RBSE Solutions for Class 9 Maths Chapter 6 Rectilinear Figures Additional Questions is part of RBSE Solutions for Class 9 Maths. Here we have given Rajasthan Board RBSE Class 9 Maths Solutions Chapter 6 Rectilinear Figures Additional Questions.

Board |
RBSE |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 6 |

Chapter Name |
Rectilinear Figures |

Exercise |
Additional Questions |

Number of Questions Solved |
34 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 9 Maths Solutions Chapter 6 Rectilinear Figures Additional Questions

**Multiple Choice Questions**

Question 1.

The measure of an interior angle (RBSESolutions.com) of a regular pentagon is:

(A) 72°

(B) 108°

(C) 118°

(D) 540°

Solution:

(B) 108°

Question 2.

Each interior angle of a regular polygon measures 135°, then the polygon is:

(A) a parallelogram

(B) a hexagon

(C) an octagon

(D) a decagon

Solution:

(C) an octagon

Question 3.

The sum of all the interior (RBSESolutions.com) angles of a hexagon is:

(A) 5 right angles

(B) 6 right angles

(C) 8 right angles

(D) 12 right angles

Solution:

(C) 8 right angles

Question 4.

The sum of the exterior angles of a regular heptagon is:

(A) 180°

(B) 1080°

(C) 480°

(D) 360°

Solution:

(D) 360°

Question 5.

In a ∆PQR, if ∠P + ∠Q = ∠R then type (RBSESolutions.com) of triangle is:

(A) scalene triangle

(B) equilateral triangle

(C) isosceles triangle

(D) right-angled triangle

Solution:

(D) right-angled triangle

Question 6.

Every triangle must have at least two:

(A) acute angles

(B) obtuse angle

(C) right angles

(D) None of the above

Solution:

(A) acute angles

Question 7.

In a right angle triangle (RBSESolutions.com) one angle is 24°, then third angle will be:

(A) 66°

(B) 156°

(C) 114°

(D) None

Solution:

(A) 66°

Question 8.

Three angles of quadrilateral are 47°, 102° and 111°, then the fourth angle is equal to:

(A) 102°

(B) 100°

(C) 360°

(D) 260°

Solution:

(B) 100°

Question 9.

The number of diagonals in (RBSESolutions.com) a hexagon is equal to:

(A) 6

(B) 7

(C) 8

(D) 9

Solution:

(D) 9

Question 10.

In the adjoining figure, MN || BC. If ∠BAC = 80°, ∠CAN = 49°, then ∠ABC is equal to:

(A) 60°

(B) 50°

(C) 80°

(D) 40°

Solution:

(A) 60°

**Very Short Answer Type Questions**

Question 1.

In the given ∆ABC, the angle (RBSESolutions.com) bisectors of ∠B and ∠C meet at O. If ∠A = 70°, then find ∠BOC.

Solution.

As we know that

∠BOC = 90° + \(\frac { 1 }{ 2 }\) ∠A

= 90° + \(\frac { 1 }{ 2 }\) x 70°

= 90° + 35°

= 125°

Question 2.

Can a triangle have two right angles?

Solution.

No, if a triangle have two right angles, then sum (RBSESolutions.com) of two angles will be 180° which is not possible.

Question 3.

In the given figure, ∠ACD = 110°, ∠BAC = 45°, then find ∠ABC.

Solution.

∠A + ∠B = ∠ACD

Reason: ext. angle is equal to sum of opposite interior angles

∠B = 110° – 45° = 65°.

Question 4.

Determine the number of sides the polygon (RBSESolutions.com) for which the ratio of the sum of the interior angles to the sum of the exterior angle is 5 : 1.

Solution.

Suppose number of sides be n, then

\(\frac { sum of interior angles }{ sum of exterior angles }\) = \(\frac { 5 }{ 1 }\)

Question 5.

Evaluate x from the figure.

Solution.

\(\frac { 3x }{ 2 }\) + x = 180° (Linear pair axiom)

⇒ 5x = 360°

⇒ x = 72°

Question 6.

If a polygon has 10 sides. Find the (RBSESolutions.com) number of diagonals.

Solution.

No. of diagonals = \(\frac { 10(10-1) }{ 2 }\) – 10 = 45 – 10 = 35

Question 7.

State whether the following statement is true or false, give reason. The angles of a certain quadrilateral are 50°, 60°, 112°, 130°.

Solution.

Sum of the four angles of a quadrilateral is 50° + 60° + 112° + 130° = 352°

But sum of all the four angles of a quadrilateral is always 360°.

Hence given statement is false.

Question 8.

In any regular polygon interior angle is double (RBSESolutions.com) of exterior angle. Find the sides of the polygon.

