SEBA Class 7 Mathematics Chapter 6 The triangle and its properties

SEBA Class 7 Mathematics Chapter 6 The triangle and its properties Solutions English Medium, SEBA Class 7 Maths Notes in English Medium, SEBA Class 7 Mathematics Chapter 6 The triangle and its properties Notes to each chapter is provided in the list so that you can easily browse throughout different chapter Assam Board SEBA Class 7 Mathematics Chapter 6 The triangle and its properties Solutions in English and select needs one.

SEBA Class 7 Mathematics Chapter 6 The triangle and its properties

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Also, you can read the SCERT book online in these sections Solutions by Expert Teachers as per SCERT (CBSE) Book guidelines. SEBA Class 7 General Maths Textual Question Answer. These solutions are part of SCERT All Subject Solutions. Here we have given SEBA Class 7 Mathematics Chapter 6 The triangle and its properties Solutions for All Subject, You can practice these here.

The triangle and its properties

Chapter – 6

1. How many medians are there in a triangle?

Ans: Three medians are there in a triangle.

2. How many altitudes are there in a triangle?

Ans: Three altitudes are there in a triangle.

3. Draw a triangle and display its medians.

Ans: AD. BE and CF are three medians in a triangle.

4. Draw a triangle and display its altitudes.

Ans:

5. Name the angle opposite to side LM in ΔLMN.

Ans: The angle opposite to side LM in Δ LMN is ∠LNM.

6. Name the side opposite to vertex Q of Δ PQR.

Ans: The side opposite to vertex Q of ΔPQR is PR.

7. Name the vertex opposite to side RT of ΔRST.

Ans: The vertex opposite to side RT of ΔRST is S.

8. (i) Tick (√) the correct answer.

In ∆PQR, PM is-

(a) A median.

(b) An altitude.

(c) A bisector of QR.

(d) A side.

Ans: (b) An altitude.

(ii) If D is the midpoint of side QR of ∆PQR, then PD is-

(a) perpendicular bisector on QR.

(b) altitude.

(c) median.

(d) opposite side of QR.

Ans: (c) Median.

1. Fill in the blanks:

(a) The angles in the interior of a triangle are called _____________.

Ans: Interior angles.

(b) The angles exterior to a triangle are called.

Ans: Exterior angles.

2. Find the value of x from the diagrams given below:

Ans: ∠x = 50° + 70° = 120°

(b) 

Ans: x + 50° = 100° 

⇒ x = 100° – 50° = 50°

Ans: x + 90° = 125°

 ⇒ x = 125° – 90° = 35°

Ans: x + 30° = 120°

⇒ x = 120° – 30° = 90°

(e)

Ans: x + 35° = 75°

⇒ x = 75° – 35° = 40°

Ans: x + 70° = 140°

⇒ x = 140° – 70° = 70°

3. In a triangle, an exterior angles is 70° and one of its interior opposite angles is 25 deg. Find the value of the other interior opposite angle.

Ans: 

70° = x + 25°

⇒ x = 70° – 25° 

= 45°

4. Two remote interior angles of a triangle are 60 deg and 80°. Find the value of the exterior angle.

Ans: 

∠PRS = ∠P + ∠Q

= 60° + 80° 

= 140°

5. In a triangle, an exterior angles is 114° and one of its interior opposite angle is 25° Find the value of other distant interior angle.

Ans: 

∠ACD = ∠A + ∠B 

⇒ 114° = ∠A + 25° 

⇒ ∠A = 114° – 25° 

⇒ ∠A = 89°

6. Two remote interior angles of an exterior angle of a triangle are 49º and 41°. Find the value of the exterior angle. (Remote interior angles are also called opposite angles).

