A C D ~ ∆ A B C . . . . . ( i ) and ∆ B C D ~ ∆ B A C . . . . . ( ii ) From ( i ) , ∆ A C D ~ ∆ A B C If two triangles are similar, then the ratio of their corresponding sides are equal. ⇒ A C A B = A D A C ⇒ A C 2 = A B × A D . . . . . ( iii ) Similarly, from ( ii ) , ∆ B C D ~ ∆ B A C ⇒ B C B A = B D B C ⇒ B C 2 = B A × B D . . . . ( iv ) Dividing ( iv ) by ( iii ) , we get B C 2 A C 2 = B A × B D A B × A D ⇒ B C 2 A C 2 = B D A D Hence, proved

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Mathematics Class - 10>Geometry>Chapter 11 - Similarity in Geometrical Shapes>Exercise – 5

Chhattisgarh Board Solutions for Exercise 14: Exercise – 5

Author:Chhattisgarh Board

Chhattisgarh Board Mathematics Solutions for Exercise - Chhattisgarh Board Solutions for Exercise 14: Exercise – 5

Attempt the practice questions from Exercise 14: Exercise – 5 with hints and solutions to strengthen your understanding. Mathematics Class - 10 solutions are prepared by Experienced Embibe Experts.

Questions from Chhattisgarh Board Solutions for Exercise 14: Exercise – 5 with Hints & Solutions

EASY

10th Chhattisgarh Board

IMPORTANT

Mathematics Class - 10>Chapter 11 - Similarity in Geometrical Shapes>Exercise – 5>Q 3

In a right triangle $△ P Q R$ , $P$ is the right angle and $M$ is the point on $Q R$ such that $\Rightarrow R M^{2} + M Q^{2} + 2 . R M . M Q = R P^{2} + P Q^{2} . . . . . (iv)$ . Show that ${PM}^{2} = QM \times MR$ .

EASY

10th Chhattisgarh Board

IMPORTANT

Mathematics Class - 10>Chapter 11 - Similarity in Geometrical Shapes>Exercise – 5>Q 4

$ABC$ is an equilateral triangle of side $2 a$ . Find the length of each altitude.

EASY

10th Chhattisgarh Board

IMPORTANT

Mathematics Class - 10>Chapter 11 - Similarity in Geometrical Shapes>Exercise – 5>Q 5

$∆ A B C$ is an isosceles triangle in which $∠ C = 90 °$ . Prove that ${AB}^{2} = 2 {AC}^{2}$ .

EASY

10th Chhattisgarh Board

IMPORTANT

Mathematics Class - 10>Chapter 11 - Similarity in Geometrical Shapes>Exercise – 5>Q 6

If in the given figure $OR' = 2 \times OR$ , $OS' = 2 \times OS$ and $OT' = 2 \times OT$ . Prove that $△ R S T ~ △ R' S' T'$ .

Question Image

EASY

10th Chhattisgarh Board

IMPORTANT

Mathematics Class - 10>Chapter 11 - Similarity in Geometrical Shapes>Exercise – 5>Q 7

Triangle $A B C$ is right angled at $C$ . Point $' D'$ and $' E'$ are located on sides $C A$ and $C B$ , respectively. Prove that $A E^{2} + B D^{2} = A B^{2} + D E^{2}$ .

EASY

10th Chhattisgarh Board

IMPORTANT

Mathematics Class - 10>Chapter 11 - Similarity in Geometrical Shapes>Exercise – 5>Q 8

In triangle $ACB, ∠ ACB = 90 °$ and $CD ⊥ AB$ . Prove that $\frac{{BC}^{2}}{{AC}^{2}} = \frac{BD}{AD}$ .

EASY

10th Chhattisgarh Board

IMPORTANT

Mathematics Class - 10>Chapter 11 - Similarity in Geometrical Shapes>Exercise – 5>Q 9

The diameter of the earth is approximately $8000$ miles and that of sun is $864000$ miles; the distance between the sun and earth is approximately $92$ million miles.
If on paper we take the diameter of the earth as $1$ inch then what will be the diameter of the sun and the distance between the sun and earth on paper? ( $1$ million $= 10^{6}$ ).