RBSE Solutions for Class 9 Maths Chapter 6 Rectilinear Figures Ex 6.1 is part of RBSE Solutions for Class 9 Maths. Here we have given Rajasthan Board RBSE Class 9 Maths Solutions Chapter 6 Rectilinear Figures Exercise 6.1.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 6 |

Chapter Name |
Rectilinear Figures |

Exercise |
Ex 6.1 |

Number of Questions Solved |
11 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 9 Maths Solutions Chapter 6 Rectilinear Figures Ex 6.1

Question 1.

From the given figure, find the (RBSESolutions.com) three angles of the triangle ABC.

Solution.

Here ∠DCE = 53°

∴∠ACB = 53°

(Vertically opposite angle)

In ∆ABC

∠ABC + ∠ACB = ∠CAF

(By exterior angle property)

⇒ ∠ABC + 53°= 112°

⇒ ∠ABC = 112° – 53°

⇒ ∠ABC = 59°

Now in ∆ABC

∠BAC + ∠ABC + ∠BCA = 180°

(by angle sum property)

⇒ ∠BAC + 59° + 53° = 180°

⇒ ∠BAC = 68°

Hence, ∠BAC = 68°, ∠ABC = 59° and ∠ACB = 53°.

Question 2.

In figure, ∆ABC is an equilateral triangle. Find (RBSESolutions.com) the values of ∠x, ∠y and ∠z from the given figure.

Solution.

∵ ∆ABC is an equilateral triangle

∴AB = BC = CA

⇒ ∠ABC = ∠ACB = ∠BAC = 60°

60° = x + 22°

(exterior angle property)

x = 38°

Also

38° + 22° + ∠z = 180°

(by angle sum property of a ∆)

⇒ ∠z = 180° – 60° = 120°

Again by exterior angle property

⇒ ∠ACB = ∠y + 38°

⇒ 60° = ∠y + 38°

⇒ ∠y = 22°

Hence, ∠x = 38°, ∠y = 22° and ∠z = 120°

Question 3.

In the given figure, the sides AB and AC of ∆ABC are (RBSESolutions.com) produced to point E and D respectively. If the bisectors BO and CO of ∠CBE and ∠BCD respectively meet at point O, then prove

∠BOC = 90°- \(\frac { 1 }{ 2 }\)∠x.

Solution.

Ray BO is the (RBSESolutions.com) bisector of ∠CBE.

Therefore, ∠CBO = \(\frac { 1 }{ 2 }\)∠CBE

⇒ ∠CBO = \(\frac { 1 }{ 2 }\)(180° – y) = 90° – \(\frac { y }{ 2 }\)…(i)

Similarly, ray CO is the bisector of ∠BCD

Therefore, ∠BCO = \(\frac { 1 }{ 2 }\)∠BCD

= \(\frac { 1 }{ 2 }\)(180° – z)

= 90°- \(\frac { z }{ 2 }\) …(ii)

Now in ∆BOC,

∠CBO + ∠BCO + ∠BOC = 180° …(iii)

(angle sum property of a triangle)

Using (i) and (ii) in (iii), we get

⇒ 90°- \(\frac { y }{ 2 }\) + 90°- \(\frac { z }{ 2 }\) + ∠BOC = 180°

⇒ ∠BOC = \(\frac { 1 }{ 2 }\) – (y + z) …(iv)

But in ∆ABC,

x + y + z = 180°

(angle sum property of a triangle)

⇒ y + z = 180° – x

⇒ 2∠BOC = 180° – x

[using relation (iv)]

⇒ ∠BOC = 90° – \(\frac { 1 }{ 2 }\) ∠x

Hence proved.

Question 4.

In figure, ∠P = 52°, ∠PQR = 64°. If QO and RO are the (RBSESolutions.com) bisectors of ∠PQR and ∠PRQ respectively of ∆PQR, find ∠x and

Solution.

Given: QO and RO are the (RBSESolutions.com) bisectors of ∠PQR and ∠PRQ respectively of ∆PQR and ∠P = 52°, ∠PQO = 64°.

To find: ∠x and ∠y

In ∆PQR,

∠P + ∠PQR + ∠PRQ = 180°

(angle sum property of a triangle)

⇒ 52° + 64° + ∠PRQ = 180°

⇒ ∠PRQ = 180° – 116°

⇒ ∠PRQ = 64°

⇒ ∠y = 32°

(as RD is bisector of ∠PRQ)

In ∆OQR,

∠OQR + ∠ORQ + ∠x = 180°

(reason as above)

⇒ 32° + 32° + ∠x= 180°

[QO and RO are bisector (RBSESolutions.com) of ∠PQR and ∠PRQ]

⇒ ∠x = 180° – 64°

⇒ ∠x = 116°

Hence, ∠x = 116°, ∠y = 32°.

Question 5.

In figure, AB || DE, ∠BAC = 35° and ∠CDE = 53°, find ∠DCE.

Solution.

We are given that AB || DE

and ∠BAC = 35° and ∠CDE = 53°

AB || DE (given)

⇒ ∠BAE = ∠AED = 35°

(alternate angles)

Now in ∆CDE,

∠CDE + ∠E + ∠DCE = 180°

(angle sum property of a triangle)

⇒ 53° + 35° + ∠DCE = 180°

⇒ ∠DCE = 180° – 88° = 92°

⇒ ∠DCE = 92°

Hence, ∠DCE = 92°.

Question 6.

In the adjoining figure if lines PQ and RS intersect (RBSESolutions.com) at point T, such that ∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°, find ∠SQT.

