## RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.4

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.4

**Other Exercises**

- RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.1
- RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.2
- RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.3
- RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.4
- RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5
- RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles VSAQS
- RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles MCQS

**Question 1.**

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

**Solution:**

Arc AB, subtends ∠AOB at the centre and ∠APB at the remaining part of the circle

∴∠AOB = 2∠APB = 2 x 50° = 100°

Join AB

∆AOB is an isosceles triangle in which

OA = OB

∴ ∠OAB = ∠OBA But ∠AOB = 100°

∴∠OAB + ∠OBA = 180° – 100° = 80°

⇒ 2∠OAB = 80°

80°

∴∠OAB = \(\frac { { 80 }^{ \circ } }{ 2 }\) = 40°

**Question 2.**

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

**Solution:**

In the circle with centre O

∠AOB = 80° and ∠AOC =110°

∴ ∠BOC = ∠AOB + ∠AOC

= 80°+ 110°= 190°

∴ Reflex ∠BOC = 360° – 190° = 170°

Now arc BEC subtends ∠BOC at the centre and ∠BAC at the remaining part of the circle.

∴ ∠BOC = 2∠BAC

⇒ 170° = 2∠BAC

⇒ ∠BAC = \(\frac { { 170 }^{ \circ } }{ 2 }\) = 85°

∴ ∠BAC = 85°

**Question 3.**

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

**Solution:**

**(i)** A circle with centre O

∠AOC = 135°

But ∠AOC + ∠COB = 180° (Linear pair)

⇒ 135° + ∠COB = 180°

⇒ ∠COB = 180°- 135° = 45°

Now arc BC subtends ∠BOC at the centre and ∠BPC at the remaining part of the circle

∴ ∠BOC = 2∠BPC

⇒ ∠BPC = \(\frac { 1 }{ 2 }\)∠BOC = \(\frac { 1 }{ 2 }\) x 45° = \(\frac { { 45 }^{ \circ } }{ 2 }\)

∴ ∠BPC = 22 \(\frac { 1 }{ 2 }\)° or x = 22 \(\frac { 1 }{ 2 }\)°

**(ii)** ∵ CD and AB are the diameters of the circle with centre O

∠ABC = 40°

But in ∆OBC,

OB = OC (Radii of the circle)

∠OCB = ∠OBC – 40°

Now in ABCD,

∠ODB + ∠OCB + ∠CBD = 180° (Angles of a triangle)

⇒ x + 40° + 90° = 180°

⇒ x + 130° = 180°

⇒ x = 180° – 130° = 50°

∴ x = 50°

**(iii)** In circle with centre O,

∠AOC = 120°, AB is produced to D

∵ ∠AOC = 120°

and ∠AOC + convex ∠AOC = 360°

⇒ 120° + convex ∠AOC = 360°

∴ Convex ∠AOC = 360° – 120° = 240°

∴ Arc APC Subtends ∠AOC at the centre and ∠ABC at the remaining part of the circle

∴ ∠ABC = \(\frac { 1 }{ 2 }\)∠AOC = \(\frac { 1 }{ 2 }\)x 240° = 120°

But ∠ABC + ∠CBD = 180° (Linear pair)

⇒ 120° + x = 180°

⇒ x = 180° – 120° = 60°

∴ x = 60°

**(iv)** A circle with centre O and ∠CBD = 65°

But ∠ABC + ∠CBD = 180° (Linear pair)

⇒ ∠ABC + 65° = 180°

⇒ ∠ABC = 180°-65°= 115°

Now arc AEC subtends ∠x at the centre and ∠ABC at the remaining part of the circle

∴ ∠AOC = 2∠ABC

⇒ x = 2 x 115° = 230°

∴ x = 230°

**(v)** In circle with centre O

AB is chord of the circle, ∠OAB = 35°

In ∆OAB,

OA = OB (Radii of the circle)

∠OBA = ∠OAB = 35°

But in ∆OAB,

∠OAB + ∠OBA + ∠AOB = 180° (Angles of a triangle)

⇒ 35° + 35° + ∠AOB = 180°

⇒ 70° + ∠AOB = 180°

⇒ ∠AOB = 180°-70°= 110°

∴ Convex ∠AOB = 360° -110° = 250°

But arc AB subtends ∠AOB at the centre and ∠ACB at the remaining part of the circle.

