RD Sharma 2020 solution class 9 chapter 15 Circles Exercise 15.5

Exercise 15.5

Page-15.100

Question 1:

In the given figure, ΔABC is an equilateral triangle. Find m∠BEC.
 

Answer 1:

It is given that, is an equilateral triangle

We have to find

Since is an equilateral triangle.

So

And

…… (1)

Since, quadrilateral BACE  is a cyclic qualdrilateral

So ,                                         (Sum of opposite angles of cyclic quadrilateral is.)

Hence

Question 2:

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

Answer 2:

Disclaimer: Figure given in the book was showing m∠PQR as m∠SQR. 

It is given that ΔPQR is an isosceles triangle with PQ = PR and m∠PQR = 35°

 

We have to find the m∠QSR and m∠QTR

Since ΔPQR is an isosceles triangle

So ∠PQR = ∠PRQ = 35° 

Then

Since PQTR is a cyclic quadrilateral

So

In cyclic quadrilateral QSRT we have

Hence,

and  

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Question 3:

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

Answer 3:

It is given that O is centre of the circle and ∠BOD = 160°

             

We have to find the values of x and y.

As we know that 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.

Therefore,

Since, quadrilateral ABCD is a cyclic quadrilateral.

So,

x + y = 180°              (Sum of opposite angles of a cyclic quadrilateral is 180°.)

Hence and

Question 4:

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

Answer 4:

It is given that ∠BCD = 100° and ∠ABD = 70°

We have to find the ∠ADB

We have

∠A + ∠C = 180°                     (Opposite pair of angle of cyclic quadrilateral)

So,

Now in is and

Therefore,

Hence,

Question 5:

If ABCD is a cyclic quadrilateral in which AD || BC (In the given figure). Prove that ∠B = ∠C.
 

Answer 5:

It is given that, ABCD is cyclic quadrilateral in which AD || BC

We have to prove

Since, ABCD is a cyclic quadrilateral

So,

and               ..… (1)

and              (Sum of pair of consecutive interior angles is 180°) …… (2)

From equation (1) and (2) we have

…… (3)

…… (4)

Hence Proved

Question 6:

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

Answer 6:

It is given that,

We have to find

Since,    (Given)

So,

                     (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,

               (Opposite pair of angle of cyclic quadrilateral)

So,

…… (1)

      (Linear pair)

         ()

Hence

Question 7:

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

Answer 7:

It is given that, AB and CD are diameter with center O and

We have to find  

Construction: Join the point A and D to form line AD

Clearly arc AD subtends at B and at the centre.

Therefore,  $\angle AOD=2\angle ABD=100°$ …… (1)

Since CD is a straight line then
        
$\angle DOA+\angle AOC=180° \left(\mathrm{Linear} \mathrm{pair}\right)$

     

Hence

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).

Answer 8:

It is given that, as diameter, is centre and

We have to find and

Since angle in a semi-circle is a right angle therefore

In we have

(Given)

    (Angle in semi-circle is right angle)

Now in we have

                                        

Hence and  

Question 9:

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

Answer 9:

It is given that, ABCD is a cyclic quadrilateral such that AB || CD and

Sum of pair of opposite angles of cyclic quadrilateral is 180°.

                  ( given)

So,

Also AB || CD and BC transversal

So,

Now

               

Question 10:

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

Answer 10:

It is given that

ABCD is cyclic quadrilateral and

We have to find

Since ABCD is cyclic quadrilateral and sum of opposite pair of cyclic quadrilateral is 180°.

So

And

Therefore

Hence

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Question 11:

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

Answer 11:

It is given that, O is the centre of the circle and $\angle DAB=50°$.

   

We have to find  the values of x and y.