Solution.

In ∆PRT,

∠RPT + ∠PRT + ∠RTP = 180°

(angle sum property of a triangle)

⇒ 95° + 40° + ∠RTP = 180°

⇒ 135° + ∠RTP = 180°

⇒ ∠RTP = 180° – 135°

= 45°

Now ∠RTP = ∠STQ = 45°

(vertically opposite angles)

In ∆STQ,

∠STQ + ∠TSQ + ∠SQT = 180°

(angle sum property of a triangle)

⇒ 45° + 75° + ∠SQT = 180°

⇒ ∠SQT = 180° – (45° + 75°)

= 180° – 120° = 60°

Hence, ∠SQT = 60°.

Question 7.

In figure, sides QP and RQ of ∆PQR are produced (RBSESolutions.com) to points S and T respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.

Solution.

We are given that ∠SPR = 135° and ∠PQT =110°.

At point Q:

∠PQT + ∠PQR = 180°

(linear pair of angles)

⇒ 110° + ∠PQR = 180°

⇒ ∠PQR = 180° – 110°

= 70° …(i)

At point P:

∠QPR + ∠SPR = 180°

(linear pair of angles)

⇒ ∠QPR + 135°= 180°

⇒ ∠QPR = 45° …(ii)

In ∆PQR,

∠PQR + ∠PRQ + ∠QPR = 180°

⇒ 70° + ∠PRQ + 45° = 180°

⇒ ∠PRQ = 180° – 115°

⇒ ∠PRQ = 65°

Hence, ∠PRQ = 65°.

Question 8.

In figure, if PQ ⊥ PS, PQ || SR, ∠SQR = 28° and ∠QRT = 65°, then find (RBSESolutions.com) the values of x and y.

Solution.

We are given that PQ ⊥ PS, PQ || SR

∠SQR = 28° and ∠QRT = 65°

SR and a transversal QR intersect them

∠PQR = ∠QRT = 65°

(alternate angles)

x + 28° = 65°

x = 65° – 28°

x = 37°

Now, in right angled triangle SPQ

x + y + ∠SPQ = 180°

(by angle sum property of a triangle)

x + y + 90° = 180°

⇒ 37° + y + 90°= 180°

⇒ 127° + y = 180°

⇒ y = 180 – 127°

⇒ y = 53°

Hence, x = 37° and y = 53°.

Question 9.

In the given figure, the side QR and APQR is produced to (RBSESolutions.com) a point S. If the bisector of ∠PQR and ∠PRS meet at a point T, then prove that ∠QTR = \(\frac { 1 }{ 2 }\)∠QPR.

Solution.

Side QR of ∆PQR is produced to S,

Exterior ∠PRS = ∠P + ∠Q …(i)

(Exterior angle is equal to sum of its opposite interior angles)

Dividing of both side (RBSESolutions.com) relation (i) by 2, we get

\(\frac { 1 }{ 2 }\)∠PRS = \(\frac { 1 }{ 2 }\)∠P + \(\frac { 1 }{ 2 }\)∠Q

⇒ x = \(\frac { 1 }{ 2 }\)∠P + y …(ii)

[∵ RT and QT are bisectors of ∠PRS and ∠PQS respectively]

Also, in ∆QRT,

Exterior ∠TRS = ∠T + ∠y

⇒ x = ∠T + y …(iii)

From (i) and (iii), we get

\(\frac { 1 }{ 2 }\) ∠P + y = ∠T + y

\(\frac { 1 }{ 2 }\)∠P = ∠T

∠QTR = \(\frac { 1 }{ 2 }\)∠QPR

Hence proved.

Question 10.

In ∆ABC, ∠A = 90°, AL ⊥ BC, prove that ∠BAL = ∠ACB.

Solution.

Given: In ∆ABC, ∠A = 90° and AL ⊥ BC

To prove: ∠BAL = ∠ACB

Proof: Suppose ∠BAL = ∠1, ∠CAL = ∠2, ∠ABL = 3 and ∠ACL = ∠4

Now in ∆ABC

∠A + ∠B + ∠C = 180°

(angle sum property of a triangle)

⇒ ∠90° + ∠3 + ∠4 = 180°

(∵ ∠A = 90 given)

⇒ ∠3 + ∠4 = ∠90°

⇒ ∠4 = ∠90° – ∠3 …(i)

Now in ∠BAL

∠1 + ∠3 + ∠ALB = 180°

(Angle sum property of a triangle)

⇒ ∠1 + ∠3 + 90° = 180°

⇒ ∠1 + ∠3 = 90°

⇒ ∠1 = 90° – ∠3 …(ii)

From (i) and (ii), we get

∠1 = ∠4 => ∠BAL = ∠ACB

Hence proved.

Question 11.

The angles of a triangle are the ratio 2 : 3 : 4. Find all the (RBSESolutions.com) three angles of a triangle.

Solution.

Let angles of a triangle are 2x, 3x, 4x

2x + 3x + 4x = 180°

⇒ 9x = 180°

⇒ x = 20°

Angles are

2x = 2 x 20° = 40°,

3x = 3 x 20° = 60°,

4x = 4 x 20° = 80°.

We hope the given RBSE Solutions for Class 9 Maths Chapter 6 Rectilinear Figures Ex 6.1 will help you. If you have any query regarding Rajasthan Board RBSE Class 9 Maths Solutions Chapter 6 Rectilinear Figures Exercise 6.1, drop a comment below and we will get back to you at the earliest.

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