Solution.

Suppose exterior angle be x Interior angle = 2x

⇒ x + 2x = 180°

⇒ 3x = 180°

⇒ x = 60°

sides = \(\frac { 360 }{ 60 }\) = 6

Question 9.

In one angle of a triangle is obtuse, what can you say about the remaining two angles.

Solution.

The other two angles must be acute.

Question 10.

In a triangle ABC, if ∠A = 58°, ∠B = 67°, then find ∠C.

Solution.

∠C = 180° – (58° + 67°) = 180° – 125° = 55°

**Short Answer Type Questions**

Question 1.

An exterior angle of a triangle is 110° and one (RBSESolutions.com) of the interior opposite angle is 30°. Find the other two angles of the triangle.

Solution.

In ∆ABC,

∠ACD = ∠ABC + ∠BAC

110° = 30° + ∠BAC

∠BAC = 80°

Now, ∠ACB = 180° – 110° = 70°

Question 2.

Two angles of a triangle are equal and (RBSESolutions.com) the third angle is greater than each of the equal angles by 21°. Find the measure of the angles.

Solution.

Let equal angles be x each and third angle = x + 21°

Then according to question, we have

x + x + (x + 21°) = 180° (angle sum property)

⇒ 3x + 21° = 180°

⇒ 3x = 159°

⇒ x = 53°

Measures of each angles are 53°, 53° and 74°.

Question 3.

Find the number of sides of (RBSESolutions.com) a polygon which has 14 diagonals.

Solution.

Number of diagonals of a polygon of n sides = \(\frac { n(n-1) }{ 2 }\) – n

⇒ 14 = \(\frac { n(n-1) }{ 2 }\) – n

⇒ 28 = n^{2} – n – 2n

⇒ n^{2} – 3n – 28 = 0

⇒ n^{2} – 7n + 4n – 28 = 0

⇒ n(n – 7) + 4(n – 7) = 0

⇒ (n – 7) (n + 4) = 0

⇒ n = 7

Question 4.

The sum of the base angles of a triangle is 130° and their difference is 40°. Find the measures of all its angles.

Solution.

Suppose base angles are B and C

According to problem

∠B + ∠C = 130° (given)

∠B – ∠C = 40° (given)

⇒ 2 ∠B = 170°

⇒ ∠B = 85°

∠C = 130° – 85° = 45°

Hence, ∠B = 85°, ∠C = 45° and ∠A = 50°.

Question 5.

In ΔPQR, if 2∠P = 3∠Q = 6∠R. calculate (RBSESolutions.com) measures of each of the angles.

Solution.

Let 2∠P = 3∠Q = 6∠R = x

Question 6.

In an isosceles triangle, the measure (RBSESolutions.com) of the greatest angle is 96°. Find the measure of each of the equal angles.

Solution.

According to question

x + x + 96° = 180° (by angle sum property of a triangle)

⇒ 2x + 96° = 180°

⇒ 2x = 84°

⇒ x = 42°

Measures of each equal angles of an isosceles triangle is 42°.

Question 7.

In figure, find ∠x, ∠y and ∠ACD where line BA || FE.

Solution.

BA || FE

∠x = 42° (alternate angle)

∠ACD = ∠x + 66° (ext. angle is equal to sum of opposite interior angles)

∠ACD = 42° + 66° = 108°

But ∠y + ∠ACD = 180°

∠y = 180° – 108° (linear pair axiom)

∠y = 72°

∠x = 42° and ∠y = 72°

Question 8.

The sum of two angles is 123° and their (RBSESolutions.com) difference is 13°. Find all the angles of the triangle.

Solution.

Let two angles are x and y, then according to question

∠x + ∠y = 123° …(i)

and ∠x – ∠y = 13° …(ii)

Adding (i) and (ii), we get

2∠x = 136° ⇒ ∠x = 68°

Now substituting ∠x = 68° in equation (i), we get

68° + ∠y = 123°

⇒ ∠y = 123° – 68° = 55°

Third angle = 180° – 123° = 57°

Hence angles of a triangle are 68°, 55°, 57°.

Question 9.

In ΔABC, ∠A = 2∠B and 2∠C = ∠A + ∠B. Find (RBSESolutions.com) the angles.

Solution.

∠A + ∠B + ∠C = 180° …(i) (by angle sum property of a triangle)

But ∠A + ∠B = 2∠C (given)

2∠C + ∠C = 180°

⇒ 3∠C = 180°

⇒ ∠C = 60°

From (i), ∠A + ∠B + 60° = 180°

⇒ ∠A + ∠B = 120°

But ∠A = 2 ∠B (given)

2∠B + ∠B = 120°

⇒ 3∠B = 120°

⇒ ∠B = 40°

∠A = 2 x 40 = 80°

Hence angles of a ΔABC are 80°, 40°, 60°.

