# The Triangle and its Properties

Solved exercise on triangle and its properties

# Exercise

1. In ∆PQR, D is the mid-pointof QR

PM is

PD is

If QM = MR?

Solution:

PM is altitude.

PD is median.

No, QM ≠ MR.

2. Draw rough sketches for the following

(a) In ∆ABC, BE is a median.

(b) In ∆PQR, PQ and PR are altitudes of the triangle.

(c) In ∆XYZ, YL is an altitude in the exterior of the triangle.

3. Verify by drawing a diagram if the median and altitude of an isosceles triangle can be same.

∆ABC is an isosceles triangle in which AB = AC

Draw AD as the median of the triangle.

Measure the angle ADC with the help of protractor we find, ∠ADC = 90°

Thus, AD is the median as well as the altitude of the ∆ABC. Hence Verified.

**Exercise 6.2**

1. Find the value of the unknown exterior angle x in the following diagrams

**Solution**

(i)∠x = 50° + 70°

= 120°

(Exterior angle is equal to sum of its interior opposite angles)

(ii) ∠x = 65°+ 45°

= 110°

(Exterior angle is equal to sum of its interior opposite angles)

(iii) ∠x = 30° + 40°

= 70°

(Exterior angle is equal to sum of its interior opposite angles)

(iv) ∠x = 60° + 60°

= 120°

(Exterior angle is equal to sum of its interior opposite angles)

(v) ∠x = 50° + 50°

=100°

(Exterior angle is equal to sum of its interior opposite angles)

(vi) ∠x = 30° + 60°

= 90°

(Exterior angle is equal to sum of its interior opposite angle)

**2. Find the value of the unknown interior angle x in the following figures:**

(i) ∠x + 50° = 115° (Exterior angle of a triangle)

∴ ∠x = 115°- 50° = 65°

(ii) ∠x + 70° = 110° (Exterior angle of a triangle)

∴ ∠x = 110° – 70° = 40°

(iii) ∠ x + 90° = 125° (Exterior angle of a right triangle)

∴ ∠x = 125° – 90° = 35°

(iv) ∠x + 60° = 120° (Exterior angle of a triangle)

∴ ∠x = 120° – 60° = 60°

(v)∠ X + 30° = 80° (Exterior angle of a triangle)

∴ ∠x = 80° – 30° = 50°

(vi) ∠ x + 35° = 75° (Exterior angle of a triangle)

∴ ∠ x = 75° – 35° = 40°

**Exercise 6.3**

1. Find the value of the unknown x in the following diagrams

**Solution:**i) By angle sum property of a triangle, we have

∠x + 50° + 60° = 180°

⇒ ∠x + 110° = 180°

∴ ∠x = 180° – 110° = 70°

(ii) By angle sum property of a triangle, we have

∠x + 90° + 30 = 180° [∆ is right angled triangle]

⇒ ∠x + 120° = 180°

∴ ∠x – 180° – 120° = 60°

(iii) By angle sum property of a triangle, we have

∠x + 30° + 110° – 180°

⇒ ∠x + 140° = 180°

∴ ∠x = 180° – 140° = 40°

**Find the value of the unknown x in the following diagrams**

Solution:(iv ) By angle sum property of a triangle, we have

∠x + ∠x + 50° = 180°

⇒ 2x + 50° = 180°

⇒ 2x = 180° – 50°

⇒ 2x = 130°

∴ x=130/2=65∘

(v) By angle sum property of a triangle, we have

∠x + ∠x +∠x =180°

⇒ 3 ∠x = 180°

∴ ∠x=180/3=60∘

(vi) By angle sum property of a triangle, we have

x + 2 x + 90° = 180° (∆ is right angled triangle)

⇒ 3x + 90° = 180°

⇒ 3x = 180° – 90°

⇒ 3x = 90°

∴ x=90/3=30∘

2. Find the values of the unknowns x and y in the following diagrams:

(i) ∠x + 50° = 120° (Exterior angle of a triangle)

∴ ∠x = 120°- 50° = 70°

∠x + ∠y + 50° = 180° (Angle sum property of a triangle)

70° + ∠y + 50° = 180°

∠y + 120° = 180°

∠y = 180° – 120°

∴ ∠y = 60°

Thus ∠x = 70 and ∠y – 60°

(ii) ∠y = 80° (Vertically opposite angles are same)

∠x + ∠y + 50° = 180° (Angle sum property of a triangle)

