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 15.4

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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