ABCD is a cyclic quadrilateral and

So,

50° + y = 180°
y = 180° − 50°
y = 130°
 

Clearly is an isosceles triangle with OA = OB and

Then,

                        (Since)

So,

x + $\angle AOB$ = 180°     (Linear pair)

Therefore, x =  180° − 80° = 100°

Hence,

and

 

Question 12:

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

Answer 12:

It is given that, and

We have to find the

In given we have

$$\begin{gathered} \angle ABC+\angle BCA+\angle BAC=180° \left(\mathrm{Angle} \mathrm{sum} \mathrm{property}\right) \\ \Rightarrow \angle ABC=180°-\left(60°+20°\right)=100° \end{gathered}$$

In cyclic quadrilateral we have

         (Sum of pair of opposite angles of a cyclic quadilateral is 180º)

Then,

Hence

Question 13:

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

Answer 13:

It is given that, ABC is an equilateral triangle

We have to find and

Since is an equilateral triangle

So,

And is cyclic quadrilateral

So      (Sum of opposite pair of angles of a cyclic quadrilateral is 180°.)

Then,

Similarly BECD is also cyclic quadrilateral

So,

Hence, and  .

Question 14:

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

Answer 14:

It is given that, O is the centre of the circle and

We have to find the value of x, y and z.

Since, angle in the same segment are equal

So

And z = 30°               

As 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.
Since

Then,

y = 2z
   = 2 × 30°
   = 60°

Since, the sum of opposite pair of angles of a cyclic quadrilateral is 180°.

z + x = 180°
x = 180° − 30°
   = 150°

Hence,
x = 150°, y = 60° and z = 30°

Question 15:

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

Answer 15:

It is given that, and, are cyclic quadrilateral

We have to find the value of x and y.

Since , is a cyclic quadrilateral

So      (Opposite angle of a cyclic quadrilateral are supplementary)

        ()

     ..… (1)

x = 180° − 102°
   = 78°

Now in cyclic quadrilateral DCFE
x + y = 180°    (Opposite angles of a cyclic quadrilateral are supplementary)
y = 180° − 78°
   = 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°

Answer 16:

It is given that ∠A – ∠C = 60° and ABCD is a cyclic quadrilateral.

We have to prove that smaller of two is 60°

Since ABCD is a cyclic quadrilateral

So ∠A + ∠C = 180°                 (Sum of opposite pair of angles of cyclic quadrilateral is 180°)   ..… (1)

And,

∠A – ∠C = 60°               (Given)            ..… (2)

Adding equation (1) and (2) we have

So, ∠C = 60°

Hence, smaller of two is 60°.

 

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Question 17:

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

Answer 17:

Here, ABCD is a cyclic quadrilateral, we need to find x.

In cyclic quadrilateral the sum of opposite angles is equal to 180°.

Therefore,

 $$\begin{gathered} \angle ADC+\angle ABC=180° \\ \Rightarrow 180°-80°+180°-x=180° \\ \Rightarrow x=100° \end{gathered}$$

Hence, the value of x is 100°.

Question 18:

ABCD is a cyclic quadrilateral in which:

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

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

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

Answer 18:

(i) It is given that , and

   

We have to find

In cyclic quadrilateral ABCD

      ..… (1)

      ..… (2)

Since, 

So,

Therefore in ,

So ,                   ..… (3)

Now,                   ( and is transversal)

(ii) It is given that , and

We have to find

                    _{(Angle in the same segment are equal)}

_Hence,

(iii) It is given that, ∠BCD = 100° and ∠ABD = 70°

As we know that sum of the opposite pair of angles of cyclic quadrilateral is 180°.

$$\begin{gathered} \angle DAB+\angle BCD=180° \\ \Rightarrow \angle DAB=180°-100° \\ =80° \end{gathered}$$

In ΔABD we have,

$$\begin{gathered} \angle DAB+\angle ABD+\angle BDA=180° \\ \Rightarrow \angle BDA=180°-150°=30° \end{gathered}$$

_Hence, 

Question 19:

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

Answer 19:

Here, ABCD is a rhombus; we have to prove the four circles described on the four sides of any rhombus ABCD pass through the point of intersection of its diagonals AC and BD.

Let the diagonals AC and BD intersect at O.

We know that the diagonals of a rhombus intersect at right angle.

Therefore,

Now, means that circle described on AB as diameter passes through O.

Similarly the remaining three circles with BC, CD and AD as their diameter will also pass through O.

Hence, all the circles with described on the four sides of any rhombus ABCD pass through the point of intersection of its diagonals AC and BD.

