The DAV Maths Book Class 7 Solutions Pdf and **DAV Class 7 Maths Chapter 8 Worksheet 1** Solutions of Triangle and Its Properties offer comprehensive answers to textbook questions.

## DAV Class 7 Maths Ch 8 WS 1 Solutions

Question 1.

Look at the figure carefully and complete the statements.

(a) ∠BAY = ∠ABC +

Answer:

∠ACB

(b) ∠CBZ = ∠____ + ∠______

Answer:

∠BAC

(c) ___________ = ∠CAB + ∠ABC

Answer:

∠ACB

Question 2.

In the given figure, find

(a) ∠YRQ

Answer:

∠YRQ = ∠P + ∠Q

= 60° + 70°

= 130°

(b) ∠PRQ

Answer:

∠PRQ = 180° – (60° + 70°)

= 180° -130°

= 50°

Question 3.

One of the exterior angles of a A ABC measures 150°. If one of the interior opposite angle is 75°, find the other interior opposite angles. What type of triangle is this?

Answer:

Let the other exterior opposite angle be x°

Exterior angle BCD = 150°

∠A = 75°

∠BCD = ∠A + ∠B

150° = 75° + ∠B

∠B = 150° – 75° = 75

Hence ∠B = 75°

As ∠A = ∠B = 75°

∆ABC is an isosceles triangle.

Question 4.

One of the exterior angles of a triangle is 100°. The interior opposite angles are equal to each other. Find the measure of these equal interior opposite angles.

Answer:

Let A and B be two equal interior opposite angles for an exterior angle ACD

∠A = ∠B

∠ACD = ∠A + ∠B

100° = ∠A + ∠A [∠A = ∠B]

⇒ 100° = 2∠A

Hence ∠A = ∠B = 50°.

Question 5.

In a triangle one of the exterior angles is 105 75°. Find the measure of all the angles of the triangle.

Answer:

∠ACD = ∠A + ∠B [Exterior angle of a

=> 105° = ∠A + 75° triangle]

=> 105° – 75° = ∠A

∠A = 30°

Hence ∠A = 30°

Question 6.

In the given triangle, find the measure of ∠PRQ.

Answer:

Here, ∠RQS = 140°

∠RPQ = 90°

∠RQS = ∠P + ∠R [Exterior angle]

140° = 90° + ∠R

∠R = 140° – 90° = 50°

Hence ∠R = 50°.

### DAV Class 8 Maths Chapter 8 Worksheet 1 Notes

Scalene Triangle:

The triangle which have its all sides of different length is called scalene triangle.

In the given Fig. 8.1, AB, BC and AC are different.

Isosceles Triangle:

The triangle in which any two sides are equal. In the given Fig. 8.2,

AB = AC, BC is the base and ¿B – ¿C.

Equilateral Triangle:

The triangle whose all sides and all angles are equal. AB = AC = BC and

∠A = ∠B = C (Fig. 8.3).

Exterior Angle:

- Exterior angle of a triangle is equal to the sum of its interior opposite angles.
- In Fig. 8.4, ∠ACD is exterior angle and ∠ACD = ∠A + ∠B.
- Exterior angle is always greater than its interior opposite angles.
- Sum of the angles of a triangle is 180°. (Angle sum property of a triangle).
- The sum of the lengths of any two sides of a triangle is greater than the third side. Triangle inequality.

Pythagoras Theorem:

In a right triangle, the square of the hypotenuse is equal to the sum of the squares of its other two sides.

In Fig. 8.5, ∠B = 900, AC^{2} = AB^{2 }+ BC^{2}

Converse of Pythagoras Theorem:

If the square of the longer side is equal to the sum of the squares of other two sides, then the angle opposite to the longer side is 90°.

In a right triangle, the hypotenuse is the longest side.

Median of a Triangle:

- A median of a triangle is a line segment joining a vertex to the mid-point of the side opposite to the vertex.
- In the given figure, D is the mid-point of the side BC of the ABC and A is the vertex opposite to it. So, AD is median from the vertex A.
- Since a triangle has three vertices, it will have three medians, one from each vertex.

Centroid of a Triangle:

The point of intersection of the medians of a triangle is called the centroid of the triangle. Centroid may be inside or outside of the triangle, in Fig. (8.6) AE, BF and CD are the medians and O is the centroid.

Altitude of a Triangle:

- An altitude of a triangle is a line segment drawn perpendicular from the vertex to the opposite side.
- In the given figure, ∆ABC is an acute angled triangle and AD is the altitude.
- Every triangle has three altitudes from each vertex.

Orthocentre of a Triangle:

The point of intersection of the altitudes of a triangle is called orthocentre. In Fig. (8.7) AD, BE and CF are the altitudes and O is the Orthocentre.

Circumcentre of a Triangle:

The point of intersection of perpendicular bisectors of the sides of a triangle is called circumcentre. In Fig. (8.8) OD, EO and FO are the perpendicular bisectors of BC, AC and AB respectively, O is the circumcentrc.

The circurncircle passes through vertices of the triangle.

Incentre of a Triangle:

- The point of intersection of the angle bisectors of a triangle is called incentre. In Fig. (8.9), AO, BO and CO are the angle bisectors and O is the incentre.
- Incentre always lies inside the triangle.
- The incircie touches each side of a triangle internally.
- Incentre is equidistant from the sides and circumcentre is equidistant from vertices of the triangle.