∴∠ACB = \(\frac { 1 }{ 2 }\)∠AOB

⇒ x = \(\frac { 1 }{ 2 }\) x 250° = 125°

∴ x= 125°

**(vi)** In the circle with centre O,

BOC is its diameter, ∠AOB = 60°

Arc AB subtends ∠AOB at the centre of the circle and ∠ACB at the remaining part of the circle

∴ ∠ACB = \(\frac { 1 }{ 2 }\) ∠AOB

= \(\frac { 1 }{ 2 }\) x 60° = 30°

But in ∆OAC,

OC = OA (Radii of the circle)

∴ ∠OAC = ∠OCA = ∠ACB

⇒ x = 30°

**(vii)** In the circle, ∠BAC and ∠BDC are in the same segment

∴ ∠BDC = ∠BAC = 50°

Now in ABCD,

∠DBC + ∠BCD + ∠BDC = 180° (Angles of a triangle)

⇒ 70° + x + 50° = 180°

⇒ x + 120° = 180° ⇒ x = 180° – 120° = 60°

∴ x = 60°

**(viii) **In circle with centre O,

∠OBD = 40°

AB and CD are diameters of the circle

∠DBA and ∠ACD are in the same segment

∴ ∠ACD = ∠DBA = 40°

In AOAC, OA = OC (Radii of the circle)

∴ ∠OAC = ∠OCA = 40°

and ∠OAC + ∠OCA + ∠AOC = 180° (Angles in a triangle)

⇒ 40° + 40° + x = 180°

⇒ x + 80° = 180° ⇒ x = 180° – 80° = 100°

∴ x = 100°

**(ix)** In the circle, ABCD is a cyclic quadrilateral ∠ADB = 32°, ∠DAC = 28° and ∠ABD = 50°

∠ABD and ∠ACD are in the same segment of a circle

∴ ∠ABD = ∠ACD ⇒ ∠ACD = 50°

Similarly, ∠ADB = ∠ACB

⇒ ∠ACB = 32°

Now, ∠DCB = ∠ACD + ∠ACB

= 50° + 32° = 82°

∴ x = 82°

**(x)** In a circle,

∠BAC = 35°, ∠CBD = 65°

∠BAC and ∠BDC are in the same segment

∴ ∠BAC = ∠BDC = 35°

In ∆BCD,

∠BDC + ∠BCD + ∠CBD = 180° (Angles in a triangle)

⇒ 35° + x + 65° = 180°

⇒ x + 100° = 180°

⇒ x = 180° – 100° = 80°

∴ x = 80°

**(xi)** In the circle,

∠ABD and ∠ACD are in the same segment of a circle

∴ ∠ABD = ∠ACD = 40°

Now in ∆CPD,

∠CPD + ∠PCD + ∠PDC = 180° (Angles of a triangle)

110° + 40° + x = 180°

⇒ x + 150° = 180°

∴ x= 180°- 150° = 30°

**(xii)** In the circle, two diameters AC and BD intersect each other at O

∠BAC = 50°

In ∆OAB,

OA = OB (Radii of the circle)

∴ ∠OBA = ∠OAB = 52°

⇒ ∠ABD = 52°

But ∠ABD and ∠ACD are in the same segment of the circle

∴ ∠ABD = ∠ACD ⇒ 52° = x

∴ x = 52°

**Question 4.**

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

**Solution:**

**Given** **:** O is the circumcentre of ∆ABC.

OD ⊥ BC

OB is joined

**To prove :** ∠BOD = ∠A

**Construction :** Join OC.

**Proof :** Arc BC subtends ∠BOC at the centre and ∠BAC at the remaining part of the circle

∴ ∠BOC = 2∠A …(i)

In right ∆OBD and ∆OCD Side OD = OD (Common)

Hyp. OB = OC (Radii of the circle)

∴ ∆OBD ≅ ∆OCD (RHS criterion)

∴ ∠BOD = ∠COD = \(\frac { 1 }{ 2 }\) ∠BOC

⇒ ∠BOC = 2∠BOD …(ii)

From (i) and (ii)

2∠BOD = 2∠A

∴∠BOD = ∠A

**Question 5.**

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

**Solution:**

**Given :** In the figure, a circle with centre O OB is the bisector of ∠ABC

**To prove :** AB = BC

**Construction :** Draw OL ⊥ AB and OM ⊥ BC

**Proof:** In ∆OLB and ∆OMB,

∠1 = ∠2 (Given)

∠L = ∠M (Each = 90°)

OB = OB (Common)

∴ ∆OLB ≅ ∆OMB (AAS criterion)

∴ OL = OM (c.p.c.t.)

But these are distance from the centre and chords equidistant from the centre are equal

∴ Chord BA = BC

Hence AB = BC

**Question 6.**

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

**Solution:**

In the figure, two circles with centres O and O’ intersect each other at B and C.

ACD is a line, ∠AOB = 130°

Arc AB subtends ∠AOB at the centre O and ∠ACB at the remaining part of the circle.

