## RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5

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 15.5

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, ∆ABC is an equilateral triangle. Find m ∠BEC.

Solution:

∵ ∆ABC is an equilateral triangle

∴ A = 60°

∵ ABEC is a cyclic quadrilateral

∴ ∠A + ∠E = 180° (Sum of opposite angles)

⇒ 60° + ∠E = 180°

⇒ ∠E = 180° – 60° = 120°

∴ m ∠BEC = 120°

Question 2.

In the figure, ∆PQR is an isosceles triangle with PQ = PR and m ∠PQR = 35°. Find m ∠QSR and m ∠QTR.

Solution:

In the figure, ∆PQR is an isosceles PQ = PR

∠PQR = 35°

∴ ∠PRQ = 35°

But ∠PQR + ∠PRQ + ∠QPR = 180° (Sum of angles of a triangle)

⇒ 35° + 35° + ∠QPR = 180°

⇒ 70° + ∠QPR = 180°

∴ ∠QPR = 180° – 70° = 110°

∵ ∠QSR = ∠QPR (Angle in the same segment of circles)

∴ ∠QSR = 110°

But PQTR is a cyclic quadrilateral

∴ ∠QTR + ∠QPR = 180°

⇒ ∠QTR + 110° = 180°

⇒ ∠QTR = 180° -110° = 70°

Hence ∠QTR = 70°

Question 3.

In the figure, O is the centre of the circle. If ∠BOD = 160°, find the values of x and y.

Solution:

In the figure, O is the centre of the circle ∠BOD =160°

ABCD is the cyclic quadrilateral

∵ Arc BAD subtends ∠BOD is the angle at the centre and ∠BCD is on the other part of the circle

∴ ∠BCD = \(\frac { 1 }{ 2 }\) ∠BOD

⇒ x = \(\frac { 1 }{ 2 }\) x 160° = 80°

∵ ABCD is a cyclic quadrilateral,

∴ ∠A + ∠C = 180°

⇒ y + x = 180°

⇒ y + 80° = 180°

⇒ y =180°- 80° = 100°

∴ x = 80°, y = 100°

Question 4.

In the figure, ABCD is a cyclic quadrilateral. If ∠BCD = 100° and ∠ABD = 70°, find ∠ADB.

Solution:

In a circle, ABCD is a cyclic quadrilateral ∠BCD = 100° and ∠ABD = 70°

∵ ABCD is a cyclic quadrilateral,

∴ ∠A + ∠C = 180° (Sum of opposite angles)

⇒ ∠A + 100°= 180°

∠A = 180°- 100° = 80°

Now in ∆ABD,

∠A + ∠ABD + ∠ADB = 180°

⇒ 80° + 70° + ∠ADB = 180°

⇒ 150° +∠ADB = 180°

∴ ∠ADB = 180°- 150° = 30°

Hence ∠ADB = 30°

Question 5.

If ABCD is a cyclic quadrilateral in which AD || BC. Prove that ∠B = ∠C.

Solution:

Given : ABCD is a cyclic quadrilateral in which AD || BC

To prove : ∠B = ∠C

Proof : ∵ AD || BC

∴ ∠A + ∠B = 180°

(Sum of cointerior angles)

But ∠A + ∠C = 180°

(Opposite angles of the cyclic quadrilateral)

∴ ∠A + ∠B = ∠A + ∠C

⇒ ∠B = ∠C

Hence ∠B = ∠C

Question 6.

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

Solution:

Arc AC subtends ∠AOC at the centre and ∠APC at the remaining part of the circle

∴ ∠APC = \(\frac { 1 }{ 2 }\) ∠AOC

= \(\frac { 1 }{ 2 }\) x 100° = 50°

∵ APCB is a.cyclic quadrilateral,

∴ ∠APC + ∠ABC = 180°

⇒ 50° + ∠ABC = 180° ⇒ ∠ABC =180°- 50°

∴ ∠ABC =130°

But ∠ABC + ∠CBD = 180° (Linear pair)

⇒ 130° + ∠CBD = 180°

⇒ ∠CBD = 180°- 130° = 50°

∴ ∠CBD = 50°

Question 7.