⇒ ∠x + 80° + 50° = 180°

⇒ ∠x + 130° = 180°

∴ ∠x = 180° – 130° = 50°

Thus, ∠x = 50° and ∠y = 80°

**Find the values of the unknowns x and y in the following diagrams:**

(iii) ∠y + 50° + 60° = 180° (Angle sum property of a triangle)

∠y + 110° = 180°

∴ ∠y = 180°- 110° = 70°

∠x + ∠y = 180° (Linear pairs)

⇒ ∠x + 70° = 180°

∴ ∠x = 180° – 70° = 110°

Thus, ∠x = 110° and y = 70°

(iv) ∠x = 60° (Vertically opposite angles)

∠x + ∠y + 30° = 180° (Angle sum property of a triangle)

⇒ 60° + ∠y + 30° = 180°

⇒ ∠y + 90° = 180°

⇒ ∠y = 180° – 90° = 90°

Thus, ∠x = 60° and ∠y = 90°

Find the values of the unknowns x and y in the following diagrams:

(v) ∠y = 90° (Vertically opposite angles)

∠x + ∠x + ∠y = 180° (Angle sum property of a triangle)

⇒ 2 ∠x + 90° = 180°

⇒ 2∠x = 180° – 90°

⇒ 2∠x = 90°

∴ ∠x=90/2=45∘

Thus, ∠x = 45° and ∠y = 90°

(vi) From the given figure, we have

Adding both sides, we have

∠y + ∠1 + ∠2 = 3∠x

⇒ 180° = 3∠x (Angle sum property of a triangle)

∴ ∠x=180/ 3=60∘

∠x = 60°, ∠y = 60°

**Exercise 6.4 & 6.5**

1. Is it possible to have a triangle with the following sides?

(i) 2 cm, 3 cm, 5 cm

(ii) 3 cm, 6 cm, 7 cm

(iii) 6 cm, 3 cm, 2 cm

Solution:We know that for a triangle, the sum of any two sides must be greater than the third side.

(i) Given sides are 2 cm, 3 cm, 5 cm

Sum of the two sides = 2 cm + 3 cm = 5 cm Third side = 5 cm

We have, Sum of any two sides = the third side i.e. 5 cm = 5 cm

Hence, the triangle is not possible.

(ii) Given sides are 3 cm, 6 cm, 7 cm

Sum of the two sides = 3 cm + 6 cm = 9 cm Third side = 7 cm

We have sum of any two sides > the third

side. i.e. 9 cm > 7 cm

Hence, the triangle is possible.

(iii) Given sides are 6 cm, 3 cm, 2 cm

Sum of the two sides = 3 cm + 2 cm – 5 cm Third side = 6 cm

We have, sum of any two sides < the third sid6 i.e. 5 cm < 6 cm Hence, the triangle is not possible.

2. Take any point O in the interior of a triangle PQR . Is

i) OP + OQ > PQ?

(ii) OQ + OR > QR?

(iii) OR + OP > RP?

Solution:(i) Yes, In ∆ OPQ, we have

OP + OQ > PQ

[Sum of any two sides of a triangle is greater than the third side]

(ii) Yes, In ∆OQR, we have OQ + OR > QR

[Sum of any two sides of a triangle is greater than the third side]

(iii) Yes, In ∆OPR, we have OR + OP > RP

[Sum of any two sides of a triangle is greater than the third side]

3. AM is a median of a triangle ABC.

Is AB + BC + CA > 2AM ?

(Consider the sides of triangles ∆ABM and ∆AMC)

**Solution:**Yes, In ∆ ABM, we have

AB + BM > AM …(i)

[Sum of any two sides of a triangle is greater than the third side]

In ∆AMC, we have

AC + CM > AM …(ii)

[Sum of any two sides of a triangle is greater than the third side]

Adding eq (i) and (ii), we have

AB + AC + BM + CM > 2AM

AB + AC + BC + > 2AM

AB + BC + CA > 2AM

Hence, proved.

4. ABCD is a quadrilateral.

Is AB + BC + CD + DA > AC + BD?

**Solution:**

In ∆ABC, we have

AB + BC > AC …(i)

[Sum of any two sides is greater than the third side]

In ∆BDC, we have

BC + CD > BD …(ii)

[Sum of any two sides is greater than the third side]

In ∆ADC, we have

CD + DA > AC …(iii)

[Sum of any two sides is greater than the third side]

In ∆DAB, we have

DA + AB > BD …(iv)

[Sum of any two sides is greater than the third side]

Adding eq. (i), (ii), (iii) and (iv), we get

2AB + 2BC + 2CD + 2DA > 2AC + 2BD or

AB + BC + CD + DA > AC + BD [Dividing both sides by 2]

Hence, proved.

