RBSE Solutions for Class 7 Maths Chapter 8 Triangle and its Properties Additional Questions

RBSE Solutions for Class 7 Maths Chapter 8 Triangle and its Properties Additional Questions is part of RBSE Solutions for Class 7 Maths. Here we have given Rajasthan Board RBSE Class 7 Maths Chapter 8 Triangle and its Properties Additional Questions.

Board	RBSE
Textbook	SIERT, Rajasthan
Class	Class 7
Subject	Maths
Chapter	Chapter 8
Chapter Name	Triangle and its Properties
Exercise	Additional Questions
Number of Questions	24
Category	RBSE Solutions

Rajasthan Board RBSE Class 7 Maths Chapter 8 Triangle and its Properties Additional Questions

Multiple Choice Question

Question 1
How many sides are (RBSESolutions.com) there in a triangle?
(A) 3
(B) 4
(C) 2
(D) 5

Question 2
How many parts are there in a triangle?
(A) 2
(B) 6
(C) 5
(D) 3

Question 3
How many maximum (RBSESolutions.com) medians may be in a triangle?
(A) 1
(B) 2
(C) 3
(D) 4

Question 4
The sides of a triangle are 3, 4 and 5 cm, then the triangle will be :
(A) Right angle triangle
(B) Obtuse angle triangle
(C) acute angle triangle
(D) Reflex angle triangle

Question 5
The sum of two interior angles of a triangle is 110°. Then the (RBSESolutions.com) value of its opposite exterior angle will be :
(A) 120°
(B) 110°
(C) 55°
(D) 220°

Question 6
The lengths of the medians drawn in an equilateral triangle are related each other as :
(A) equal
(B) unequal
(C) equal – unequal
(D) none of these

Question 7
The largest side in (RBSESolutions.com) a right angled triangle will be:
(A) Hypotenuse
(B) Base
(C) Perpendicular
(D) Median

Question 8
In an equilateral triangle, which of the following may be on some line segment?
(A) Median
(B) Altitude
(C) Median and Altitude both
(D) None of these

Question 9
The sum of all three (RBSESolutions.com) angles of a triangle will be:
(A) 180°
(B) 90°
(C) 360°
(D) None of these

Question 10
If two angles in a triangle are 60° and 40° then the third angle will be :
(A) 110°
(B) 80°
(C) 40°
(D) 60°
Answer:
1. (A), 2. (B), 3. (C), 4. (A), 5. (B), 6. (A), 7. (A), 8. (C), 9. (A), 10. (B).

Fill in the blanks

(i) The sides of an (RBSESolutions.com) equilateral triangle are ………… .
(ii) (Hypotenuse)² = (Perpendicular)² + (Base)² is called
(iii) A Triangles has ……….. medians
Answer:
(i) 3
(ii) Pythagoraous Theorem
(iii) 3

True/False
(i) The sides of an (RBSESolutions.com) equilateral triangles are 4.
(ii) The sum of all the angles in a triangle is 180°.
(iii) The largest side of the right angled triangle is hypotenuse,
Answer:
(i) False
(ii) True
(iii) True

Very Short Answer Type Questions

Question 1
In a ∆ABC, ∠B is a right angle. If AB = 5 cm and BC = 12 cm then (RBSESolutions.com) find the length of AC.
Solution:
By Pythagorous Theorem

AC² = AB² + BC²
= 5² + 12²
= 25 + 144
= 169
= 13²
or AC² = 13² = AC = 13 cm
Thus, length of AC is 13 cm.

Question 2
Name the given triangle (RBSESolutions.com) after finding the value of x.
Solution:
In ∆DEF,

x° + 2x° + 3x° = 180°
⇒ 6° = 180°
⇒ x° = 30°
∴ ∠E = x° = 30°
∠D = 2x° = 60°
and ∠F = 3x° = 90°
∆DEF is a right angled triangle.

Question 3
How many medians (RBSESolutions.com) may be in a triangle?
Solution:
A triangle has three vertices and three sides. Therefore each vertex makes a median after joining the mid point of its opposite sides. Therefore there may be three medians.

Question 4
Does a median lies inside the triangle completely? If you think, it is not true, then draw a figure showing that position.
Solution:
Yes, a median lies inside the triangle completely.

Question 5
Find the value of x in (RBSESolutions.com) the given figure.
Solution:
Sum of opposite interior angles = Exterior angle

60° + x = 110°
⇒ x = 50°

Short Answer Type Questions

Question 1
The construction of a triangle, on (RBSESolutions.com) the basis of the measures of the sides given below, is possible or not ? Find.
(a) 7.5 cm, 3.5 cm, 5 cm
(b) 4.3 cm, 3.7 cm, 8 cm
(c) 3.5 cm, 6.5 cm, 2.7 cm
Solution:
(a) 7.5 cm + 3.5 cm > 5 cm
3.5 cm + 5 cm > 7.5 cm
5 cm + 7.5 cm > 3.5 cm
∴ Construction of triangle is possible.
(b) 4.3 cm + 3.7 cm = 8 cm.
∴ The construction of triangle is not possible.
(c) 3.5 cm + 2.7 cm < 6.5 cm
∴ The construction of triangle is not possible.

