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

Board |
RBSE |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 6 |

Chapter Name |
Rectilinear Figures |

Exercise |
Miscellaneous Exercise |

Number of Questions Solved |
32 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 9 Maths Solutions Chapter 6 Rectilinear Figures Miscellaneous Exercise

**Multiple Choice Questions (Q1 to Q12)**

Question 1.

If two angles of a triangle (RBSESolutions.com) measures 90° and 30°, the third angle is equal to:

(A) 90°

(B) 30°

(C) 60°

(D) 120°

Answer.

(C) 60°

Question 2.

The three angles of a triangle are in the ratio of 2 : 3 : 4, its greatest angle measures:

(A) 80°

(B) 60°

(C) 40°

(D) 180°

Answer.

(A) 80°

Question 3.

Each angle of an equilateral (RBSESolutions.com) triangle measures:

(A) 90°

(B) 30°

(C) 45°

(D) 60°

Answer.

(D) 60°

Question 4.

The angles of quadrilateral are in the ratio 1 : 2 : 3: 4, its smallest angle measures.

(A) 120°

(B) 36°

(C) 18°

(D) 10°

Answer.

(C) 18°

Question 5.

In figure, the side BC of a ΔABC has (RBSESolutions.com) been extended to D. If ∠A = 55° and ∠B = 60° then ∠ACD is:

(A) 120°

(B) 110°

(C) 115°

(D) 125°

Answer.

(C) 115°

Question 6.

The sum of interior angles of a hexagon is:

(A) 720°

(B) 360°

(C) 540°

(D) 1080°

Answer.

(A) 720°

Question 7.

The sum of n exterior (RBSESolutions.com) angles of an n-sided polygon is:

(A) n right angle

(B) 2n rigjjt angle

(C) (n – 4) right angle

(D) 4 right angle

Answer.

(D) 4 right angle

Question 8.

The number of sides in a regular polygon is n. Then measure of each interior angle is:

(A) \(\frac { 360 }{ n }\) degree

(B) (\(\frac { 2n – 4 }{ n }\)) right angle

(C) n right angle

(D) 2n right angle

Answer.

(B) (\(\frac { 2n – 4 }{ n }\)) right angle

Question 9.

If one angle of a triangle is (RBSESolutions.com) equal to the sum of the other two angles then the triangle is:

(A) Isosceles triangle

(B) Obtuse angled triangle

(C) Equilateral triangle

(D) Right angled triangle

Answer.

(D) Right angled triangle

Question 10.

One of the exterior angle of a triangle is 105° and its two opposite interior angles are equal, then each of the equal angle is:

(A) 37\(\frac { 1 }{ 2 }\)°

(B) 52\(\frac { 1 }{ 2 }\)°

(C) 72\(\frac { 1 }{ 2 }\)°

(D) 75°

Answer.

(B) 52\(\frac { 1 }{ 2 }\)°

Question 11.

The angles of a triangle are (RBSESolutions.com) in the ratio 5 : 3 : 7 then the triangle is:

(A) Acute angled triangle

(B) Obtuse angled triangle

(C) Right angled triangle

(D) Isosceles triangle

Answer.

(A) Acute angled triangle

Question 12.

If one of the angle of a triangle is 130° then the angle between the bisectors of the remaining two angles will be

(A) 50°

(B) 65°

(C) 145°

(D) 155°

Answer.

(D) 155°

Question 13.

In figure, find ∠A.

Solution.

∠ACB = 180° – 120° = 60° (due to linear pair)

Exterior ∠B = ∠A + ∠ACB

112° = ∠A + 60°

⇒ ∠A = 52°

Question 14.

In figure, ∠B = 60° and ∠C = 40°. Find the (RBSESolutions.com) measure of ∠A.

Solution.

∠A + ∠B + ∠C = 180°

⇒ ∠A + 60° + 40° = 180°

⇒ ∠A = 80°

Question 15.

In figure, m || QR then find ∠QPR.

Solution

∠QPR = 180° – (50° + 45°) = 180° – 95° = 85°

Question 16.

In figure, find ∠A.

Solution.