Ans:

∠P + ∠Q = ∠PRS

⇒ 49° + 41° = ∠PRS

⇒ ∠PRS = 90°

1. Find the value of x from the figs given below.

Ans: x + 50° + 60° = 180° 

⇒ x = 180° – 110° = 70°

Ans: x + 90° + 30° = 180°

⇒ x = 180°- 120° = 60°

Ans: ∵ Isosceles triangle

∴ ∠A = ∠C = 45° 

∴ ∠A + ∠B + ∠C = 180° 

⇒ 45° + x + 45° = 180°

⇒ x = 180° – 90° = 90°

Ans: ∵ Isosceles triangle x = 40°

Ans: ∵ ∆ABC is an isosceles right angled triangle

∴ AB = CB

∠A = ∠C = 45°

⇒ x = 45°

2. Find the values of x and y from the figs given below:

Ans: y = 80° [∴ Vertically opp. angles]

∴ x + 50° + y = 180° 

⇒ x + 50° + 80° = 180° 

⇒ x + 180° – 130° = 50°

Ans: x = 60° [Vertically opp. angles]

Now, x + y + 30 = 180° 

⇒ 60° + y + 30° = 180° 

⇒ y = 180° – 90° = 90°

(c) 

Ans: All three triangles have equal angles

∴ x + x + x = 180° [∴ x = y]

⇒ 3x = 180°

∴ x = 180°/3 = 60°

⇒ y = 60°

3. In a triangle, the measure of one angle is 60°. What will be the measure of the other two angles?

(a) 50°, 40°

Ans: 150° + 40° + 60° = 150° ≠ 180°

(b) 40°, 60°

Ans: 40° + 60° + 60° = 160° ≠ 180°

(c) 60°, 70°

Ans: 60° + 70° + 60° = 190° ≠ 180°

(d) 50°, 70°

Ans: 50° + 70° + 60° = 180°

∴ 50°, 70°

4. Find the value of ∠P from the figures given below-

Ans: ∠P = ∠R + ∠Q

∠P = 52° + 47° = 99°

5. Two angles of a triangle are 30° and 80°. Find the measure of the third angle.

Ans: Let the third angle be x°

∴ x° + 30° + 80° = 180°

⇒ x° = 180° – 110° = 70°

6. One angle of a triangle is 80° and the other two angles are equal. What are the measures the other tow angles?

Ans: One angle of a triangle is 80° and the other two angles are equal. 

(Let ∠B = ∠C = X )

∴ ∠A + ∠ B + ∠C = 180° 

⇒ 80° + ∠B + ∠C = 180°

⇒ 80° + x + x = 180°

⇒ 2x = 180° – 80° = 100°

 ⇒ x = (100°/2) = 50°

∴ ∠B = ∠C = 50°

7. The ratio of three angles of a triangle is 1:2:1. Find the measures of the angles.

Ans: The ratio of three angles of a triangle is 1:2:1, let three angles are x, 2x and x. 

∴ x + 2x + x = 180°

⇒ 4x = 180°

⇒ x = (180°)/4 = 45°

∴ The measure of the angles are 45°, 90°, 45°

8. The angles of a triangle are (x + 21°) (x – 20°) and (2x – 45°) Find x.

Ans: Three angles of a Delta are, (x + 21°), (x – 20°) and (2x – 45°)

∴ (x + 21°) + (x – 20°) + (2x – 45°) = 180° 

⇒ 4x + (21°- 20°- 45°) = 180°

⇒ 4x = 180° + 44° 

⇒ 4x = 224°

∴ x = 224°/4 = 56°

x = 56°

9. The ratio of three angles of a traingle is 1:2:3. Find the measures of the angles.

Ans: The ratio of three angles of a triangle is 1:2:3. Let three measures of angles are x, 2x, 3x. 

∴ x + 2x + 3x = 180°

⇒ 6x = 180° 

⇒ x = 180°/6 = 30°

∴ The measures of the angles are 30°, 60° and 90°.