∴ ∠ACB =\(\frac { 1 }{ 2 }\)∠AOB

= \(\frac { 1 }{ 2 }\) x 130° = 65°

But ∠ACB + ∠BCD = 180° (Linear pair)

⇒ 65° + ∠BCD = 180°

⇒ ∠BCD = 180°-65°= 115°

Now, arc BD subtends reflex ∠BO’D at the centre and ∠BCD at the remaining part of the circle

∴ ∠BO’D = 2∠BCD = 2 x 115° = 230°

But ∠BO’D + reflex ∠BO’D = 360° (Angles at a point)

⇒ x + 230° = 360°

⇒ x = 360° -230°= 130°

Hence x = 130°

**Question 7.**

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

**Solution:**

Arc AB subtend ∠ACB and ∠ADB in the same segment of a circle

∴ ∠ACB = ∠ADB = 40°

In ∆PDB,

∠DPB + ∠PBD + ∠ADB = 180° (Sum of angles of a triangle)

⇒ 120° + ∠PBD + 40° = 180°

⇒ 160° + ∠PBD = 180°

⇒ ∠PBD = 180° – 160° = 20°

⇒ ∠CBD = 20°

**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.**

**Solution:**

A circle with centre O, a chord AB = radius of the circle C and D are points on the minor and major arcs of the circle

∴ ∠ACB and ∠ADB are formed Now in ∆AOB,

OA = OB = AB (∵ AB = radii of the circle)

∴ ∆AOB is an equilateral triangle,

∴ ∠AOB = 60°

Now arc AB subtends ∠AOB at the centre and ∠ADB at the remainder part of the circle.

∴ ∠ADB = \(\frac { 1 }{ 2 }\) ∠AOB = \(\frac { 1 }{ 2 }\)x 60° = 30°

Now ACBD is a cyclic quadrilateral,

∴ ∠ADB + ∠ACB = 180° (Sum of opposite angles of cyclic quad.)

⇒ 30° + ∠ACB = 180°

⇒ ∠ACB = 180° – 30° = 150°

∴ ∠ACB = 150°

Hence angles are 150° and 30°

**Question 9.**

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

**Solution:**

In circle with centre O and ∠AOC = 150°

But ∠AOC + reflex ∠AOC = 360°

∴ 150° + reflex ∠AOC = 360°

⇒ Reflex ∠AOC = 360° – 150° = 210°

Now arc AEC subtends ∠AOC at the centre and ∠ABC at the remaining part of the circle.

Reflex ∠AOC = 2∠ABC

⇒ 210° = 2∠ABC

∴ ∠ABC = \(\frac { { 210 }^{ \circ } }{ 2 }\) = 105°

**Question 10.**

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

**Solution:**

**Given :** In circle, O is centre

**To prove :** ∠x = ∠y + ∠z

**Proof :** ∵ ∠3 and ∠4 are in the same segment of the circle

∴ ∠3 = ∠4 …(i)

∵ Arc AB subtends ∠AOB at the centre and ∠3 at the remaining part of the circle

∴ ∠x = 2∠3 = ∠3 + ∠3 = ∠3 + ∠4 (∵ ∠3 = ∠4) …(ii)

In ∆ACE,

Ext. ∠y = ∠3 + ∠1

(Ext. is equal to sum of its interior opposite angles)

⇒ ∠3 – ∠y – ∠1 …(ii)

From (i) and (ii),

∠x = ∠y – ∠1 + ∠4 …(iii)

Similarly in ∆ADF,

Ext. ∠4 = ∠1 + ∠z …(iv)

From (iii) and (iv)

∠x = ∠y-∠l + (∠1 + ∠z)

= ∠y – ∠1 + ∠1 + ∠z = ∠y + ∠z

Hence ∠x = ∠y + ∠z

**Question 11.**

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

**Solution:**

In the figure, O is the centre of the circle,

PQ is the diameter and ∠ROS = 40°

Now we have to find ∠RTS

Arc RS subtends ∠ROS at the centre and ∠RQS at the remaining part of the circle

∴ ∠RQS = \(\frac { 1 }{ 2 }\) ∠ROS

= \(\frac { 1 }{ 2 }\) x 40° = 20°

∵ ∠PRQ = 90° (Angle in a semi circle)

∴ ∠QRT = 180° – 90° = 90° (∵ PRT is a straight line)

Now in ∆RQT,

∠RQT + ∠QRT + ∠RTQ = 180° (Angles of a triangle)

⇒ 20° + 90° + ∠RTQ = 180°

⇒ ∠RTQ = 180° – 20° – 90° = 70° or ∠RTS = 70°

Hence ∠RTS = 70°

Hope given RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.4 are helpful to complete your math homework.

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