At point B,

100° + ∠ABC = 180° (linear pair angles)

∠ABC = 80°

At point C,

∠DCB + 95° = 180°

⇒ ∠DCB = 85°

Similarly at D,

82° + ∠CDA = 180° (linear pair axiom)

∠CDA = 180° – 82° = 98°

Now in quadrilateral ABCD

∠A + 80° + 85° + 98° = 360° (by angle sum property of a quadrilateral)

⇒ ∠A = 360° – 263°

⇒ ∠A = 97°.

Question 17.

The four angles of (RBSESolutions.com) a pentagon are 40°, 75°, 125°, and 135° respectively. Find the fifth angle.

Solution.

Suppose fifth angle be x

x + 40° + 75° + 125° + 135° = 540° (sum of interior angles of a pentagon)

⇒ x + 375° = 540°

⇒ x = 540° – 375°

⇒ x = 165°

Question 18.

Each exterior angle of a regular (RBSESolutions.com) polygon is 45°. Find the number of sides in the polygon.

Solution.

Measure of each exterior angle = \(\frac { { 360 }^{ 0 } }{ n }\)

n = \(\frac { 360 }{ 45 }\) = 8

Hence number of sides in the polygon = 8.

Question 19.

A regular polygon has 12 sides, find the measure of each of its interior angles.

Solution.

Question 20.

The sum of the interior angles (RBSESolutions.com) of a polygon is 10 right angles. Find the number of sides.

Solution.

Sum of the interior angles of a polygon = 10 right angles = 10 x 90° = 900°

But sum of the interior angles of the polygon = (2n – 4) right angles

i.e. 900 = (2n – 4) x 90°

⇒ 180n – 360 = 900

⇒ 180n = 900 + 360

⇒ 180n = 1260

⇒ n = 7

Hence the required number of sides = 7.

Question 21.

Examine, if it is possible to have a regular (RBSESolutions.com) polygon whose each interior angle is 110°.

Solution.

Here, it is given that the measure of each interior angle of a regular polygon =110° (if possible)

Hence, measure of each exterior angle = 180° – 110° = 70°.

Suppose number of sides of the polygon = n

But sum of all the n exterior angles = 360°

⇒ n x 70° = 360°

⇒ n = \(\frac { 360 }{ 70 }\)

= 5\(\frac { 1 }{ 7 }\) ≠ a whole number.

Hence, a regular polygon having measure of each interior angle 110° can not exist.

Question 22.

In a ∆ABC, ∠A + ∠B = ∠C , then find the (RBSESolutions.com) greatest angle of the triangle ABC.

Solution.

∠A + ∠B + ∠C = 180° (due to angle sum property)

But ∠A + ∠B = ∠C

⇒ 2∠C = 180°

⇒ ∠C = 90°

Hence greatest angle = 90° which is right angle.

Question 23.

Find the sum of the (RBSESolutions.com) interior angles of an octagon.

Solution.

The sum of the interior angles of a polygon = (2n – 4) x 90°

Here n = 8

The sum of the interior angles = (2 x 8 – 4) x 90° = (16 – 4) x 90° = 12 x 90° = 1080°

Question 24.

Find the measure of each interior angle of a 10-sided regular polygon.

Solution.

Question 25.

The exterior angles of a triangle (RBSESolutions.com) obtained by producing, the sides in order are 110°, 130° and x. Find x.

Solution.

The sum of the exterior angles in a polygon = 360°

110° + 130° + x = 360°

⇒ x = 360° – 240°

⇒ x = 120°

Question 26.

One interior angle of (RBSESolutions.com) a hexagon is 165° and the remaining interior angles are x° each.

Solution.

Sum of the interior angles of a hexagon = 720°

165° + x + x + x + x + x = 720°

⇒ 165° + 5x = 720°

⇒ 5x = 720° – 165° = 555°

⇒ x = \(\frac { 555 }{ 11 }\) = 111°

Question 27.

In the given figure, if AB || DC then find ∠x, ∠y and ∠z.

Solution.