In the figure, AB and CD are diameiers of a circle with centre O. If ∠OBD = 50°, find ∠AOC.

Solution:

Two diameters AB and CD intersect each other at O. AC, CB and BD are joined

∠DBA = 50°

∠DBA and ∠DCA are in the same segment

∴ ∠DBA = ∠DCA = 50°

In ∆OAC, OA = OC (Radii of the circle)

∴ ∠OAC = ∠OCA = ∠DCA = 50°

and ∠OAC + ∠OCA + ∠AOC = 180° (Sum of angles of a triangle)

⇒ 50° + 50° + ∠AOC = 180°

⇒ 100° + ∠AOC = 180°

⇒ ∠AOC = 180° – 100° = 80°

Hence ∠AOC = 80°

Question 8.

On a semi circle with AB as diameter, a point C is taken so that m (∠CAB) = 30°. Find m (∠ACB) and m (∠ABC).

Solution:

A semicircle with AB as diameter

∠ CAB = 30°

∠ACB = 90° (Angle in a semi circle)

But ∠CAB + ∠ACB + ∠ABC = 180°

⇒ 30° + 90° + ∠ABC – 180°

⇒ 120° + ∠ABC = 180°

∴ ∠ABC = 180°- 120° = 60°

Hence m ∠ACB = 90°

and m ∠ABC = 60°

Question 9.

In a cyclic quadrilateral ABCD, if AB || CD and ∠B = 70°, find the remaining angles.

Solution:

In a cyclic quadrilateral ABCD, AB || CD and ∠B = 70°

∵ ABCD is a cyclic quadrilateral

∴ ∠B + ∠D = 180°

⇒ 70° + ∠D = 180°

⇒ ∠D = 180°-70° = 110°

∵ AB || CD

∴ ∠A + ∠D = 180° (Sum of cointerior angles)

∠A+ 110°= 180°

⇒ ∠A= 180°- 110° = 70°

Similarly, ∠B + ∠C = 180°

⇒ 70° + ∠C- 180° ‘

⇒ ∠C = 180°-70°= 110°

∴ ∠A = 70°, ∠C = 110°, ∠D = 110°

Question 10.

In a cyclic quadrilateral ABCD, if m ∠A = 3(m ∠C). Find m ∠A.

Solution:

In cyclic quadrilateral ABCD, m ∠A = 3(m ∠C)

∵ ABCD is a cyclic quadrilateral,

∴ ∠A + ∠C = 180°

⇒ 3 ∠C + ∠C = 180° ⇒ 4∠C = 180°

⇒ ∠C = \(\frac { { 180 }^{ \circ } }{ 4 }\) = 45°

∴ ∠A = 3 x 45°= 135°

Hence m ∠A =135°

Question 11.

In the figure, O is the centre of the circle and ∠DAB = 50°. Calculate the values of x and y.

Solution:

In the figure, O is the centre of the circle ∠DAB = 50°

∵ ABCD is a cyclic quadrilateral

∴ ∠A + ∠C = 180°

⇒ 50° + y = 180°

⇒ y = 180° – 50° = 130°

In ∆OAB, OA = OB (Radii of the circle)

∴ ∠A = ∠OBA = 50°

∴ Ext. ∠DOB = ∠A + ∠OBA

x = 50° + 50° = 100°

∴ x= 100°, y= 130°

Question 12.

In the figure, if ∠BAC = 60° and ∠BCA = 20°, find ∠ADC.

Solution:

In ∆ABC,

∠BAC + ∠ABC + ∠ACB = 180° (Sum of angles of a triangle)

60° + ∠ABC + 20° = 180°

∠ABC + 80° = 180°

∴ ∠ABC = 180° -80°= 100°

∵ ABCD is a cyclic quadrilateral,

∴ ∠ABC + ∠ADC = 180°

100° + ∠ADC = 180°

∴ ∠ADC = 180°- 100° = 80°

Question 13.

In the figure, if ABC is an equilateral triangle. Find ∠BDC and ∠BEC.

