RD Sharma Solutions for Class 9 Maths Chapter 16 Circles Exercise 16.4

The RD Sharma Class 9 Solutions for the chapter “Circle” Exercise 16.4 are given here. In this exercise, students will learn about arcs and angles subtended by them. All the questions are solved by subject experts using a step-by-step problem- solving approach, which can help students understand this topic in a better and easier way.

RD Sharma Solutions for Class 9 Maths Chapter 16 Circles Exercise 16.4

Download PDF

carouselExampleControls111

Previous



Next

Access Answers to Maths RD Sharma Solutions for Class 9 Chapter 16 Circles Exercise 16.4 Page number 16.60

Question 1: In figure, O is the centre of the circle. If ∠APB = 50⁰, find ∠AOB and ∠OAB.

Solution:

∠APB = 50⁰ (Given)

By degree measure theorem: ∠AOB = 2∠APB

∠AOB = 2 × 50⁰ = 100⁰

Again, OA = OB [Radius of circle]

Then ∠OAB = ∠OBA [Angles opposite to equal sides]

Let ∠OAB = m

In ΔOAB,

By angle sum property: ∠OAB+∠OBA+∠AOB=180⁰

=> m + m + 100⁰ = 180⁰

=>2m = 180⁰ – 100⁰ = 80⁰

=>m = 80⁰/2 = 40⁰

∠OAB = ∠OBA = 40⁰

Question 2: In figure, it is given that O is the centre of the circle and ∠AOC = 150⁰. Find ∠ABC.

Solution:

∠AOC = 150⁰ (Given)

By degree measure theorem: ∠ABC = (reflex∠AOC)/2 …(1)

We know, ∠AOC + reflex(∠AOC) = 360⁰ [Complex angle]

150⁰ + reflex∠AOC = 360⁰

or reflex ∠AOC = 360⁰−150⁰ = 210⁰

From (1) => ∠ABC = 210 ^o /2 = 105^o

Question 3: In figure, O is the centre of the circle. Find ∠BAC.

Solution:

Given: ∠AOB = 80⁰ and ∠AOC = 110⁰

Therefore, ∠AOB+∠AOC+∠BOC=360⁰ [Completeangle]

Substitute given values,

80⁰ + 100⁰ + ∠BOC = 360⁰

∠BOC = 360⁰ – 80⁰ – 110⁰ = 170⁰

or ∠BOC = 170⁰

Now, by degree measure theorem

∠BOC = 2∠BAC

170⁰ = 2∠BAC

Or ∠BAC = 170⁰/2 = 85⁰

Question 4: If O is the centre of the circle, find the value of x in each of the following figures.

(i)

Solution:

∠AOC = 135⁰ (Given)

From figure, ∠AOC + ∠BOC = 180⁰ [Linear pair of angles]

135⁰ +∠BOC = 180⁰

or ∠BOC=180⁰−135⁰

or ∠BOC=45⁰

Again, by degree measure theorem

∠BOC = 2∠CPB

45⁰ = 2x

x = 45⁰/2

(ii)

Solution:

∠ABC=40⁰ (given)

∠ACB = 90⁰ [Angle in semicircle]

In ΔABC,

∠CAB+∠ACB+∠ABC=180⁰ [angle sum property]

∠CAB+90⁰+40⁰=180⁰

∠CAB=180⁰−90⁰−40⁰

∠CAB=50⁰

Now, ∠CDB = ∠CAB [Angle is on same segment]

This implies, x = 50⁰

(iii)

Solution:

∠AOC = 120⁰ (given)

By degree measure theorem: ∠AOC = 2∠APC

120⁰ = 2∠APC

∠APC = 120⁰/2 = 60⁰

Again, ∠APC + ∠ABC = 180⁰ [Sum of opposite angles of cyclic quadrilaterals = 180 ^o ]

60⁰ + ∠ABC=180⁰

∠ABC=180⁰−60⁰

∠ABC = 120⁰

∠ABC + ∠DBC = 180⁰ [Linear pair of angles]

120⁰ + x = 180⁰

x = 180⁰−120⁰=60⁰

The value of x is 60⁰

(iv)

Solution:

∠CBD = 65⁰ (given)

From figure:

∠ABC + ∠CBD = 180⁰ [ Linear pair of angles]

∠ABC + 65⁰ = 180⁰

∠ABC =180⁰−65⁰=115⁰

Again, reflex ∠AOC = 2∠ABC [Degree measure theorem]

x=2(115⁰) = 230⁰

The value of x is 230⁰

(v)

Solution:

∠OAB = 35⁰ (Given)

From figure:

∠OBA = ∠OAB = 35⁰ [Angles opposite to equal radii]

InΔAOB:

∠AOB + ∠OAB + ∠OBA = 180⁰ [angle sum property]

∠AOB + 35⁰ + 35⁰ = 180⁰

∠AOB = 180⁰ – 35⁰ – 35⁰ = 110⁰

Now, ∠AOB + reflex∠AOB = 360⁰ [Complex angle]

110⁰ + reflex∠AOB = 360⁰

reflex∠AOB = 360⁰ – 110⁰ = 250⁰

By degree measure theorem: reflex ∠AOB = 2∠ACB

250⁰ = 2x

x = 250⁰/2=125⁰

(vi)

Solution:

∠AOB = 60^o (given)

