This chapter delves into the world of triangles, their classifications, and key properties.

**Key Concepts:**

**Triangle:**A polygon with three sides, three angles, and three vertices.**Types of triangles based on sides:**Equilateral (all sides equal), Isosceles (two sides equal), Scalene (no sides equal).**Types of triangles based on angles:**Acute-angled (all angles less than 90°), Obtuse-angled (one angle greater than 90°), Right-angled (one angle equal to 90°).**Angle Sum Property of a Triangle:**The sum of the interior angles of a triangle is always 180°.**Exterior angle property:**An exterior angle of a triangle is equal to the sum of its two interior opposite angles.**Pythagoras Theorem:**In a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.**Congruence of triangles:**Triangles that have the same shape and size.**Criteria for congruence:**Side-Side-Side (SSS), Side-Angle-Side (SAS), Angle-Side-Angle (ASA), Right Angle-Hypotenuse-Side (RHS).

By understanding these concepts, you can analyze and solve various problems related to triangles.

**Exercise 6.1**

**In ∆PQR, D is the mid-point of ****QR****,**

**PM**** is ____________**

**PD**** is ____________**

**If QM = MR?**

**Ans : **

PM is the altitude

PD is the median.

QM is not equal to 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.**

**Ans : **

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

**Ans : **

∆ABC is an isosceles triangle in which AB = AC

Therefore, in an isosceles triangle, the median from the vertex formed by the equal sides is also the altitude.

**Exercise 6.2**

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

**Ans : **

**Using the exterior angle property:**

**(i)**x = 50° + 70° = 120°**(ii)**x = 65° + 45° = 110°**(iii)**x = 40° + 30° = 70°**(iv)**x = 60° + 60° = 120°**(v)**x = 50° + 50° = 100°**(vi)**x = 30° + 60° = 90°

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

**Ans : **

(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:**

**Ans : **

**(i)**

In triangle ABC, we have:

∠A + ∠B + ∠C = 180°

Substituting the given values:

x + 50° + 60° = 180°

Combining like terms:

x + 110° = 180°

Solving for x:

x = 180° – 110°

x = 70°

**(ii)**

In triangle PQR, we have:

∠P + ∠Q + ∠R = 180°

Substituting the given values:

90° + 30° + x = 180°

Combining like terms:

120° + x = 180°

Solving for x:

x = 180° – 120°

x = 60°

**(iii)**

In triangle XYZ, we have:

∠X + ∠Y + ∠Z = 180°

Substituting the given values:

x + 30° + 110° = 180°

Combining like terms:

x + 140° = 180°

Solving for x:

x = 180° – 140°

x = 40°

**(iv)**

In the given triangle, we have two equal angles, each represented by x. The third angle is 50°.

Using the angle sum property:

x + x + 50° = 180°

Combining like terms:

2x + 50° = 180°

Solving for x:

2x = 180° – 50°

2x = 130°

x = 130° / 2

x = 65°

**(v)**

In the given triangle, all three angles are equal and represented by x.

Using the angle sum property:

x + x + x = 180°

Combining like terms:

3x = 180°

Solving for x:

x = 180° / 3

x = 60°

**(vi)**

In the given triangle, we have two equal angles, each represented by x, and a right angle (90°).

Using the angle sum property:

x + x + 90° = 180°

Combining like terms:

2x + 90° = 180°

Solving for x:

2x = 180° – 90°

2x = 90°

x = 90° / 2

x = 45°

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

**Ans : **

**(i)**

**Exterior angle property:**x = 50° + 120° = 170°**Angle sum property:**50° + y + 120° = 180°, so y = 10°

**(ii)**

**Vertically opposite angles:**x = 80°**Angle sum property:**50° + x + y = 180°, so y = 50°

**(iii)**

**Exterior angle property:**x = 50° + 60° = 110°**Angle sum property:**50° + 60° + y = 180°, so y = 70°

**(iv)**

**Vertically opposite angles:**y = 60°**Angle sum property:**30° + x + 60° = 180°, so x = 90°

**(v)**

**Vertically opposite angles:**y = 90°**Angle sum property:**y + x + x = 180°, so 2x = 90°, and x = 45°

**(vi)**

**Vertically opposite angles:**x = y**Angle sum property:**x + x + y = 180°, so 3x = 180°, and x = 60°

**Exercise 6.4**

**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**

**Ans : ****(i) 2 cm, 3 cm, 5 cm**

- 2 + 3 = 5 Since the sum of two sides is equal to the third side, a triangle cannot be formed with these side lengths.