Solution:

In a circle, ∆ABC is an equilateral triangle

∴ ∠A = 60°

∵ ∠BAC and ∠BDC are in the same segment

∴ ∠BAC = ∠BDC = 60°

∵ BECD is a cyclic quadrilateral

∴ ∠BDC + ∠BEC = 180°

⇒ 60° + ∠BEC = 180°

⇒ ∠BEC = 180°-60°= 120°

Hence ∠BDC = 60° and ∠BEC = 120°

Question 14.

In the figure, O is the centre of the circle. If ∠CEA = 30°, find the values of x, y and z.

Solution:

∠AEC and ∠ADC are in the same segment

∴ ∠AEC = ∠ADC = 30°

∴ z = 30°

ABCD is a cyclic quadrilateral

∴ ∠B + ∠D = 180°

⇒ x + z = 180°

⇒ x + 30° = 180°

⇒ x = 180° – 30° = 150°

Arc AC subtends ∠AOB at the centre and ∠ADC at the remaining part of the circle

∴ ∠AOC = 2∠D = 2 x 30° = 60°

∴ y = 60°

Hence x = 150°, y – 60° and z = 30°

Question 15.

In the figure, ∠BAD = 78°, ∠DCF = x° and ∠DEF = y°. Find the values of x and y.

Solution:

In the figure, two circles intersect each other at C and D

∠BAD = 78°, ∠DCF = x, ∠DEF = y

ABCD is a cyclic quadrilateral

∴ Ext. ∠DCF = its interior opposite ∠BAD

⇒ x = 78°

In cyclic quadrilateral CDEF,

∠DCF + ∠DEF = 180°

⇒ 78° + y = 180°

⇒ y = 180° – 78°

y = 102°

Hence x = 78°, and y- 102°

Question 16.

In a cyclic quadrilateral ABCD, if ∠A – ∠C = 60°, prove that the smaller of two is 60°.

Solution:

In cyclic quadrilateral ABCD,

∠A – ∠C = 60°

But ∠A + ∠C = 180° (Sum of opposite angles)

Adding, 2∠A = 240° ⇒ ∠A = \(\frac { { 62 }^{ \circ } }{ 2 }\) = 120° and subtracting

2∠C = 120° ⇒ ∠C = \(\frac { { 120 }^{ \circ } }{ 2 }\) = 60°

∴ Smaller angle of the two is 60°.

Question 17.

In the figure, ABCD is a cyclic quadrilateral. Find the value of x.

Solution:

∠CDE + ∠CDA = 180° (Linear pair)

⇒ 80° + ∠CDA = 180°

⇒ ∠CDA = 180° – 80° = 100°

In cyclic quadrilateral ABCD,

Ext. ∠ABF = Its interior opposite angle ∠CDA = 100°

∴ x = 100°

Question 18.

ABCD is a cyclic quadrilateral in which:

(i) BC || AD, ∠ADC =110° and ∠B AC = 50°. Find ∠DAC.

(ii) ∠DBC = 80° and ∠BAC = 40°. Find ∠BCD.

(iii) ∠BCD = 100° and ∠ABD = 70°, find ∠ADB.

Solution:

(i) In the figure,

ABCD is a cyclic quadrilateral and AD || BC, ∠ADC = 110°

∠BAC = 50°

∵ ∠B + ∠D = 180° (Sum of opposite angles)

⇒ ∠B + 110° = 180°

∴ ∠B = 180°- 110° = 70°

Now in ∆ABC,

∠CAB + ∠ABC + ∠BCA = 180° (Sum of angles of a triangle)

⇒ 50° + 70° + ∠BCA = 180°

⇒ 120° + ∠BCA = 180°

⇒ ∠BCA = 180° – 120° = 60°

But ∠DAC = ∠BCA (Alternate angles)

∴ ∠DAC = 60°

(ii) In cyclic quadrilateral ABCD,

Diagonals AC and BD are joined ∠DBC = 80°, ∠BAC = 40°

Arc DC subtends ∠DBC and ∠DAC in the same segment

∴ ∠DBC = ∠DAC = 80°

∴ ∠DAB = ∠DAC + ∠CAB = 80° + 40° = 120°

But ∠DAC + ∠BCD = 180° (Sum of opposite angles of a cyclic quad.)

