RBSE Solutions for Class 7 Maths Chapter 10 Construction of Triangles Additional Questions

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 = 6² + 8²
⇒ (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.

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