**(ii) 3 cm, 6 cm, 7 cm**

- 3 + 6 > 7
- 3 + 7 > 6
- 6 + 7 > 3 Since the sum of any two sides is greater than the third side, a triangle can be formed with these side lengths.

**(iii) 6 cm, 3 cm, 2 cm**

- 3 + 2 = 5, which is not greater than 6 Since the sum of two sides is not greater than the third side, a triangle cannot be formed with these side lengths.

**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?**

**Ans : **

**Yes, all three inequalities hold true.**

This is based on a fundamental property of triangles: **the sum of the lengths of any two sides of a triangle is greater than the length of the third side**.

**(i) OP + OQ > PQ:** This is true because triangle OPQ exists, and the sum of lengths of two sides of a triangle is always greater than the third side.

**(ii) OQ + OR > QR:** Similarly, triangle OQR exists, and the sum of lengths of two sides is greater than the third side.

**(iii) OR + OP > RP:** Triangle ORP also exists, and the same property applies.

**3. AM is a median of a triangle ABC.**

**Is AB + BC + CA > 2AM ?**

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

**Ans : **

**In triangle ABM:**

- AB + BM > AM (sum of two sides of a triangle is greater than the third side)

**In triangle AMC:**

- AC + MC > AM (sum of two sides of a triangle is greater than the third side)

Adding both inequalities:

AB + BM + AC + MC > AM + AM

AB + AC + (BM + MC) > 2AM

Since BM + MC = BC, we have:

AB + AC + BC > 2AM

**4. ABCD is a quadrilateral.**

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

**Ans : **Consider the quadrilateral ABCD with diagonals AC and BD. The diagonals divide the quadrilateral into four triangles: ΔABC, ΔBCD, ΔCDA, and ΔABD.

Applying the triangle inequality theorem to each triangle, we get:

In ΔABC: AB + BC > AC

In ΔBCD: BC + CD > BD

In ΔCDA: CD + DA > AC

In ΔABD: DA + AB > BD

Adding these inequalities, we get:

(AB + BC) + (BC + CD) + (CD + DA) + (DA + AB) > AC + BD + AC + BD

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

Dividing both sides by 2, we get:

AB + BC + CD + DA > AC + BD

**5. ABCD is a quadrilateral.**

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

**Ans : **

**Step 1: Analyze the Quadrilateral**

- Consider quadrilateral ABCD with diagonals AC and BD intersecting at point O.

**Step 2: Apply Triangle Inequality Theorem**

- In triangle AOB, we have: AO + BO > AB
- In triangle BOC, we have: BO + CO > BC
- In triangle COD, we have: CO + DO > CD
- In triangle AOD, we have: DO + AO > AD

**Step 3: Combine the Inequalities** Adding all the above inequalities, we get: 2(AO + BO + CO + DO) > AB + BC + CD + DA

**Step 4: Simplify**

- We know that AO + CO = AC and BO + DO = BD.
- Substituting these values, we get: 2(AC + BD) > AB + BC + CD + DA

**Step 5: Reverse the Inequality** To match the original statement, we reverse the inequality:

- AB + BC + CD + DA < 2(AC + BD)

**Therefore, it is proven that AB + BC + CD + DA is less than 2(AC + BD).**

**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?**

**Ans : **

**Triangle Inequality Theorem:** The sum of the lengths of any two sides of a triangle is greater than the length of the third side.

Let the third side be x cm.

**Condition 1:**x < 12 + 15

- x < 27 cm

**Condition 2:**x > 15 – 12

- x > 3 cm

Therefore, the length of the third side should be **greater than 3 cm and less than 27 cm**.

In other words, the third side lies 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.**

**Ans : **

Since triangle PQR is right-angled at P, we can apply the **Pythagoras theorem**.

**Pythagoras theorem:** In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

In this case, QR is the hypotenuse.

So, QR² = PQ² + PR²

Substituting the given values:

QR² = 10² + 24² QR² = 100 + 576 QR² = 676

Taking the square root of both sides:

QR = √676 QR = 26 cm

**Therefore, the length of QR is 26 cm.**

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

**Ans : **

**Pythagoras Theorem:** In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

In this case, AB is the hypotenuse.