⇒ 120° +∠BCD = 180°

⇒ ∠BCD = 180°- 120° = 60°

(iii) In the figure, ABCD is a cyclic quadrilateral BD is joined

∠BCD = 100°

and ∠ABD = 70°

∠A + ∠C = 180° (Sum of opposite angles of cyclic quad.)

∠A+ 100°= 180°

⇒ ∠A= 180°- 100°

∴ ∠A = 80°

Now in ∆ABD,

∠A + ∠ABD + ∠ADB = 180° (Sum of angles of a triangle)

⇒ 80° + 70° + ∠ADB = 180°

⇒ 150° +∠ADB = 180°

⇒ ∠ADB = 180°- 150° = 30°

∴ ∠ADB = 30°

Question 19.

Prove that the circles described on the four sides of a rhombus as diameter, pass through the point of intersection of its diagonals.

Solution:

Given : ABCD is a rhombus. Four circles are drawn on the sides AB, BC, CD and DA respectively

To prove : The circles pass through the point of intersection of the diagonals of the rhombus ABCD

Proof: ABCD is a rhombus whose diagonals AC and BD intersect each other at O

∵ The diagonals of a rhombus bisect each other at right angles

∴ ∠AOB = ∠BOC = ∠COD = ∠DOA = 90°

Now when ∠AOB = 90°

and a circle described on AB as diameter will pass through O

Similarly, the circles on BC, CD and DA as diameter, will also pass through O

Question 20.

If the two sides of a pair of opposite sides of a cyclic quadrilateral are equal, prove that is diagonals are equal.

Solution:

Given : In cyclic quadrilateral ABCD, AB = CD

AC and BD are the diagonals

To prove : AC = BC

Proof: ∵ AB = CD

∴ arc AB = arc CD

Adding arc BC to both sides, then arc AB + arc BC = arc BC + arc CD

⇒ arc AC = arc BD

∴ AC = BD

Hence diagonal of the cyclic quadrilateral are equal.

Question 21.

Circles are described on the sides of a triangle as diameters. Prove that the circles on any two sides intersect each other on the third side (or third side produced).

Solution:

Given : In ∆ABC, circles are drawn on sides AB and AC

To prove : Circles drawn on AB and AC intersect at D which lies on BC, the third side

Construction : Draw AD ⊥ BC

Proof: ∵ AD ⊥ BC

∴ ∠ADB = ∠ADC = 90°

So, the circles drawn on sides AB and AC as diameter will pass through D

Hence circles drawn on two sides of a triangle pass through D, which lies on the third side.

Question 22.

ABCD is a cyclic trapezium with AD || BC. If ∠B = 70°, determine other three angles of the trapezium.

Solution:

In the figure, ABCD is a trapezium in which AD || BC and ∠B = 70°

∵ AD || BC

∴ ∠A + ∠B = 180° (Sum of cointerior angles)

⇒ ∠A + 70° = 180°

⇒ ∠A= 180°- 70° = 110°

∴ ∠A = 110°

But ∠A + ∠C = 180° and ∠B + ∠D = 180° (Sum of opposite angles of a cyclic quadrilateral)

∴ 110° + ∠C = 180°

⇒ ∠C = 180°- 110° = 70°

and 70° + ∠D = 180°

⇒ ∠D = 180° – 70° = 110°

∴ ∠A = 110°, ∠C = 70° and ∠D = 110°

Question 23.

In the figure, ABCD is a cyclic quadrilateral in which AC and BD are its diagonals. If ∠DBC = 55° and ∠BAC = 45°, find ∠BCD.

Solution:

In the figure, ABCD is a cyclic quadrilateral whose diagonals AC and BD are drawn ∠DBC = 55° and ∠BAC = 45°

∵ ∠BAC and ∠BDC are in the same segment

∴ ∠BAC = ∠BDC = 45°

Now in ABCD,

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

⇒ 55° + 45° + ∠BCD = 180°

⇒ 100° + ∠BCD = 180°

⇒ ∠BCD = 180° – 100° = 80°

Hence ∠BCD = 80°

Question 24.