Question 20:

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

Answer 20:

To prove: AC = BD

Proof: We know that equal chords subtend equal at the centre of circle and the angle subtended by a chord at the centre is twice the angle subtended by it at remaining part of the circle.

$$\begin{gathered} \angle AOD=\angle BOC \left(O \mathrm{is} \mathrm{the} \mathrm{centre} \mathrm{of} \mathrm{the} \mathrm{circle}\right) \\ \angle AOD=2\angle ACD \\ \mathrm{and} \angle BOC=2\angle BDC \\ \mathrm{Since}, \angle AOD=\angle BOC \\ \Rightarrow \angle ACD=\angle BDC .....\left(1\right) \\ \angle ACB=\angle ADB .....\left(2\right) \left(\mathrm{Angle} \mathrm{in} \mathrm{the} \mathrm{same} \mathrm{segment} \mathrm{are} \mathrm{equal}\right) \\ \mathrm{Adding} \left(1\right) \mathrm{and} \left(2\right) \\ \angle BCD=\angle ADC .....\left(3\right) \\ \mathrm{In} △ACD \mathrm{and} △BDC \\ CD=CD \left(\mathrm{common}\right) \\ \angle BCD=\angle ADC \left[\mathrm{Using}\left(3\right)\right] \\ AD=BC \left(\mathrm{given}\right) \\ \mathrm{Hence}, △ACD\cong BDC \left(\mathrm{SAS} \mathrm{congruency} \mathrm{criterion}\right) \\ \therefore AC=BD \left(\mathrm{cpct}\right) \\ \mathrm{Hence} \mathrm{Proved} \end{gathered}$$

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).

Answer 21:

$$\begin{gathered} \angle ADB=90° \left(\mathrm{Angle} \mathrm{in} \mathrm{a} \mathrm{semicircle}\right) \\ \angle ADC=90° \left(\mathrm{Angle} \mathrm{in} \mathrm{a} \mathrm{semicircle}\right) \\ \mathrm{So}, \angle ADB+\angle ADC=90°+90°=180° \\ \mathrm{Therefore}, BDC \mathrm{is} \mathrm{a} \mathrm{line}. \\ \mathrm{Hence}, \mathrm{the} \mathrm{point} \mathrm{of} \mathrm{intersection} \mathrm{of} \mathrm{two} \mathrm{circles} \mathrm{lie} \mathrm{on} \mathrm{the} \mathrm{third} \mathrm{side}. \end{gathered}$$

 

Question 22:

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

Answer 22:

If in cyclic quadrilateral , then we have to find the other three angles.

Since, AD is parallel to BC, So,
  (Alternate interior angles)

Now, since ABCD is cyclic quadrilateral, so

And,

$\mathrm{Hence}, \angle A=110°, \angle C=70° \mathrm{and} \angle D=110°$.

Question 23:

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

Answer 23:

It is given that is a cyclic quadrilateral with and as its diagonals.

We have to find

Since angles in the same segment of a circle are equal

So

Since (Opposite angle of cyclic quadrilateral)

Hence

Question 24:

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

Answer 24:

To prove: Perpendicular bisector of side AB, BC, CD and DA are concurrent i.e, passes through the same point.

Proof:

We know that the perpendicular bisector of every chord of a circle always passes through the centre.

Therefore, Perpendicular bisectors of chord AB, BC, CD and DA pass through the centre which means they all passes through the same point.

Hence, the perpendicular bisector of AB, BC, CD and DA are concurrent.

Question 25:

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

Answer 25:

Here, ABCD is a cyclic rectangle; we have to prove that the centre of the corresponding circle is the intersection of its diagonals.

Let O be the centre of the circle.

We know that the angle formed in the semicircle is 90°.

Since, ABCD is a rectangle, So

Therefore, AC and BD are diameter of the circle.

We also know that the intersection of any two diameter is the centre of the circle.

Hence, the centre of the circle circumscribing the cyclic rectangle ABCD is the point of intersection of its diagonals.

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.

