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

AC^{2} = AB^{2} + BC^{2}

= 5^{2} + 12^{2}

= 25 + 144

= 169

= 13^{2}

or AC^{2} = 13^{2} = 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

AC^{2} = AB^{2} + BC^{2}

= 12^{2} + 5^{2}

= 144 + 25 = 169

AC^{2} = 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) PQ^{2} + QR^{2} = RP^{2}

(ii) PQ^{2} + RP^{2} = QR^{2}

(iii) RP^{2} + QR^{2} = PQ^{2}

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 PR^{2} + PQ^{2} = QR^{2}

= QR^{2}

∴ (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

BC^{2} = AC^{2} – AB^{2} = 5.5^{2} – 4.5^{2}

= (5.5 + 4.5) (5.5 – 4.5) = 10 x 1

BC^{2} = 10

BC= 3.16 m

∴ Perimeter of rectangle = 2(AB + BC)

= 2(4.5 + 3.16) cm

= 2 x 7.66

= 15.32 cm

We hope the RBSE Solutions for Class 7 Maths Chapter 8 Triangle and its Properties Additional Questions will help you. If you have any query regarding Rajasthan Board RBSE Class 7 Maths Chapter 8 Triangle and its Properties Additional Questions, drop a comment below and we will get back to you at the earliest.

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