10. In ∆ ABC, ∠A + ∠B = 116°, ∠B + ∠C = 126 ° Find out the measures of the interior angles of the triangle.

Ans: ∴ ∠A + ∠B + ∠C = 180° 

⇒ 116° + ∠C = 180°

⇒ ∠C = 180° – 116° = 64°

∴ ∠B + ∠C = 126° 

⇒ ∠B + 64° = 126°

⇒ ∠B = 126° – 64° = 62° 

∴ ∠A + ∠B + ∠C = 180° 

⇒ ∠A + 62° + 64° = 180°

⇒ ∠A + 126° = 180°

⇒ ∠A = 180° – 126° = 54°

∴ ∠A = 54°, ∠B = 62° and ∠C = 64°

11. In ∆ABC, 2 ∠A = 3 ∠B = 6 ∠C. Find out ∠A, ∠B, ∠C

Ans: ∆ABC,2 ∠A = 3 ∠B = 6 ∠C 

Now, 3 ∠B = 6 ∠C 

⇒ ∠B = 6/3 ∠C = 2 ∠C 

2∠A = 6∠C

∴ ∠A = 6/2 ∠C = 3 ∠C 

Now, ∠A + ∠B + ∠C = 180° 

⇒ 3 ∠C  + ∠A + 3∠C = 180°

⇒ 6 ∠C = 180°

⇒ ∠C = 30°

∴ ∠A = 3 × 30° = 90° 

∠B = 2 × 30° = 60° 

∠C = 30°

12. In the figure given below ∠CAB = 40°, AC = AB and BC = BD Find out

(a) ∠ACB and

(b) ∠CDB 

Ans: In the figure, ∠CAB = 40°

AC = AB 

∴ ∠C = ∠B 

Now, ∠A + ∠B + ∠C = 180° 

⇒ 40° + 2 ∠B = 180°

⇒ 2 ∠B = 180° – 40°

∴ ∠B = 140°/2 = 70°

∴ ∠C = 70°

∴ ∠CBD = 40° + 70° = 110°

That, ∠ACB = 70° 

Again, BC = BD 

∴ ∠BCD = ∠CDB 

∴ ∆BCD get from, ∠CBD + ∠BCD + ∠CDB = 180°

∠CBD + ∠BCD +∠CBD = 180

⇒ ∠110° + 2 ∠CDB = 180°

⇒ 2∠CDB = 180° – 110°

⇒ 2∠CDB = 70° 

∴ ∠CDB = 70°/2 = 35°

∴ (a) ∠ACB = 70°, (b) ∠CDB = 35°

1. In the following figure AB = 10cm AC = 17 cm and AD = 8cm. Determine BC.

Ans: In the figure, AB = 10 cm

BC = 17 cm and AD = 8 cm

∴ BD² = AB² – AD² 

= 100² – 8² = 100 – 64 = 36

∴ CD = BC – BD = (17 – 6) cm = 11 cm

∴ AC² = AD² + CD²

= 8² + 11² = 64 + 121 = 185

2. In the perimeter of a triangle is 15 cm and two sides are 5 cm and 7 cm then what is the length of the third side.

Ans: Perimeter = 15cm, two sides are 5 cm and 7 cm.

∴ Measure of third side

= 15 – (5 + 7) cm

= (15-12) cm 

= 3cm.

3. In a rectangle the lengths of two adjacent sides are 16 cm and 12 cm. Find out the lengths of two diagonals.

Ans: AD = 16cm, 

CD = 12cm

∴ ∆ACD- [In ∆ACD] 

AC² = AD²  + CD²

⇒ AC² = 16² + 12²

⇒ AC² = 256 + 244 = 400

⇒ AC² = √(400) = 20

∴ Length of each diagonals = 20cm 

4. O is an external point of ∆ABC. Show that 2( OA + OB+OC) > AB + BC + CA.

Ans: ∆ABC, ‘O’ An external point

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

Now, ∆ΟΑΒ, OA + OB > AB ………….. (i)

∆ОАС, OA + OC > AC ………. (ii)

∆ΟBC ,OB + OC > BC ……….. (iii)

From eqn (i), (ii) and (iii) we get, 2(OA + OB + OC) > AB + BC + CA

5. Will the sides of the following measures from right angled triangle?

(a) 5, 12, 13

Ans: 5, 12, 13,

∴  5² + 12² = 13²

⇒ 25 + 144 = 169 

⇒ 25 + 144 = 169

⇒ 169 = 169

(b) 3, 4, 5

Ans: 3, 4, 5

∴ 3² + 4² = 5 2 

⇒ 9 + 16 = 25°

⇒ 25 = 25

(c) 6, 8, 10

Ans: 6, 8, 10

∴  6² + 8² = 10²

⇒ 36 + 64 = 100

⇒ 100 = 100

(d) 6, 7, 8

Ans: 6, 7, 8

∴  6²  + 7²  = 8²  

⇒ 36 + 49 = 64 

⇒ 85 ≠ 64

∴ (a), (b), (c) the sides of measures from right angled ∆.

