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RBSE Solutions for Class 10 Maths Chapter 12 Circle Ex 12.4

April 12, 2019 by Fazal Leave a Comment

RBSE Solutions for Class 10 Maths Chapter 12 Circle Ex 12.4 is part of RBSE Solutions for Class 10 Maths. Here we have given Rajasthan Board RBSE Class 10 Maths Chapter 12 Circle Exercise 12.4.

Rajasthan Board RBSE Class 10 Maths Chapter 12 Circle Ex 12.4

Question 1.
An angle of a cyclic (RBSESolutions.com) quadrilateral is given find its opposite angle.
(i) 70°
(ii) 135°
(iii) 112\(\frac { 1 }{ 2 } \circ \)
(iv) \(\frac { 3 }{ 5 }\) right angle
(v) 165°
Solution :
(i) We know that sum of opposite angles of a cyclic quadrilateral is 180°. Let opposite angle be x°, then
70° + x° = 180°
x° = 180° – 70° = 1 10°
(ii) ∵ Sum of opposite angles of a cyclic quadrilateral is 180°
Let other angle be x°
x° + 135° = 180°
x° = 180°- 13° = 45°
(iii) ∵ Sum of opposite angles of a cyclic quadrilateral is 180°
Let other angle be x°
RBSE Solutions for Class 10 Maths Chapter 12 Circle Ex 12.4 1
(iv) Given : first angle = \(\frac { 3 }{ 5 }\) rectangle = \(\frac { 3 }{ 5 }\) × 90° =3 × 18 = 54°
∵ Sum of opposite (RBSESolutions.com) angles of a cyclic quadrilateral is 180°
Let other angle be x°
x° + 540 = 180°
x° = 180° – 54° = 126°
(v) Let other angle = x°
Sum of opposite angles of a cyclic quadrilateral is 180°
x° + 165° = 180°
x° = 180° – 165° = 15°

RBSE Solutions

Question 2.
Find opposite angle of cyclic (RBSESolutions.com) quadrilateral if one angle is (i) \(\frac { 2 }{ 7 }\) of other
(ii) \(\frac { 11 }{ 4 }\) of other.
Solution :
(i) Let other opposite angle
So, according to question, first opposite = \(\frac { 2 }{ 7 }\)x
We know that sum of opposite angles of cyclic quadrilateral is 180°
RBSE Solutions for Class 10 Maths Chapter 12 Circle Ex 12.4 2
Thus, opposite angles of cyclic quadrilateral will be 40°, 140°.
(ii) Let other opposite ∠ = x°
Thus according to question., (RBSESolutions.com) first opposite ∠ = x × \(\frac { 11 }{ 4 }\) = \(\frac { 11x }{ 4 }\)
We know that sum of opposite angles of cyclic quadrilateral is 180°
RBSE Solutions for Class 10 Maths Chapter 12 Circle Ex 12.4 3
Thus, opposite angles of cyclic quadrilateral will be 132°, 48°.

Question 3.
In figure, find all the (RBSESolutions.com) four angles of cyclic.
RBSE Solutions for Class 10 Maths Chapter 12 Circle Ex 12.4 4
Solution :
We know that sum of opposite angles of cyclic quadrilateral is 180°
RBSE Solutions for Class 10 Maths Chapter 12 Circle Ex 12.4 5

Question 4.
In figure, some angles are (RBSESolutions.com) marked a, b, c and d, find them.
RBSE Solutions for Class 10 Maths Chapter 12 Circle Ex 12.4 6
Solution :
ABEF and BCDE arc cyclic quadrilateral
∵ We know that sum of opposite angles of cyclic quadrilateral is 180°
Thus, ∠FAB + ∠BEF = 180°
65° + ∠BEF = 180°
∠BEF = 180° – 65° = 115°
and ∠EBA + ∠AFE = 180°
∠EBA + 91° = 180°
∠EBA = 180° – 91° = 89°
We know that exterior angle of (RBSESolutions.com) cyclic quadrilateral is equal to internal opposite angle
Thus ∠EBC = ∠AFE = 91°
∴ ∠EBC = c = 91°
∠DEB = ∠BAF = 65° = a = 65°
∠EBA = ∠CDE = 89°
∠CDE = b = 89°
∴ ∠BEF = ∠BCD= 115°
∴ ∠BCD = d = 115°
Thus, ∠a = 65°, ∠b = 89°, ∠c = 91°, ∠d = 115°.

Question 5.
If in cyclic quadrilateral ABCD, AD || BC, then Prove that ∠A = ∠D.
Solution :
Given :
ABCD is a cyclic quadrilateral on which AD || BC
RBSE Solutions for Class 10 Maths Chapter 12 Circle Ex 12.4 7
To Proved :
∠A = ∠D
Proof : ABCD is a cyclic quadrilateral
We know that sum of opposite (RBSESolutions.com) angles of a cyclic quadrilateral is 180°
∠A + ∠C = 180°
and ∠B + ∠D = 180° ….(ii)
but ∠A + ∠B = 180° (co-interior corresponding angles) (iii)
From equation (ii) and (iii)
⇒ 2∠A + ∠B + ∠C = ∠B + ∠C + 2∠D
⇒ 2∠A = 2∠D
⇒ ∠A = ∠D

RBSE Solutions

Question 6.
ABCD is a cyclic quadrilateral. (RBSESolutions.com) produced AB and DC meet at E. Prove that ∆EBNC and ∆EOA arc similar.
Solution :
Given :
ABCD is a cyclic quadrilateral. Produced AB and DC meet at F.
RBSE Solutions for Class 10 Maths Chapter 12 Circle Ex 12.4 8
To Prove : ∆EBC ~ ∆EDA
In ∆EBC and ∆EDA
∠EBC = ∠EDA [∵ Exterior of cyclic quadrilateral is equal to interior opposite angle.]
∠ECB = ∠EAD [∵ Exterior of cyclic (RBSESolutions.com) quadrilateral is equal to interior opposite angle.]
∠E = ∠E (Common angle)
Thus, by AAA similarly
∆EBC ~ ∆EDA

Question 7.
Prove that angle bisectors of cyclic quadrilaterals formed a cyclic quadrilateral.
Solution :
Given :
A cyclic quadrilateral ABCD.
AP, BP, CR and DR are respectively bisector of
∠A, ∠B, ∠C, and ∠D.
RBSE Solutions for Class 10 Maths Chapter 12 Circle Ex 12.4 9
To Proved : PQRS is a cyclic quadrilateral.
Proof : To Prove PQRS a (RBSESolutions.com) cyclic quadrilateral, we should prove
∠APB + ∠CRD = 180°
We know that sum of opposite angles of a cyclic quadrilateral is 180°
Now in ∆APB and ∆CRD
RBSE Solutions for Class 10 Maths Chapter 12 Circle Ex 12.4 10
RBSE Solutions for Class 10 Maths Chapter 12 Circle Ex 12.4 11
⇒ One pair of opposite angles of (RBSESolutions.com) quadrilateral PQRS arc supplementary.
⇒ PQRS is a cyclic quadrilateral.

RBSE Solutions

We hope the given RBSE Solutions for Class 10 Maths Chapter 12 Circle Ex 12.4 will help you. If you have any query regarding Rajasthan Board RBSE Class 10 Maths Chapter 12 Circle Exercise 12.4, drop a comment below and we will get back to you at the earliest.

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