# R.D. Sharma Solutions Class 9th: Ch 10 Congruent Triangles Exercise 10.6

#### Chapter 10 Congruent Triangles R.D. Sharma Solutions for Class 9th Exercise 10.6

**Exercise 10.6**

1. 1. In ΔABC, if ∠A = 40° and ∠B = 60°. Determine the longest and shortest sides of the triangle.

**Solution**

2. In a ΔABC, if ∠B = ∠C = 45°, which is the longest side?

**Solution**

3. In ΔABC, 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**

4. Is a possible to draw a triangle with sides of length 2 cm, 3 cm and 7 cm?

**Solution**

As we know that a triangle can only be formed if

The sum of two sides is greater than the third side.

Here we have 2 cm, 3 cm and 7 cm as sides.

If we add 2+3 = 5

5<7 (Since 5 is less than 7)

Hence, the sum of two sides is less than the third sides

So, the triangle will not exist.

5. In Δ ABC, ∠B = 35°, ∠C = 65° and the bisector of ∠BAC meets BC in P. Arrange AP, BP and CP in descending order.

**Solution**

6. O is any point in the interior of △ABC. Prove that:

(i) AB+AC > OB+OC

(ii) AB + BC + CA > OA + OB + OC

(iii) OA + OB + OC > 1/2 (AB+BC+CA)

**Solution**

So, we have

In ΔABC

AB + BC > AC

BC + AC > AB

AC + AB > BC

In ΔOBC

OB + OC > BC …….(i)

In ΔOAC

OA + OC > AC ……(ii)

In ΔOAB

OA + OB > AB ……(iii)

Now, extend (or) produce BO to meet AC in D.

Now, in ΔABD, we have

AB + AD > BD

AB + AD > BO + OD …….(iv) [BD = BO + OD]

Similarly in ΔODC, we have

OD + DC > OC ……(v)

(i) Adding (iv) and (v), we get

AB + AD + OD + DC > BO + OD + OC

AB + (AD + DC) > OB + OC

AB + AC > OB + OC ……(vi)

Similarly, we have

BC + BA > OA + OC ……(vii)

and CA+ CB > OA + OB ……(viii)

(ii) Adding equation (vi), (vii) and (viii), we get

AB + AC + BC + BA + CA + CB > OB + OC + OA + OC + OA + OB

⇒ 2AB + 2BC + 2CA > 2OA + 2OB + 2OC

⇒ 2(AB + BC + CA) > 2(OA + OB + OC)

⇒AB + BC + CA > OA + OB + OC

(iii) Adding equations (i), (ii) and (iii)

OB + OC + OA + OC + OA + OB > BC + AC + AB

2OA + 2OB + 2OC > AB + BC + CA

We get,

2(OA+OB+OC) > AB + BC +CA

(OA+OB + OC) > (1/2)(AB + BC +CA)

7. Prove that the perimeter of a triangle is greater than the sum of its altitudes.

**Solution**

8. Prove that in a quadrilateral the sum of all the sides is greater than the sum of its diagonals.

**Solution**

9. In Fig. 10.131, prove that:

(i) CD + DA + AB + BC > 2AC

(ii) CD + DA + AB > BC

**Solution**

10. Which of the following statements are true (T) and which are false (F)?

(i) Sum of the three sides of a triangle is less than the sum of its three altitudes.

(ii) Sum of any two sides of a triangle is greater than twice the median drawn to the third side.

(iii) Sum of any two sides of a triangle is greater than the third side.

(iv) Difference of any two sides of a triangle is equal to the third side.

(v) If two angles of a triangle are unequal, then the greater angle has the larger side opposite to it.

(vi) Of all the line segments that can be drawn from a point to a line not containing it, the perpendicular line segment is the shortest one.

**Solution**

(i) False

(ii) True

(iii) True

(iv) False

(v) True

(vi) True

11. Fill in the blanks to make the following statements true.

(i) In a right triangle the hypotenuse is the ...... side.

(ii) The sum of three altitudes of a triangle is ....... than its perimeter.

(iii) The sum of any two sides of a triangle is ..... than the third side.

(iv) If two angles of a triangle are unequal, then the smaller angle has the ..... side opposite to it.

(v) Difference of any two sides of a triangle is ...... than the third side.

(vi) If two sides of a triangle are unequal, then the larger side has ...... angle opposite to it.

**Solution**

(i) Largest

(ii) Less

(iii) Greater

(iv) Smaller

(v) Less

(vi) Greater