**Long Answer Type Questions**

Question 1.

In a ΔABC the bisector of ∠ABC and ∠BCA intersect (RBSESolutions.com) at the point O. Prove that ∠BOC = 90° + ∠\(\frac { A }{ 2 }\).

Solution.

In ΔABC, we have

∠A + ∠B + ∠C = 180° (sum of the angles of a triangle)

⇒ \(\frac { 1 }{ 2 }\) ∠A + \(\frac { 1 }{ 2 }\) ∠B + \(\frac { 1 }{ 2 }\) ∠C = 90°

⇒ \(\frac { 1 }{ 2 }\) ∠A + ∠x + ∠y = 90° ….(i)

Again in ΔOBC

∠OBC + ∠OCB + ∠BOC = 180° (by angle sum property of a triangle)

⇒ ∠x + ∠y + ∠BOC = 180° …(ii)

Substituting ∠x + ∠y = 90° – \(\frac { 1 }{ 2 }\)∠A from (i) in eqn (ii), we get

90° – \(\frac { 1 }{ 2 }\) ∠A + ∠BOC = 180°

⇒ ∠BOC = 180° – 90° + \(\frac { 1 }{ 2 }\) ∠A

⇒ ∠BOC = 90° + \(\frac { 1 }{ 2 }\) ∠A

Hence proved.

Question 2.

Prove that the sum of the exterior angles of a triangle (RBSESolutions.com) formed by producing the sides of the triangle in order is equal to 4 right angles.

Solution.

In ΔABC,

∠y = ∠1 + ∠3 …(i) (due to ext. angle property)

and ∠z = ∠1 + ∠2 …(ii) (due to ext. angle property)

also ∠x = ∠2 + ∠3 …(iii) (due to ext. angle property)

Adding equation (i), (ii) and (iii), we get

∠x + ∠y + ∠z = ∠1 + ∠3 + ∠1 + ∠2 + ∠2 + ∠3

⇒ ∠x + ∠y + ∠z = 2∠1 + 2∠2 + 2∠3

⇒ ∠x + ∠y + ∠z = 2(∠1 + ∠2 + ∠3) = 2 x 180° = 360°

Hence, ∠x + ∠y + ∠z = 4 right angles.

Question 3.

The exterior angle of a regular polygon is one-third (RBSESolutions.com) of its interior angle. How many sides has the polygon?

Solution.

Question 4.

In the figure, AO and BO are the bisectors (RBSESolutions.com) of two adjacent angles A and B of a quad. ABCD. Prove that 2∠AOB = ∠C + ∠D

Solution.

Given: In quad. ABCD the bisectors of ∠A and ∠B intersect at O.

To prove: 2∠AOB = ∠C + ∠D

Proof: In quad. ABCD

∠A + ∠B + ∠C + ∠D = 360° …(i) (angle sum property of a quad.)

Also in ΔAOB

∠OAB + ∠OBA + ∠AOB = 180° (by angle sum property of a Δ)

⇒ \(\frac { 1 }{ 2 }\) ∠A + \(\frac { 1 }{ 2 }\) ∠B + ∠AOB = 180°

⇒ ∠A + ∠B + 2∠AOB = 360°…(ii)

From (i) and (ii), we get

∠A + ∠B + ∠C + ∠D = ∠A + ∠B + 2∠AOB

⇒ 2∠AOB = ∠C + ∠D

Hence proved.

Question 5.

In the figure, the side BC of ΔABC is produced to D. The (RBSESolutions.com) bisector of ∠A meets BC at P. Prove that: ∠ABC + ∠ACD = 2 ∠APC.

Solution.

Given: The side BC of ΔABC is produced to D.

The bisector of ∠A meets BC at P.

To prove: ∠ABC + ∠ACD = 2∠APC

Proof: ∠APC = ∠ABP + ∠BAP (the exterior angle is (RBSESolutions.com) equal to sum of opposite interior angles)

∠APC = ∠B + \(\frac { 1 }{ 2 }\) ∠A …(i)

Now in ΔABC

∠ABC + ∠BAC = ∠ACD …(ii) (reason as above)

Adding ∠ABC to both side of equation (ii), we get

∠ABC + ∠ACD = ∠ABC + ∠ABC + ∠BAC

= 2 ∠ABC + ∠BAC

= 2 ∠B + ∠A

= 2(∠B + \(\frac { 1 }{ 2 }\) ∠A)

= 2 ∠APC [from equation (i)]

∠ABC + ∠ACD = 2 ∠APC.

Hence proved.

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