Prove that the perpendicular bisectors of the sides of a cyclic quadrilateral are concurrent.

Solution:

Given : ABCD is a cyclic quadrilateral

To prove : The perpendicular bisectors of the sides are concurrent

Proof : ∵ Each side of the cyclic quadrilateral is a chord of the circle and perpendicular of a chord passes through the centre of the circle

Hence the perpendicular bisectors of each side will pass through the centre O

Hence the perpendicular bisectors of the sides of a cyclic quadrilateral are concurrent

Question 25.

Prove that the centre of the circle circumscribing the cyclic rectangle ABCD is the point of intersection of its diagonals.

Solution:

Given : ABCD is a cyclic rectangle and diagonals AC and BD intersect each other at O

To prove : O is the point of intersection is the centre of the circle.

Proof : Let O be the centre of the circle- circumscribing the rectangle ABCD

Since each angle of a rectangle is a right angle and AC is the chord of the circle

∴ AC will be the diameter of the circle Similarly, we can prove that diagonal BD is also the diameter of the circle

∴ The diameters of the circle pass through the centre

Hence the point of intersection of the diagonals of the rectangle is the centre of the circle.

Question 26.

ABCD is a cyclic quadrilateral in which BA and CD when produced meet in E and EA = ED. Prove that:

(i) AD || BC

(ii) EB = EC.

Solution:

Given : ABCD is a cyclic quadrilateral in which sides BA and CD are produced to meet at E and EA = ED

To prove :

(i) AD || BC

(ii) EB = EC

Proof: ∵ EA = ED

∴ In ∆EAD

∠EAD = ∠EDA (Angles opposite to equal sides)

In a cyclic quadrilateral ABCD,

Ext. ∠EAD = ∠C

Similarly Ext. ∠EDA = ∠B

∵ ∠EAD = ∠EDA

∴ ∠B = ∠C

Now in ∆EBC,

∵ ∠B = ∠C

∴ EC = EB (Sides opposite to equal sides)

and ∠EAD = ∠B

But these are corresponding angles

∴ AD || BC

Question 27.

Prove that the angle in a segment shorter than a semicircle is greater than a right angle.

Solution:

Given : A segment ACB shorter than a semicircle and an angle ∠ACB inscribed in it

To prove : ∠ACB < 90°

Construction : Join OA and OB

Proof : Arc ADB subtends ∠AOB at the centre and ∠ACB at the remaining part of the circle ∴ ∠ACB = \(\frac { 1 }{ 2 }\) ∠AOB But ∠AOB > 180° (Reflex angle)

∴ ∠ACB > \(\frac { 1 }{ 2 }\) x [80°

⇒ ∠ACB > 90°

Question 28.

Prove that the angle in a segment greater than a semi-circle is less than a right angle

Solution:

Given : A segment ACB, greater than a semicircle with centre O and ∠ACB is described in it

To prove : ∠ACB < 90°

Construction : Join OA and OB

Proof : Arc ADB subtends ∠AOB at the centre and ∠ACB at the remaining part of the circle

∴ ∠ACB =\(\frac { 1 }{ 2 }\) ∠AOB

But ∠AOB < 180° (A straight angle) 1

∴ ∠ACB < \(\frac { 1 }{ 2 }\) x 180°

⇒ ∠ACB <90°

Hence ∠ACB < 90°

Question 29.

Prove that the line segment joining the mid-point of the hypotenuse of a rijght triangle to its opposite vertex is half of the hypotenuse.

Solution:

Given : In a right angled ∆ABC

∠B = 90°, D is the mid point of hypotenuse AC. DB is joined.

To prove : BD = \(\frac { 1 }{ 2 }\) AC

Construction : Draw a circle with centre D and AC as diameter

Proof: ∵ ∠ABC = 90°

∴ The circle drawn on AC as diameter will pass through B

∴ BD is the radius of the circle

But AC is the diameter of the circle and D is mid point of AC

∴ AD = DC = BD

∴ BD= \(\frac { 1 }{ 2 }\) AC

Hope given RD Sharma Class 9 Solutions Chapter 15 Areas of Parallelograms and Triangles Ex 15.5 are helpful to complete your math homework.

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