Question 2
The two angles of (RBSESolutions.com) a triangle are 30° and 80°. Find the third angle of this triangle.
Solution:
Let ∆ABC is a triangle where ∠B = 30° and ∠C = 80°.
Now, ∠B + ∠C = 30° + 80° = 110°
∵∠A + ∠B + ∠C = 180° or ∠A + 110° = 180° or ∠A = 180° – 110° = 70°
Thus, third angle is 70°.

Question 3
One angle of a triangle is 80° and remaining two (RBSESolutions.com) angles are equal. Find the measure of each of the equal angles.
Solution:
Let in ∆ABC, = 80° and ∠B = ∠C
∵∠A + ∠B +∠c = 180°
⇒ 80° +∠B + ∠B = 180°
[∵ ∠A= 80° and 2C = ∠B]
⇒ 2∠B = 180° – 80°
⇒ ∠B = ( \(\frac { { 100 }^{ 0 } }{ 2 }\))= 50°
Measure of each of the equal angles = 50°

Long Answer Type Questions

Question 1
Find the value of ∠x and ∠y in each (RBSESolutions.com) of the given figures :


Solution:
(i) Interior angles ∠C + 120° = 180°
∠C = 180° – 120° = 60°
∵ ∆ABC is an isosceles (RBSESolutions.com) triangle where
AB = AC
⇒ ∠B =∠C = 60° = y
∵ ∠A + ∠B +∠C = 180°
x + 60° + 60° = 180°
x = 180° – 120° = 60°
∴ x = 60° and y = 60°

(ii) ∠ABC is an isosceles (RBSESolutions.com) triangle where BC =AC
BC = AC
⇒ ∠A = ∠B = x
and ∠A + ∠B = 90° [∵ ∠C = 90°]
⇒ x + x = 90°
⇒ 2x = 90°
⇒ x = \(\frac { { 90 }^{ 0 } }{ 2 }\) = 45°
But ∠B +1 = 180°
x + y = 180°
⇒ 45° + y = 180°
⇒ y = 180° – 45 = 135°
∴ x = 45° and y = 135°

(iii) ∆ABC is an isosceles (RBSESolutions.com) triangle where AB = AC
AB = AC
⇒ ∠B = ∠C
or ∠B =∠C = x
and ∠A = 92° [vertically opposite angles]
∠A + ∠B + ∠C = 180°
92° + x + x = 180°
⇒ 2x = 180° – 92° = 88°
⇒ x = (\(\frac { { 88 }^{ 0 } }{ 2 }\)) = 44°
and ∠C + y = 180° [Linear pair]
⇒ y = 180° – 44° [∵∠C = x = 41°]
= 136०
∴ x = 44° and y = 136°

Question 2
A tree breaks from the height of 5 m and its upper part touches (RBSESolutions.com) the ground at the distance of 12 m from it base. Find the total height of the tree.
Solution:
Let BCD is a tree which breaks from point C and D is its upper part which touches the ground at point A

∴ CD = AC
BC = 5 m
AB = 12 m
In right triangle, ∆ABC
AC² = AB² + BC²
= 12² + 5²
= 144 + 25 = 169
AC² = 169 = (13)²
AC = 13 m
∴ CD = AC = 13m
Height of the tree (BD) = BC+ CD
= 5 + 13 = 18 m

Question 3
In ∆PQR ∠Q = 25° and ∠R = 65° which of the (RBSESolutions.com) following statement is true?

(i) PQ² + QR² = RP²
(ii) PQ² + RP² = QR²
(iii) RP² + QR² = PQ²
Solution:
∵ The sum of all three angles (RBSESolutions.com) of a triangle is always 180°.
∴ ∠P + ∠Q +∠R = 180°
⇒ ∠P + 25° + 65° = 180°
⇒ ∠P + 90° = 180°
⇒ ∠P = 180° – 90° = 90°
∴ ∆PQR is a right angled triangle where ∠P = 90°
∴ From Pythagorous Theorem PR² + PQ² = QR²
= QR²
∴ (ii) is true.

Question 4
The length of a rectangle is 4.5 cm and its (RBSESolutions.com) diagonal is 5.5 cm. Find its perimeter.

Solution:
Let ABCD is a rectangle (RBSESolutions.com) where AB = 4.5 cm and AC = 5.5 cm
In right angled triangle ∆ABC, ∠B = 90°.
∴ From Pythagorous Theorem
BC² = AC² – AB² = 5.5² – 4.5²
= (5.5 + 4.5) (5.5 – 4.5) = 10 x 1
BC² = 10
BC= 3.16 m
∴ Perimeter of rectangle = 2(AB + BC)
= 2(4.5 + 3.16) cm
= 2 x 7.66
= 15.32 cm

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