RD Sharma 2020 solution class 9 chapter 15 Circles Exercise 15.4

Exercise 15.4

Page-15.72

Question 1:

In the given figure, O is the centre of the circle. If $\angle APB$= 50°, find ∠AOB and ∠OAB.

Answer 1:

This question seems to be incorrect.

Question 2:

In the given figure, O is the centre of the circle. Find ∠BAC.
 

Answer 2:

It is given that

And (given)

We have to find

In given triangle

(Given)

 OB = OA               (Radii of the same circle)

Therefore, is an isosceles triangle.

So, $\angle OBA=\angle OAB$          ..… (1)

                       (Given)

                    [From (1)]

So

Again from figure, is given triangle and

Now in , 

                  (Radii of the same circle)

$\angle OAC=\angle OCA$

               (Given that)

Then,

Since

Hence

Question 3:

If O is the centre of the circle, find the value of x in each of the following figures.

(i)


(ii)



(iii)



(iv)



(v)



(vi)



(vii)



(viii)



(ix)



(x)



(xi)



(xii)

Answer 3:

We have to find in each figure.

(i) It is given that

$\angle AOC+\angle COB=180° \left[\mathrm{Linear} \mathrm{pair}\right]$

 
As we know the angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.

Now ,$x=\frac{1}{2}\angle COB=22{\frac{1}{2}}^{°}$

Hence

(ii) As we know that = x                 [Angles in the same segment]

line is diameter passing through centre,

So,

$\angle BCA= 90° \left[\mathrm{Angle} \mathrm{inscribed} \mathrm{in} \mathrm{a} \mathrm{semicircle} \mathrm{is} \mathrm{a} \mathrm{right} \mathrm{angle} \right]$

  

 

$$\begin{gathered} \angle CAB+\angle ABC+\angle BCA=180° \left[\mathrm{Angle} \mathrm{sum} \mathrm{property}\right] \\ \Rightarrow x+40°+90°=180° \\ \Rightarrow x=50° \end{gathered}$$

(iii) It is given that

$\angle ABC=\frac{1}{2}\left(\mathrm{Reflex} \angle AOC\right)$

 

So

And

Then

Hence

(iv)

  (Linear pair)

   

And
x =
 
 
Hence,

(v) It is given that

is an isosceles triangle.
  

Therefore

And,

$$\begin{gathered} \mathrm{In} △AOB, \\ \angle AOB+\angle OBA+\angle BAO=180° \\ \Rightarrow 70°+\angle BAO=180° \\ \Rightarrow \angle BAO=110° \end{gathered}$$

$\angle AOB=2\left(\mathrm{Reflex}\angle ACB\right)$
                     

Hence,

(vi) It is given that

 

And

$$\begin{gathered} \angle COA+\angle AOB=180° \\ \Rightarrow \angle COA=180°-60° \\ \Rightarrow \angle COA=120° \end{gathered}$$

$∆OCA \mathrm{is} \mathrm{an} \mathrm{isosceles} \mathrm{triangle}.$

So

Hence,

(vii)                    (Angle in the same segment)

 

In we have

Hence

(viii)

 

As    (Radius of circle)

Therefore, is an isosceles triangle.

So          (Vertically opposite angles)

Hence,

(ix) It is given that

   

…… (1)          (Angle in the same segment)

$\angle ADB=\angle ACB=32°$  ......(2)           (Angle in the same segment)

Because and are on the same segment of the circle.

Now from equation (1) and (2) we have

Hence,

(x) It is given that

  
$\angle BAC=\angle BDC=35°$                (Angle in the same segment)

 Now in $∆BDC$ we have

$$\begin{gathered} \angle BDC+\angle DCB+\angle CBD=180° \\ \Rightarrow 35°+65°+\angle CBD=180° \\ \Rightarrow \angle CBD=180°-100°=80° \end{gathered}$$

Hence,

(xi)

 

                    (Angle in the same segment)

In we have

Hence

(xii)
 

          (Angle in the same segment)

is an isosceles triangle

So,                   (Radius of the same circle)

Then

Hence

Page-15.73

Question 4:

O is the circumcentre of the triangle ABC and OD is perpendicular on BC. Prove that ∠BOD = ∠A

Answer 4:

We have to prove that

Since, circumcenter is the intersection of perpendicular bisectors of each side of the triangle.

