Exercise E10.5 Circles NCERT Solutions Class 9

Go back to  'Circles'

Chapter 10 Ex.10.5 Question 1

In the given figure, \(A,\, B\) and \(C\) are three points on a circle with center \(O\) such that \(\begin{align}\angle {BOC}=30^{\circ} \text { and } \angle {AOB}=60^{\circ} \end {align}\)If  \(D\) is a point on the circle other than the arc \(ABC,\)  find \(\begin{align} \angle {ADC} \end{align}\)

Solution

Video Solution

What is known?

Two angles subtended by arcs at the centre.

What is unknown?

Value of  \( \angle {ADC.}\)

Reasoning:

The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

Steps:

\[\begin{align} \angle {AOC} &=\angle {AOB}+\angle {BOC} \\ &=90^\circ \end{align}\]

By Theorem \(10.8\) ,

\[\begin{align}  \angle {AOC} &=2 \angle {ADC}  \\ \angle ADC &= \frac{1}{2}\angle AOC\\\angle ADC &= \frac{1}{2} \times 90 = \,45^\circ \\\therefore \angle ADC &= \,45^\circ \end{align}\]
          

Chapter 10 Ex.10.5 Question 2

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

Video Solution

What is known?

Chord’s length is equal to the radius.

What is unknown?

Angle subtended by the chord at a point on the minor arc and also at a point on the major arc.

Reasoning:

  • The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
  • A quadrilateral \(ABCD\) is called cyclic if all the four vertices of it lie on a circle.
  • The sum of either pair of opposite angles of a cyclic quadrilateral is \(\begin{align}180^{\circ}.\end {align}\)

Steps:

Draw a circle with any radius and centre \(O\). Let \(AO\) and \(BO\) be the \(2\) radii of the circle and let \(AB\) be the chord equal to the length of radius. Join them to form a triangle.

\(OA = OB = AB\)

Hence \(\begin{align} \Delta {ABO} \end {align}\) becomes an equilateral triangle.

Draw \(2\) points \(C\) and \(D\) on the circle such that they lie on major arc and minor arc respectively.

Since \(\begin{align} \Delta {ABO} \end {align}\) is an equilateral triangle, we get \(\begin{align} \angle {ABO}=60^{\circ} \end {align}\)

For the arc \(AB,\) \(\begin {align} \angle {AOB}=2 \angle {ACB} \end {align}\) as we know that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

\[\begin{align}\angle ACB &= \frac{1}{2}\angle AOB\\&= \frac{1}{2} \times 60 \\&= 30^\circ \end{align}\]

As you can notice the points \(A, B, C\) and \(D\) lie on the circle. Hence \(A B C D\) is a cyclic quadrilateral.

We know that, the sum of either pair of opposite angles of a cyclic quadrilateral is \(\begin{align}180^\circ.\end {align}\)

Therefore,

\[\begin{align}\angle ACB + \angle ADB &= 180^\circ \\30 + \angle ADB &= 180\\\angle ADB &= 150^\circ \end{align}\]

So when the chord of a circle is equal to the radius of the circle, the angle subtended by the chord at a point on the minor arc is \(\begin{align}150^\circ.\end {align}\) and also at a point on the major arc is \(\begin{align}30^\circ.\end {align}\)

Chapter 10 Ex.10.5 Question 3

In the given figure, \(\begin {align} \angle {PQR}=100^{\circ} \end {align}\) where \(P, Q\) and \(R\)  are points on a circle with center \(O\) . Find \(\begin {align}\angle {OPR} \end {align}\).

Solution

Video Solution

What is known?

\(3\) points on the circle and

\(\begin {align}\angle {PQR}=100^{\circ} \end {align}\)

What is unknown?

Value of \(\angle {PQR} \)

Reasoning:

  • The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
  • A quadrilateral \(\begin {align} {ABCD} \end {align}\) is called cyclic if all the four vertices of it lie on a circle.
  • The sum of either pair of opposite angles of a cyclic quadrilateral is \(\begin {align} 180^{\circ}. \end {align}\)

Steps:

Mark any point on the major arc side (opposite side to point \(\begin {align} {Q}\end {align}\)) as \(\begin {align} {S.}\end {align}\)

Since all points \(\begin {align} {P, Q, R, S}\end {align}\) lie on the circle, \(\begin {align} {PQRS}\end {align}\) becomes a cyclic quadrilateral.

