RD Sharma Solutions for Class 9 Mathematics Chapter 10 Exercise 10.6 Congruent Triangles are provided here. In this exercise, students can get the detailed RD Sharma Class 9 Solutions for Chapter 10 Exercise 10.6 in an easy-to-understand format. Exercise 10.6 question-answers help students to learn about the congruence criterion of right triangles, mainly using RHS (Right-angle-Hypotenuse-Side) and its related concepts.
RD Sharma Solutions for Class 9 Maths Chapter 10 Congruent Triangles Exercise 10.6
Access Answers to RD Sharma Solutions for Class 9 Maths Chapter 10 Congruent Triangles Exercise 10.6 Page Number 10.66
Question 1: In Δ ABC, if ∠ A = 40° and ∠ B = 60°. Determine the longest and shortest sides of the triangle.
Solution: In Δ ABC, ∠ A = 40° and ∠ B = 60°
We know that the sum of angles in a triangle = 180°
∠ A + ∠ B + ∠ C = 180°
40° + 60° + ∠ C = 180°
∠ C = 180° – 100° = 80°
∠ C = 80°
Now, 40° < 60° < 80°
⇒ ∠ A < ∠ B < ∠ C
⇒ ∠ C is a greater angle and ∠ A is a smaller angle.
Now, ∠ A < ∠ B < ∠ C
We know the side opposite to the greater angle is larger, and the side opposite to the smaller angle is smaller.
Therefore, BC < AC < AB
AB is the longest and BC is the shortest side.
Question 2: In a Δ ABC, if ∠ B = ∠ C = 45°, which is the longest side?
Solution: In Δ ABC, ∠ B = ∠ C = 45°
The sum of angles in a triangle = 180°
∠ A + ∠ B + ∠ C = 180°
∠ A + 45° + 45° = 180°
∠ A = 180° – (45° + 45°) = 180° – 90° = 90°
∠ A = 90°
⇒ ∠ B = ∠ C < ∠ A
Therefore, BC is the longest side.
Question 3: In Δ ABC, side AB is produced to D so that BD = BC. If ∠ B = 60° and ∠ A = 70°.
Prove that: (i) AD > CD (ii) AD > AC
Solution: In Δ ABC, side AB is produced to D so that BD = BC.
∠ B = 60°, and ∠ A = 70°
To prove: (i) AD > CD (ii) AD > AC
Construction: Join C and D
We know the sum of angles in a triangle = 180°
∠ A + ∠ B + ∠ C = 180°
70° + 60° + ∠ C = 180°
∠ C = 180° – (130°) = 50°
∠ C = 50°
∠ ACB = 50° ……(i)
And also in Δ BDC
∠ DBC =180° – ∠ ABC = 180 – 60° = 120°
[∠DBA is a straight line]and BD = BC [given]
∠ BCD = ∠ BDC [Angles opposite to equal sides are equal]
The sum of angles in a triangle =180°
∠ DBC + ∠ BCD + ∠ BDC = 180°
120° + ∠ BCD + ∠ BCD = 180°
120° + 2∠ BCD = 180°
2∠ BCD = 180° – 120° = 60°
∠ BCD = 30°
∠ BCD = ∠ BDC = 30° ….(ii)
Now, consider Δ ADC.
∠ DAC = 70° [given]
∠ ADC = 30° [From (ii)]
∠ ACD = ∠ ACB+ ∠ BCD = 50° + 30° = 80° [From (i) and (ii)]
Now, ∠ ADC < ∠ DAC < ∠ ACD
AC < DC < AD
[Side opposite to the greater angle is longer and the smaller angle is smaller]AD > CD and AD > AC
Hence proved.
Question 4: Is it possible to draw a triangle with sides of length 2 cm, 3 cm and 7 cm?
Solution:
Lengths of sides are 2 cm, 3 cm and 7 cm.
A triangle can be drawn only when the sum of any two sides is greater than the third side.
So, let’s check the rule.
2 + 3 ≯ 7 or 2 + 3 < 7
2 + 7 > 3
and 3 + 7 > 2
Here 2 + 3 ≯ 7
So, the triangle does not exist.
RD Sharma Solutions for Class 9 Maths Chapter 10 Congruent Triangles Exercise 10.6
Class 9 Maths Chapter 10 Congruent Triangles Exercise 10.6 is based on the topic, Some Inequality Relations in a Triangle.
As we have learnt that if two sides of a triangle are equal, the angles opposite to them are also equal, and vice-versa. What happens when two angles of a triangle are unequal? Confused? Do not worry! RD Sharma Solutions Chapter 10 will answer all your questions in the form of theorems. Students can download this study material now for free and clear all their doubts.
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