RBSE Solutions for Class 7 Maths Chapter 10 Construction of Triangles Additional Questions is part of RBSE Solutions for Class 7 Maths. Here we have given Rajasthan Board RBSE Class 7 Maths Chapter 10 Construction of Triangles Additional Questions.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 7 |

Subject |
Maths |

Chapter |
Chapter 10 |

Chapter Name |
Construction of Triangles |

Exercise |
Additional Questions |

Number of Questions |
15 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 7 Maths Chapter 10 Construction of Triangles Additional Questions

**Multiple Choice Questions**

Question 1

The sum of all three (RBSESolutions.com) internal angles of a triangle is :

(A) 80°

(B) 180°

(C) 100°

(D) 90°

Question 2

The sides of an equilateral triangle are:

(A) equal

(B) unequal

(C) different

(D) smaller – greater

Question 3

In figure find ∠X, if XY = YZ and (RBSESolutions.com) value of ∠Y = 58° :

(A) 71°

(B) 60°

(C) 61°

(D) 90°

Question 4

(Hypotenuse)^{2} = (Perpendicular)^{2} + (Base)^{2} is the (RBSESolutions.com) theorem of :

(A) Pythagorous

(B) Euclid

(C) Newton

(D) Dekart

Question 5

The sides of a right angle triangle are 3 cm and 4 cm. Then the measure of third side will be :

(A) 8

(B) 10

(C) 6

(D) 5

Answer:

1. (B), 2. (A), 3. (C), 4. (A), 5. (D).

**Fill in the blanks**

(i) The sum of two acute angles (RBSESolutions.com) in a right angled triangle is ………….. .

(ii) The sum of two sides of a triangle is greater then the …………. .

(iii) In a triangle, there are ………….. and ………… .

(iv) The meausre of exterior angle in a triangle is equal to the ………… of the two opposite interior angles.

Answer:

(i) 90°,

(ii) third side

(iii) 3 sides, 3 angles,

(iv) Sum.

**True/False**

(i) Triangle has four sides.

(ii) A triangle has three altitudes.

(iii) The difference (RBSESolutions.com) of any two sides of a triangle is smaller than the third side.

(iv) Three sides are equal in an isosoceles triangle.

Answer:

(i) False

(ii) True

(iii) True

(iv) False

**Very Short Answer Type Questions**

Question 1

In ∆ABC ∠A = 50°, ∠B = 70°, then find (RBSESolutions.com) the value of ∠C.

Solution:

Given ∠A = 50° and ∠B = 70°

Sum of all three angles of a triangle is 180°.

∠A + ∠B + ∠C = 180°

⇒ 50° + 70° + ∠C = 180°

⇒ 120° + ∠C = 180°

⇒∠C = 180° – 120° = 60°

Question 2

In a right angled (RBSESolutions.com) triangle base = 8 cm, perpendicular = 6 cm, then find the value of hypotenuse.

Solution:

In a right angled triangle Base = 8 cm,

perpendicular = 6 cm

By Pythagorous theorem

∵ (Hypotenuse)^{2} = (Perp)^{2} + (Base)^{2}

(Hypotenuse)^{2} = 6^{2} + 8^{2}

⇒ (Hypotenuse)^{2} =36 + 64

⇒ (Hypotenuse)^{2} = 100,

∴ Hypotenuse = \(\sqrt { 100 }\) = 10 cm.

Question 3

In a ∆ABC, AB = 3 cm, BC = 2 cm and CA = 10 cm. Is it possible to construct (RBSESolutions.com) a triangle on the basis of given measures ? If not, why ?

Solution:

In ∆ABC

AB = 3 cm, BC = 2 cm and CA = 10 cm

∵ 3 + 2 ⊁ 10, ⇒ AB + BC ⊁ AC

∴ ∆ is not possible because sum of any two side of a triangle is less than third side.

Question 4

Construct a scalene triangle.

Solution:

Let in ∆ABC, AB = 2 cm, BC = 6 cm, CA = 7.2 cm which are different in lengths.

Thus, ∆ABC a scalene triangle.

**Short and Long Answer Type Questions**

Question 1

Construct a right angled triangle (RBSESolutions.com) whose side BC = 4 cm, hypotenus AC = 5 cm and ∠B = 90°.

Solution:

Steps of construction :

1. Draw a line segment BC = 4 cm.

2. Make an angle 90° at point B.

3. Make an arc of radius 5 cm from point C which cuts the line making ∠90° at point A.

4. Join AC

Thus ∆ABC is required triangle.

Question 2

Construct an isosceles triangle whose (RBSESolutions.com) base is 3 cm and one of the other two sides is 5 cm.

Solution:

Steps of construction :

Given a ∆ABC where BC = 3 cm and AC = AB = 5 cm

Steps of construction :

1. Draw a line BC = 3 cm as base

2. Make an arc of radius 5 cm from point B and also same arc from point C which intersect the prior arc at point A.

3. Thus ∆ABC is required (RBSESolutions.com) isosceles triangle.

Question 3

Construct a ∆PQR where QR = 8 cm, ∠Q = 120° and ∠R = 30°.

Solution:

Steps of construction:

1. Draw a line QR = 8 cm.

2. Make an angle 120° ∠AQR at point Q with the help of compass.

3. Make ∠BRQ = 30° at point R with (RBSESolutions.com) the help of compass.

4. Line AQ and BR intersect at point P each other.

Thus ∆PQR is required triangle.

Question 4

Construct a ∆ABC where ∠A = 90°, side AC = 5.4 cm and hypotenuse BC = 10 cm.

Solution:

Given in ∆ABC where ∠A = 90°, AC = 5.4 cm and hypotenuse BC = 10 cm

Steps of construction :

1. Draw a line (RBSESolutions.com) segment AD of any length.

2. Make ∠EAD at point A with the help of compass.

3. Make an arc of radius 5.4 cm from point A which cuts AE at point C.

4. Now taking C as centre, make an arc of radius 10 cm which cuts AD at point B.

5. Join C and B.

Thus ∆ABC is required right angled triangle.

We hope the RBSE Solutions for Class 7 Maths Chapter 10 Construction of Triangles Additional Questions will help you. If you have any query regarding Rajasthan Board RBSE Class 7 Maths Chapter 10 Construction of Triangles Additional Questions, drop a comment below and we will get back to you at the earliest.

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