Answer 26:

(i) If ABCD is a cyclic quadrilateral in which AB and CD when produced meet in E such that EA =  ED, then we have to prove the following, AD || BC

(ii) EB = EC

 

 

(i) It is given that EA = ED, so

$\angle EAD=\angle EDA=x$

Since, ABCD is cyclic quadrilateral

Now,

Therefore, the adjacent angles andare supplementary

Hence, AD || BC

(ii) Since, AD and BC are parallel to each other, so,

$$\begin{gathered} \angle ECB=\angle EDA \left(\mathrm{Corresponding} \mathrm{angles}\right) \\ \angle EBC=\angle EAD \left(\mathrm{Corresponding} \mathrm{angles}\right) \\ \mathrm{But}, \angle EDA=\angle EAD \\ \mathrm{Therefore}, \angle ECB=\angle EBC \\ \Rightarrow EC=EB \\ \mathrm{Therefore}, △ECB \mathrm{is} \mathrm{an} \mathrm{isosceles} \mathrm{triangle}. \end{gathered}$$

 

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Question 27:

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

Answer 27:

$$\begin{gathered} \stackrel{⏜}{QP} \mathrm{is} \mathrm{a} \mathrm{major} \mathrm{arc} \mathrm{and} \angle PSQ \mathrm{is} \mathrm{the} \mathrm{angle} \mathrm{formed} \mathrm{by} \mathrm{it} \mathrm{in} \mathrm{the} \mathrm{alternate} \mathrm{segment}. \\ \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{the} \mathrm{angle} \mathrm{subtended} \mathrm{by} \mathrm{an} \mathrm{arc} \mathrm{at} \mathrm{the} \mathrm{centre} \mathrm{is} \mathrm{twice} \mathrm{the} \mathrm{angle} \mathrm{subtended} \mathrm{by} \mathrm{it} \mathrm{at} \mathrm{any} \mathrm{point} \mathrm{of} \mathrm{the} \mathrm{alternate} \mathrm{segment} \mathrm{of} \mathrm{the} \mathrm{circle}. \\ \therefore 2\angle PSQ=m\left(\stackrel{⏜}{QP}\right) \\ \Rightarrow 2\angle PSQ=360°-m\left(\stackrel{⏜}{PQ}\right) \\ \Rightarrow 2\angle PSQ=360°-\angle POQ \\ \Rightarrow 2\angle PSQ=360°-180° \left(\because \angle POQ<180°\right) \\ \Rightarrow 2\angle PSQ>180° \\ \Rightarrow \angle PSQ>90° \end{gathered}$$

Thus, the angle in a segment shorter than a semi-circle is greater than a right angle.

Question 28:

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

Answer 28:

$$\begin{gathered} \mathrm{To} \mathrm{prove}: \angle ABC \mathrm{is} \mathrm{an} \mathrm{acute} \mathrm{angle} \\ \mathrm{Proof}: \\ AD \mathrm{being} \mathrm{the} \mathrm{diameter} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{circle} \\ \Rightarrow \angle ACD=90° \left[\mathrm{Angle} \mathrm{in} \mathrm{a} \mathrm{semicircle} \mathrm{is} \mathrm{a} \mathrm{right} \mathrm{angle}\right] \\ \mathrm{Now}, \mathrm{in} △ACD, \angle ACD=90° \mathrm{which} \mathrm{means} \mathrm{that} \angle ADC \mathrm{is} \mathrm{an} \mathrm{acute} \mathrm{angle}. .....\left(1\right) \\ \mathrm{Again}, \angle ABC=\angle ADC \left[\mathrm{Angle} \mathrm{in} \mathrm{a} \mathrm{same} \mathrm{segment} \mathrm{are} \mathrm{always} \mathrm{equal}\right] \\ \Rightarrow \angle ABC \mathrm{is} \mathrm{also} \mathrm{an} \mathrm{acute} \mathrm{angle}. \left[\mathrm{Using}\left(1\right)\right] \\ \mathrm{Hence} \mathrm{proved} \end{gathered}$$

Question 29:

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

Answer 29:

We have to prove that

Let be a right angle at B and P be midpoint of AC

Draw a circle with center at P and AC diameter

Since therefore circle passing through B

So

Hence

Proved.