Now according to figure A, B, C are the vertices of ΔABC

In , is perpendicular bisector of BC

So, BD = CD                 

OB = OC             (Radius of the same circle)

And,

OD = OD         (Common)

Therefore,
$△BDO\cong △CDO \left(\mathrm{SSS} \mathrm{congruency} \mathrm{criterion}\right)$

$\angle BOD=\angle COD \left(\mathrm{by} \mathrm{cpct}\right)$

We know that angle formed any chord of the circle at the center is twice of the angle formed at the circumference by same chord

Therefore,

Therefore,

Hence proved

Question 5:

In the given figure, O is the centre of the circle, BO is the bisector of ∠ABC. Show that AB = AC.

Answer 5:

It is given that,∠ABC is on circumference of circle BD is passing through centre.

Construction: Join A and C to form AC and extend BO to D such that BD be the perpendicular bisector of AC.

Now in $△BDA \mathrm{and} △BDC$ we have

AD = CD           (BD is the perpendicular bisector) 

So $\angle BDA=\angle BDC=90°$

  (Common)

$△BDA\cong △BDC \left(\mathrm{SAS} \mathrm{congruency} \mathrm{criterion}\right)$

Hence     (by cpct)

Question 6:

In the given figure, O and O' are centres of two circles intersecting at B and C. ACD is a straight line, find x.

Answer 6:

It is given that

Two circles having center O and O' and ∠AOB = 130°

And AC is diameter of circle having center O

 

We have

So

Now, reflex

So

$x°=360°-230°=130°$

Hence,

Page-15.74

Question 7:

In the given figure, if ∠ACB = 40°, ∠DPB = 120°, find ∠CBD.

Answer 7:

It is given that ∠ACB = 40° and ∠DPB = 120°

Construction: Join the point A and B

           (Angle in the same segment)

Now in $△BDP$ we have

$$\begin{gathered} \angle DPB+\angle PBD+\angle BDP=180° \\ \Rightarrow 120°+\angle PBD+40°=180° \\ \Rightarrow \angle PBD=20° \end{gathered}$$

Hence

Question 8:

A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.

Answer 8:

We have to find and

Construction: - O is centre and r is radius and given that chord is equal to radius of circle

Now in we have

AO = OB = BA       ( It is given that chord is equal to radius of circle)

So, is an equilateral triangle

So, $\angle AOB=2\angle ADB$                (The angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle)

Then

So

$$\begin{gathered} \angle AEB=\frac{1}{2}\left(\mathrm{Reflex}\angle AOB\right) \\ =\frac{1}{2}\left(360°-60°\right) \\ =150° \end{gathered}$$

Therefore,

and

Hence, the angle subtended by the chord at a point on the minor arc is 150° and also at a point on the major arc is 30°.

 

Question 9:

In the given figure, it is given that O is the centre of the circle and ∠AOC = 150°. Find ∠ABC.
 

Answer 9:

It is given that O is the centre of circle and A, B and C are points on circumference.

  (Given)

We have to find ∠ABC

The angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.

$$\begin{gathered} \angle ABC=\frac{1}{2}\left(\mathrm{reflex} \angle AOC\right) \\ =\frac{1}{2}\left(360°-150°\right) \\ =\frac{1}{2}\times 210° \\ =105° \end{gathered}$$

Hence,

Question 10:

In the given figure, O is the centre of the circle, prove that ∠x = ∠y + ∠z.

Answer 10:

It is given that, O is the center of circle and A, B and C are points on circumference on triangle

  

We have to prove that ∠x = ∠y + ∠z

∠4 and ∠3 are on same segment

So, ∠4 = ∠3           

               (Angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle)

…… (1)

               (Exterior angle is equal to the sum of two opposite interior angles)         …… (2)

$\angle 4=\angle z+\angle 1$               (Exterior angle is equal to the sum of two opposite interior angles)  

                 …… (3)

Adding (2) and (3)

……(4)

From equation (1) and (4) we have 

Question 11:

In the given figure, O is the centre of a circle and PQ is a diameter. If ∠ROS = 40°, find ∠RTS.
 

Answer 11:

It is given that O is the centre and $\angle ROS=40°$

We have

In right angled triangle RQT we have

$$\begin{gathered} \angle RQT+\angle QTR+\angle TRQ=180° \\ \Rightarrow 20°+\angle QTR+90°=180° \\ \Rightarrow \angle QTR=70° \end{gathered}$$

$\mathrm{Hence}, \angle RTS=70°$