6. Is it possible to four traingles with sides of the following measures?

(a) 3 cm., 4 cm, 5 cm.

Ans: We know that the sum of two sides of a triangle is greater than the third side.

Now, (a) 3 cm. + 4 cm.

= 7 cm. > 5 cm.

∴ A triangle can be obtained with the given measurements.

(b) 5 cm, 7 cm., 12 cm.

Ans: 5 cm. + 7 cm. = 12 cm.

∴ A triangle cannot be found with the given measurements.

(c) 3.4 cm., 2 cm., 5.8 cm.

Ans: 3.4 cm + 2 cm = 5.4 cm >5.8 cm

∴ A triangle cannot be found with the given measurements.

(d) 6 cm., 7 cm., 14 cm.

Ans: 6 cm + 7 cm = 13 cm ≯ 14 cm

∴ A triangle cannot be found with the given measurements.

7. Prove that in a quadrilateral ABCD.(AB + BC + CD + DA) > (AC + BD)

Ans: Given,

ABCD is a quadrilateral.

We know it

The measures of two sides of a triangle are larger than the third side.

∴ From ∆ABC 

AB + BC > AC → (i)

Again, ∆ABD 

AD + AB > BD → (ii)

From, ∆ADC 

AD + CD > AC → (iii)

And from ∆BDC 

BC + CD > BD → (iv)

Now, (i) + (ii) + (iii) + (iv) 

⇒ AB + BC + AD + AB + AD + CD + BC + CD > AC + BD + AC + BD

⇒ 2AB + 2BC + 2AD + 2CD > 2AC + 2BD

⇒ 2(AB + BC + AD + CD) > 2(AC + BD)

∴ AB + BC + AD + CD > AC + BD

Find out the correct answers for Q. No. 1 to Q. No. 12.

Q. 1. Value of x in the adjoining triangle is-

(a) 40°

(b) 60°

(c) 35°

(d) 180°

Ans: (b) 60°

2. Value of ‘x’ in the adjoining figure is–

(a) 180°

(b) 55°

(c) 90°

(d) 60°

Ans: (d) 60°

3. In ∆ABC, if ∠A = 35°, ∠B = 65° then ∠C is

(a) 50°

(b) 80°

(c) 30°

(d) 60°

Ans: (b) 80°

4. Hypotenuse of a right-angled triangle is 17cm. If one side is measures 8 cm, then the measure of the third side is-

(a) 15 cm

(b) 12 cm

(c) 13 cm

(d) 25 cm

Ans: (a) 15 cm

5. In ∆ABC if ∠A = 72°, ∠B = 63° then ∠C is

(a) 45°

(b) 80°

(c) 30°

(d) 60°

Ans: (a) 45°

6. In a right-angled triangle, one of the acute angles is 36° then the other acute angle is-

(a) 55°

(b) 54°

(c) 51°

(d) 52°

Ans: (b) 54°

7. The value of ‘x’ in the adjoining figure is-

(a) 5 cm

(b) 7 cm

(c) 3 cm

(d) 4 cm

Ans: (a) 5 cm

৪. The value of ‘x’ in the figure is-

(a) 15 cm

(b) 17cm

(c) 13 cm

(d) 14 cm

Ans: (b) 17 cm

9. In a right-angled triangle ABC, if ∠C = 90°, AC = 5cm, BC = 12 cm then AB is-

(a) 7 cm

(b) 17 cm

(c) 13 cm

(d) 14 cm

Ans: (c) 13 cm

10. In triangle PQR, if ∠P = 90°, PQ = 3 cm, PR = 4 cm, then QR is-

(a) 7 cm

(b) 17 cm

(c) 5 cm

(d) 13 cm

Ans: (c) 5 cm

11. The value of ‘x’ in the figure is-

(a) 90°

(b) 60°

(c) 80°

(d) 40°

Ans: (b) 60°

12. Pythagoras Property is satisfied if the triangle is…

(a) Obtuse-angled.

(b) Right-angled.

(c) Acute-angled.

Ans: (b) Right-angled.

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