RD Sharma class 9 Mathematics chapter 10 exercise 10.6 Congruent Triangles solutions 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 way to understand. Exercise 10.6 question-answers help students to learn about congruence criterion of right triangles mainly using RHS (Right angle-Hypotenuse-Side) and its related concepts.

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**Question 1: In Î” ABC, if âˆ A = 40Â° and âˆ B = 60Â°. Determine the longest and shortest sides of the triangle.**

**Solution:** In Î” ABC, âˆ A = 40Â° and âˆ B = 60Â°

We know, sum of angles in a triangle = 180Â°

âˆ A + âˆ B + âˆ C = 180Â°

40Â° + 60Â° + âˆ C = 180Â°

âˆ C = 180Â° â€“ 100Â° = 80Â°

âˆ C = 80Â°

Now, 40Â° < 60Â° < 80Â°

â‡’ âˆ A < âˆ B < âˆ C

â‡’ âˆ C is greater angle and âˆ A is smaller angle.

Now, âˆ A < âˆ B < âˆ C

We know, side opposite to greater angle is larger and side opposite to smaller angle is smaller.

Therefore, BC < AC < AB

AB is longest and BC is shortest side.

**Question 2: In a Î” ABC, if âˆ B = âˆ C = 45Â°, which is the longest side?**

**Solution:** In Î” ABC, âˆ B = âˆ C = 45Â°

Sum of angles in a triangle = 180Â°

âˆ A + âˆ B + âˆ C = 180Â°

âˆ A + 45Â° + 45Â° = 180Â°

âˆ A = 180Â° â€“ (45Â° + 45Â°) = 180Â° â€“ 90Â° = 90Â°

âˆ A = 90Â°

â‡’ âˆ B = âˆ C < âˆ A

Therefore, BC is the longest side.

**Question 3: In Î” ABC, side AB is produced to D so that BD = BC. If âˆ B = 60Â° and âˆ A = 70Â°. **

**Prove that: (i) AD > CD (ii) AD > AC**

** Solution:** In Î” ABC, side AB is produced to D so that BD = BC.

âˆ B = 60Â°, and âˆ A = 70Â°

To prove: (i) AD > CD (ii) AD > AC

Construction: Join C and D

We know, sum of angles in a triangle = 180Â°

âˆ A + âˆ B + âˆ C = 180Â°

70Â° + 60Â° + âˆ C = 180Â°

âˆ C = 180Â° â€“ (130Â°) = 50Â°

âˆ C = 50Â°

âˆ ACB = 50Â° â€¦â€¦(i)

And also in Î” BDC

âˆ DBC =180Â° â€“ âˆ ABC = 180 â€“ 60Â° = 120Â°

[âˆ DBA is a straight line]and BD = BC [given]

âˆ BCD = âˆ BDC [Angles opposite to equal sides are equal]

Sum of angles in a triangle =180Â°

âˆ DBC + âˆ BCD + âˆ BDC = 180Â°

120Â° + âˆ BCD + âˆ BCD = 180Â°

120Â° + 2âˆ BCD = 180Â°

2âˆ BCD = 180Â° â€“ 120Â° = 60Â°

âˆ BCD = 30Â°

âˆ BCD = âˆ BDC = 30Â° â€¦.(ii)

Now, consider Î” ADC.

âˆ DAC = 70Â° [given]

âˆ ADC = 30Â° [From (ii)]

âˆ ACD = âˆ ACB+ âˆ BCD = 50Â° + 30Â° = 80Â° [From (i) and (ii)]

Now, âˆ ADC < âˆ DAC < âˆ ACD

AC < DC < AD

[Side opposite to greater angle is longer and 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, letâ€™s check the rule.

2 + 3 â‰¯ 7 or 2 + 3 < 7

2 + 7 > 3

and 3 + 7 > 2

Here 2 + 3 â‰¯ 7

So, the triangle does not exit.

## 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 to 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.