By degree measure theorem: reflex∠AOB = 2∠OAC

60 ^o = 2∠ OAC

∠OAC = 60 ^o / 2 = 30 ^o [Angles opposite to equal radii]

Or x = 30⁰

(vii)

Solution:

∠BAC = 50⁰ and ∠DBC = 70⁰ (given)

From figure:

∠BDC = ∠BAC = 50⁰ [Angle on same segment]

Now,

In ΔBDC:

Using angle sum property, we have

∠BDC+∠BCD+∠DBC=180⁰

Substituting given values, we get

50⁰ + x⁰ + 70⁰ = 180⁰

x⁰ = 180⁰−50⁰−70⁰=60⁰

or x = 60^o

(viii)

Solution:

∠DBO = 40⁰ (Given)

Form figure:

∠DBC = 90⁰ [Angle in a semicircle]

∠DBO + ∠OBC = 90⁰

40⁰+∠OBC=90⁰

or ∠OBC=90⁰−40⁰=50⁰

Again, By degree measure theorem: ∠AOC = 2∠OBC

or x = 2×50⁰=100⁰

(ix)

Solution:

∠CAD = 28, ∠ADB = 32 and ∠ABC = 50 (Given)

From figure:

In ΔDAB:

Angle sum property: ∠ADB + ∠DAB + ∠ABD = 180⁰

By substituting the given values, we get

32⁰ + ∠DAB + 50⁰ = 180⁰

∠DAB=180⁰−32⁰−50⁰

∠DAB = 98⁰

Now,

∠DAB+∠DCB=180⁰ [Opposite angles of cyclic quadrilateral, their sum = 180 degrees]

98⁰+x=180⁰

or x = 180⁰−98⁰=82⁰

The value of x is 82 degrees.

(x)

Solution:

∠BAC = 35⁰ and ∠DBC = 65⁰

From figure:

∠BDC = ∠BAC = 35⁰ [Angle in same segment]

In ΔBCD:

Angle sum property, we have

∠BDC + ∠BCD + ∠DBC = 180⁰

35⁰ + x + 65⁰ = 180⁰

or x = 180⁰ – 35⁰ – 65⁰ = 80⁰

(xi)

Solution:

∠ABD = 40⁰, ∠CPD = 110⁰ (Given)

Form figure:

∠ACD = ∠ABD = 40⁰ [Angle in same segment]

In ΔPCD,

Angle sum property: ∠PCD+∠CPO+∠PDC=180⁰

400 + 110⁰ + x = 180⁰

x=180⁰−150⁰ =30⁰

The value of x is 30 degrees.

(xii)

Solution:

∠BAC = 52⁰ (Given)

From figure:

∠BDC = ∠BAC = 52⁰ [Angle in same segment]

Since OD = OC (radii), then ∠ODC = ∠OCD [Opposite angle to equal radii]

So, x = 52⁰

Question 5: O is the circumcentre of the triangle ABC and OD is perpendicular on BC. Prove that ∠BOD = ∠A.

Solution:

In ΔOBD and ΔOCD:

OB = OC [Radius]

∠ODB = ∠ODC [Each 90⁰]

OD = OD [Common]

Therefore, By RHS Condition

ΔOBD ≅ ΔOCD

So, ∠BOD = ∠COD…..(i)[By CPCT]

Again,

By degree measure theorem: ∠BOC = 2∠BAC

2∠BOD = 2∠BAC [Using(i)]

∠BOD = ∠BAC

Hence proved.

Question 6: In figure, O is the centre of the circle, BO is the bisector of ∠ABC. Show that AB = AC.

Solution:

Since, BO is the bisector of ∠ABC, then,

∠ABO = ∠CBO …..(i)

From figure:

Radius of circle = OB = OA = OB = OC

∠OAB = ∠OCB …..(ii) [opposite angles to equal sides]

∠ABO = ∠DAB …..(iii) [opposite angles to equal sides]

From equations (i), (ii) and (iii), we get

∠OAB = ∠OCB …..(iv)

In ΔOAB and ΔOCB:

∠OAB = ∠OCB [From (iv)]

OB = OB [Common]

∠OBA = ∠OBC [Given]

Then, By AAS condition : ΔOAB ≅ ΔOCB

So, AB = BC [By CPCT]

Question 7: In figure, O is the centre of the circle, then prove that ∠x = ∠y + ∠z.

Solution:

From the figure:

∠3 = ∠4 ….(i) [Angles in same segment]

∠x = 2∠3 [By degree measure theorem]

∠x = ∠3 + ∠3

∠x = ∠3 + ∠4 (Using (i) ) …..(ii)

Again, ∠y = ∠3 + ∠1 [By exterior angle property]

or ∠3 = ∠y − ∠1 …..(iii)

∠4 = ∠z + ∠1 …. (iv) [By exterior angle property]

Now, from equations (ii) , (iii) and (iv), we get

∠x = ∠y − ∠1 + ∠z + ∠1

or ∠x = ∠y + ∠z + ∠1 − ∠1

or x = ∠y + ∠z

Hence proved.

RD Sharma Solutions for Class 9 Maths Chapter 16 Circles Exercise 16.4

Class 9 Maths Chapter 16 Circles Exercise 16.4 is based on the topic – Arcs and angles subtended by them. Students can practise more Maths concepts using RD Sharma Solutions and clear their doubts.