So, AB² = AC² + BC²

Substituting the given values:

25² = 7² + BC² 625 = 49 + BC² BC² = 625 – 49 BC² = 576

Taking the square root of both sides:

BC = √576 BC = 24 cm

**Therefore, the length of BC is 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.**

**Ans : **

According to the Pythagoras theorem:

(15)^2 = (12)^2 + a^2

Calculating the squares:

225 = 144 + a^2

Subtracting 144 from both sides:

225 – 144 = a^2

Simplifying:

81 = a^2

Taking the square root of both sides:

a = 9

**Therefore, the distance of the foot of the ladder from the wall is 9 meters.**

**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**

**Ans : **

**Pythagorean Theorem:**

- a² + b² = c²

where:

- a and b are the lengths of the legs
- c is the length of the hypotenuse (the longest side)

Let’s check each set of numbers:

**i) 2.5 cm, 6.5 cm, 6 cm**

- Longest side (hypotenuse) = 6.5 cm
- Checking if it’s a right triangle: 6.5² = 2.5² + 6²
- 42.25 = 6.25 + 36
- 42.25 = 42.25 Since the equation holds true, these sides can form a right triangle.

**ii) 2 cm, 2 cm, 5 cm**

- Longest side (hypotenuse) = 5 cm
- Checking if it’s a right triangle: 5² = 2² + 2²
- 25 = 4 + 4
- 25 ≠ 8 Since the equation doesn’t hold true, these sides cannot form a right triangle.

**iii) 1.5 cm, 2 cm, 2.5 cm**

- Longest side (hypotenuse) = 2.5 cm
- Checking if it’s a right triangle: 2.5² = 1.5² + 2²
- 6.25 = 2.25 + 4
- 6.25 = 6.25 Since the equation holds true, these sides can form a right triangle.

Therefore, the sets of sides **(i) 2.5 cm, 6.5 cm, 6 cm** and **(iii) 1.5 cm, 2 cm, 2.5 cm** can form right triangles.

**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.**

**Ans : **

**Pythagorean Theorem:** In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Let the length of the broken part be ‘x’.

So, x² = 5² + 12²

x² = 25 + 144

x² = 169

x = √169

x = 13 meters

Therefore, the length of the broken part (hypotenuse) is 13 meters.

The original height of the tree is the sum of the height of the break point and the length of the broken part.

Original height = 5 meters + 13 meters = 18 meters

So, the original height of the tree was 18 meters.

**6. Angles Q and R of a APQR are 25° and 65°. Write which of the following is true.**

**(i) PQ****2**** + QR****2**** = RP****2**

**(ii) PQ****2**** + RP****2**** = QR****2**

**(iii) RP****2**** + QR****2**** = PQ****2**

**Ans : **

**Given:**

- ∠Q = 25°
- ∠R = 65°

**To find:**

- Which of the given options is true

**Solution:**

- First, let’s find the third angle ∠P using the angle sum property of a triangle: ∠P + ∠Q + ∠R = 180° ∠P + 25° + 65° = 180° ∠P + 90° = 180° ∠P = 90°
- Since ∠P is 90°, triangle PQR is a right-angled triangle with the right angle at P.

**Applying Pythagoras Theorem:** In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Therefore, **PQ² + PR² = QR²**

**Hence, option (ii) is true.**

**7. Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.**

**Ans : **

**Step 1: Find the breadth** A rectangle forms a right-angled triangle with its diagonal as the hypotenuse. So, we can use the Pythagorean theorem:

- Diagonal² = Length² + Breadth²
- 41² = 40² + Breadth²
- 1681 = 1600 + Breadth²
- Breadth² = 1681 – 1600
- Breadth² = 81
- Breadth = √81
- Breadth = 9 cm

**Step 2: Find the perimeter**

Perimeter of a rectangle = 2 × (Length + Breadth)

= 2 × (40 cm + 9 cm)

= 2 × 49 cm

= 98 cm

**Therefore, the perimeter of the rectangle is 98 cm.**

**8. The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.**

**Ans : **

**Key properties of a rhombus:**

- All sides are equal in length.
- Diagonals bisect each other at right angles.

**Finding the side length:**

**Diagonals bisect each other:**Since the diagonals of a rhombus bisect each other at right angles, we can form four right-angled triangles within the rhombus.**Using Pythagoras theorem:**Consider one of these right-angled triangles. Let ‘s’ be the length of one side of the rhombus.- s² = (16/2)² + (30/2)²
- s² = 8² + 15²
- s² = 64 + 225
- s² = 289
- s = √289
- s = 17 cm

**Finding the perimeter:**

- Perimeter of a rhombus = 4 × side length
- Perimeter = 4 × 17 cm = 68 cm

**Therefore, the perimeter of the rhombus is 68 cm.**