5. ABCD is a quadrilateral.

Is AB + BC + CD + DA < 2(AC + BD)?

Solution:we have a quadrilateral ABCD.

In ∆AOB, we have AB < AO + BO …(i)

[Any side of a triangle is less than the sum of other two sides]

In ∆BOC, we have

BC < BO + CO …(ii)

[Any side of a quadrilateral is less than the sum of other two sides]

In ∆COD, we have

CD < CO + DO …(iii)

[Any side of a triangle is less than the sum of other two sides]

In ∆AOD, we have

DA < DO + AO …(iv)

[Any side of a triangle is less than the sum of other two sides]

Adding eq. (i), (ii), (iii) and (iv), we have

AB + BC + CD + DA

∠2AO + ∠BO + ∠CO + ∠DO

∠2(AO + BO + CO + DO)

∠2 [(AO + CO) + (BO + DO)]

∠2(AC + BD)

Thus, AB + BC + CD + DA < 2(AC + BD)

Hence, proved.

6. The length of two sides of a triangle are 12 cm and 15 cm. Between what two measures should the length of the third side fall?

Solution:Sum of two sides

= 12 cm + 15 cm = 27 cm

Difference of the two sides

= 15 cm – 12 cm = 3 cm

∴ The measure of third side should fall between 3 cm and 27 cm.

**Exercise 6.5**

1. PQR is a triangle, right angled at P. If PQ = 10 cm and PR = 24 cm, find QR.

Solution:In right angled triangle PQR, we have

QR^{2} = PQ^{2} + PR^{2} From Pythagoras property)

= (10)^{2} + (24)^{2}

= 100 + 576 = 676

∴ QR = √676 = 26 cm

The, the required length of QR = 26 cm.

2. ABC is a triangle, right angled at C. If AB = 25 cm and AC = 7 cm, find BC.

Solution:In right angled ∆ ABC, we have

BC^{2} + (7)^{2} = (25)^{2} (By Pythagoras property)

⇒ BC^{2} + 49 = 625

⇒ BC^{2} = 625 – 49

⇒ BC^{2} = 576

∴ BC = √576 = 24 cm

Thus, the required length of BC = 24 cm.

3. A 15 m long ladder reached a window 12 m high from the ground on placing it against a wall at a distance a. Find the distance of the foot of the ladder from the wall.

Solution:Here, the ladder forms a right angled triangle.

∴ a^{2} + (12)^{2} = (15)^{2} (By Pythagoras property)

⇒ a^{2}+ 144 = 225

⇒ a2 = 225 – 144

⇒ a^{2} = 81

∴ a = √ 81 = 9 m

Thus, the distance of the foot from the ladder = 9m

4. Which of the following can be the sides of a right triangle?

(i) 2.5 cm, 6.5 cm, 6 cm.

(ii) 2 cm, 2 cm, 5 cm.

(iii) 1.5 cm, 2 cm, 2.5 cm

Solution:i) Given sides are 2.5 cm, 6.5 cm, 6 cm.

Square of the longer side = (6.5)^{2} = 42.25 cm.

Sum of the square of other two sides

= (2.5)^{2} + (6)^{2} = 6.25 + 36

= 42.25 cm.

Since, the square of the longer side in a triangle is equal to the sum of the squares of other two sides.

∴ The given sides form a right triangle.

(ii) Given sides are 2 cm, 2 cm, 5 cm .

Square of the longer side = (5)^{2} = 25 cm Sum of the square of other two sides

= (2)^{2} + (2)^{2} =4 + 4 = 8 cm

Since 25 cm ≠ 8 cm

∴ The given sides do not form a right triangle.

(iii) Given sides are 1.5 cm, 2 cm, 2.5 cm

Square of the longer side = (2.5)^{2} = 6.25 cm Sum of the square of other two sides

= (1.5)^{2} + (2)^{2} = 2.25 + 4

Since 6.25 cm = 6.25 cm = 6.25 cm

Since the square of longer side in a triangle is equal to the sum of square of other two sides.

∴ The given sides form a right triangle.

5. A tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tree . Find the original height of the tree.