In ∆BCE

88° = 22° + ∠z (by ext. angle property)

∠z = 88° – 22°

⇒ ∠z = 66°

Also AB || DC

88° + ∠y = 180° (interior angle property)

∠y = 180° – 88°

⇒ ∠y = 92°

Again, 102° + ∠DAB = 180° (linear pair axiom)

∠DAB = 180° – 102° = 78°

AB || DC, then x + 78° = 180° (The sum of the interior angles on the same of a transversal is 180°)

⇒ x = 102°

Hence ∠x = 102°, ∠y = 92°, ∠z = 66°

Question 28.

In figure, if ∠x – ∠y = 10°, then find (RBSESolutions.com) the measure of ∠x and ∠y.

Solution.

According to figure

∠x + ∠y = 120° …(i) (by exterior angle property)

and ∠x – ∠y = 10° (given) …(ii)

2∠x = 130°

⇒ ∠x = 65°

Now, 65° + ∠y = 120°

⇒ ∠y = 120° – 65°

⇒ ∠y = 55°

Hence, ∠x = 65° and ∠y = 55°.

Question 29.

In a polygon two of its interior angles (RBSESolutions.com) are each equal to 90° and remaining angles are 150° each. Find the number of sides of the polygon.

Solution.

Suppose number of sides of the polygon be n.

According to problem

2 x 90° + (n – 2) x 150°=(2n – 4) x 90°

⇒ 180 + 150n – 300 = 180n – 360°

⇒ 360 – 300 + 180 = 180n – 150n

⇒ 240 = 30n

⇒ n = 8

Question 30.

From the given (RBSESolutions.com) figure, prove that ∠x + ∠y = ∠A + ∠C.

Solution.

Construction: Join A to C

In ∆ABC,

∠x = ∠1 + ∠2 …(i) (exterior angle is equal to sum of opposite interior angles)

Also in ∆ADC,

∠y = ∠3 + ∠4 …(ii) (reason as above)

Adding (i) and (ii), we get

∠x + ∠y = ∠1 + ∠2 + ∠3 + ∠4

⇒ ∠x + ∠y = (∠1 + ∠3) + (∠2 + ∠4)

Hence, ∠x + ∠y = ∠A + ∠C

Hence proved.

Question 31.

From figure, find ∠x. Here BO and CO are (RBSESolutions.com) bisectors of ∠B and ∠C respectively.

Solution.

We know that

∠BOC = 90° + \(\frac { 1 }{ 2 }\) ∠A

∵ ∠A = 80°

⇒ ∠BOC = 90° + \(\frac { 1 }{ 2 }\) x 80° = 90° + 40° = 130°

Question 32.

In figure, ∠Q > ∠R and PA is the (RBSESolutions.com) bisector of ∠QPR and PM ⊥ QR, then prove that ∠APM = \(\frac { 1 }{ 2 }\) (∠Q – ∠R).

Solution.

Given: In ΔPQR, PA is the bisector of ∠QPR and PM ⊥ QR.

To prove: ∠APM = \(\frac { 1 }{ 2 }\) (∠Q – ∠R)

Proof: Suppose ∠APR = ∠1, ∠APM = ∠2 and ∠QPM = ∠3

∠1 = ∠2 + ∠3 …(i) [As PA is the bisector of ∠QPR]

In ΔPMR,

∠MPR + ∠PMR + ∠PRM = 180° (angle sum property of a triangle)

⇒ (∠1 + ∠2) + 90° + ∠PRM = 180°

⇒ (∠1 + ∠2) + ∠PRM = 90° …(ii)

Similarly in ΔPQM,

∠3 + ∠Q = 90° …(iii) [As ∠PMQ = 90°]

From (ii) and (iii), we get

∠1 + ∠2 + ∠PRM = ∠3 + ∠Q

⇒ ∠1 + ∠2 + ∠R = ∠3 + ∠Q

⇒ ∠1 + ∠2 – ∠3 = ∠Q – ∠R

⇒ (∠1 – ∠3) + ∠2 = ∠Q – ∠R (as ∠Q > ∠R)

Using relation (i), we get

∠2 + ∠2 = ∠Q – ∠R

⇒ 2∠2 = ∠Q – ∠R

⇒ ∠2 = \(\frac { 1 }{ 2 }\) (∠Q – ∠R)

⇒ ∠APM = \(\frac { 1 }{ 2 }\) (∠Q – ∠R)

Hence proved.

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