Example 1:

In the given figure, ∠BCD = 130°, ∠B = 60°, find the measure of ∠A.

Answer:

∠BCD = ∠A + ∠B

(Exterior angle of a triangle is equal to the sum of its interior opposite angles]

⇒ 130° = ∠A + 60°

⇒ ∠A = 130° – 60°

∴ ∠A = 70°

Example 2:

In the given figure, ABC is an isosceless triangle in which AB = AC, ∠A = 80°, find ∠B and ∠C.

Answer:

In ABC, AB = AC.

∴ ∠B = ∠C [Base angles of an isosceles triangle]

∠A + ∠B + ∠C = 180° [Sum of the angles of a triangle is 1800]

⇒ 80° + ∠B + ∠B = 180°

⇒ 80° + 2∠B = 180°

⇒ 2∠B = 180° – 80°

⇒ 2∠B = 100°

⇒ ∠B = 50°

Hence ∠B = ∠C = 50°

Example 3:

In the given figure, AB = AC, ∠A = 70°, find a and b.

Answer:

AB = AC

∴ ∠B = ∠C [Angles opposite to equal sides are equal]

∠A + ∠B + ∠C = 180° [Angle sum property]

⇒ 70° + ∠B + ∠B = 180°

⇒ 2∠B = 180° – 70° = 110°

∴ ∠B = \(\frac{110}{2}\) = 55°

⇒ ∠B = ∠C = 55°

⇒ a + ∠B = 180° [Linear Pair]

⇒ a + 55° = 180°

⇒ a = 180° – 55°

⇒ a = 125°

⇒ b + ∠C = 180°

⇒ b + 55° = 180° [Linear Pair]

⇒ b = 180° – 55° = 125° [Linear Pair]

Hence a = 125° and b = 125°.

Example 4:

In the given figure find the values of x and y.

Answer:

y + 120° = 180° (Linear Pair)

⇒ y = 180° – 120°

⇒ y = 60°

∠A + ∠B + ∠C = 180° [Angle sum propertyl

⇒ 50° + x + 60° = 180°

⇒ x = 180° – 50° – 60° = 70°

Hence x = 70° and y = 60°

Example 5:

State which of the following are Pythagorean triplets.

(i) 8, 10, 6

(ii) 7, 8, 9

(iii) 20, 21, 29,

(iv) 5, 13, 12

Answer:

(i) The triplet is 8, 10, 6 100 = 64 +36

(8)^{2} = 64,

(10)^{2} = 100,

(6)^{2} = 36

⇒ 100 = 64 + 36

⇒ 100 = 100

Hence 8, 10, 6 is a Pythagorean triplet.

(ii) The given triplet is 7, 8, 9

(7)^{2} = 49,

(8)^{2} = 64,

(9)^{2} = 81

⇒ 81 ≠ 49 + 64

⇒ 81 ≠ 113

Hence 7, 8, 9 is not a Pythagorean triplet.

(iii) The given triplet is 20, 21, 29

(20)^{2} = 400,

(21)^{2} = 441,

(29)^{2} = 841

⇒ 841 = 400 + 441

⇒ 841 = 841

Hence 20, 21, 29 is a Pythagorean triplet.

(iv) 5, 13, 12 is the given triplet

(5)^{2}= 25,

(13)^{2} = 169,

(12)^{2} = 144

⇒ 169 = 25 + 144

⇒ 169 = 169

Hence 5, 13, 12 is a Pythagorean triplet.

Example 6:

ABC is a right triangle in which ∠B = 90°. If AB = 7 cm and BC = 24 cm, find AC.

Answer:

In Δ ABC, ∠B = 90°

∴ AB^{2} + BC^{2} = BC^{2} (Pythagoras Theorem)

⇒ AC^{2} = (7)^{2} + (24)^{2}

⇒ AC^{2} = 49 + 576

⇒ AC^{2} = 625

AC = \(\sqrt{625}\)

∴ AC = 25

Hence AC = 25 cm.

Example 7:

Find the measure of all the exterior angles of a triangle ABC.

Answer:

Let the three exterior angles of Δ ABC be x°, y° and z° respectively.

∴ ∠x = ∠B + ∠C (Exterior angle) ……………..(i)

Similarly ∠y = ∠A + ∠C ………….(ii)

and ∠z = ∠A + ∠B ………………(iii)

Adding (i), (ii) and (iii) we get,

∠x + ∠y + ∠z = 2∠A + 2∠B + 2∠C

= 2(∠A + ∠B + ∠C)

= 2 × 180° (Angle Sum Property)

Hence the sum of all the exterior angles is 360°.

Example 8.

In ABC, ∠A = (2x + 15)°, ∠B = (3x – 20)° and ∠C = (x + 65)°, find x.

Answer:

∠A + ∠B + ∠C = 180° (AngIe Sum Property)

(2x + 15)° + (3x – 20)° + (x + 65)° = 180°

2x + 15 + 3x – 20 + x + 65 = 180°

⇒ 6x + 60° = 180°

⇒ 6x = 180° – 60°

⇒ 6x = 120°

⇒ x = \(\frac{120}{6}\) = 20°

Hence x = 20°.