We know that, the sum of either pair of opposite angles of a cyclic quadrilateral is \(\begin {align} 180^\circ.\end {align}\)

Therefore,

\[\begin{align} \angle {PQR}+\angle {PSR} &=180^{\circ} \\ 100^{\circ}+\angle {PSR} &=180^{\circ} \\ \angle {PSR} &=180^{\circ}-100^{\circ} \\ &=80^{\circ} \end{align}\]

We know that, the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

Therefore,

\[\begin{aligned} \angle {POR} &=2 \angle {PSR} \\ &=2 \times 80^{\circ} \\ &=160^{\circ} \end{aligned}\]

Consider the \(\begin{aligned} \Delta {OPR.}\end{aligned}\) It is an isosceles triangle as

\[\begin{align} OP &= OR \\ &= \text{Radius of the circle.}\end{align}\]

\[\begin {align}∴ \angle {OPR}=\angle {ORP} \end {align}\]

Sum of all angles in a triangle is \(\begin {align}180^{\circ}. \end {align}\)

Therefore,

\[\begin{align}\angle {OPR}+ \!\angle {POR}+ \! \angle {ORP}&=180^{\circ} \\ \angle {OPR}+160^{\circ}+\angle {OPR}&=180^{\circ} \end{align}\]

\[\begin{align} 2 \angle {OPR}&= 180^{\circ} -160^{\circ} \\ \angle {OPR}&=10^{0}\end{align}\]

Chapter 10 Ex.10.5 Question 4

In the given figure, \(\begin {align} \angle {ABC}=69^{\circ} \end {align}\) and \(\begin {align} \angle {ACB}=31^{\circ} \end {align}\)find \(\begin {align} \angle {BDC} . \end {align}\)

Solution

Video Solution

What is known?

Two angles in a triangle.

What is unknown?

Value of \(\begin {align} \angle {BDC}\end {align}\)

Reasoning:

  • Sum of angles in a triangle is \(\begin {align} 180^{\circ}.\end {align}\)
  • Angles in the same segment are equal.

Steps:

Consider the \(\begin {align} \Delta {ABC,}\end {align}\) the sum of all angles will be \(\begin {align} 180^\circ.\end {align}\)

\[\begin{align} \angle {ABC}+ \!\!\angle {BAC}+ \!\angle {ACB} &=180^{\circ} \\ 69^{\circ}+\angle {BAC}+31^{\circ} &=180^{\circ} \end{align}\]

\[\begin{align}  \angle {BAC} &=180^{\circ}-\left(69^{\circ}+31^{\circ}\right) \\ &=180^{\circ}-100^{\circ} \\ &=80^{\circ} \end{align}\]

We know that, angles in the same segment of a circle are equal.

\[\begin {align} ∴ \angle {BDC}=\angle {BAC}=80^{\circ} \end {align}\]

Chapter 10 Ex.10.5 Question 5

In the given figure, \(A, B, C\) and \(D\)  are four points on a circle. \(AC\) and \(BD\)  intersect at a point \(\begin {align}{E} \end {align}\) such that \(\begin {align} \angle {BEC}=130^{\circ} \end {align}\) and \(\begin {align} \angle {ECD}=20^{\circ}. \end {align}\) Find \(\begin {align} \angle {BAC}\end {align}\).

Solution

Video Solution

What is known?

Two angles  \(\begin {align} \angle {BEC}=130^{\circ} \end {align}\)

and \(\begin {align} \angle {ECD}=20^{\circ} \end {align}\)

What is unknown?

Value of  \( \angle {BAC} \)

Reasoning:

  • Sum of angles in a triangle is \(180^{\circ}.\)
  • Angles in the same segment are equal.

Steps:

Consider the straight line \(\begin {align} {BD} \end {align}\). As the line \(\begin {align} {AC} \end {align}\) intersects with the line \(\begin {align} {BD,} \end {align}\) then the sum of two adjacent angles so formed is \(180^{\circ}.\)

Therefore,

\[\begin{align} \angle {BEC}+\angle {DEC} &=180^{\circ} \\ 130^{\circ}+\angle {DEC} &=180^{\circ} \\ \angle {DEC} &=180^{\circ}-130^{\circ} \\ &=50^{\circ} \end{align}\]

Consider the \(\begin{align} \Delta {DEC,} \end{align}\) the sum of all angles will be \(180^{\circ}.\)

\[\begin{align} \angle {DEC}+\angle {EDC}+\angle {ECD} &=180^{\circ} \\ 50^{\circ}+\angle {EDC}+20^{\circ} &=180^{\circ} \end{align}\]

\[\begin{align} \angle {EDC} &=180^{\circ}-70^{\circ} \\ &=110^{\circ} \\ ∴ \angle {BDC}=\angle {EDC} &=110^{\circ} \end{align}\]

We know that, angles in the same segment of a circle are equal.

\[\begin {align} ∴ \angle {BAC}=\angle {BDC}=110^{\circ}\end {align}\]

Chapter 10 Ex.10.5 Question 6

\(\begin {align} {ABCD}\end {align}\) is a cyclic quadrilateral whose diagonals intersect at a point \(\begin {align} {E. }\end {align}\)If \(\begin {align} \angle {DBC}=70^{\circ}, \quad \angle {BAC}=30^{\circ} \end {align}\) find \(\begin {align} \angle {BCD}. \end {align}\) Further if \(\begin {align} {AB=BC,} \end {align}\) find \(\begin {align} \angle {ECD.} \end {align}\)

Solution

Video Solution

What is known?

\(\begin {align} {ABCD}\end {align}\) is cyclic quadrilateral. \(\begin {align} {DBC}=70^\circ \end {align}\) and \(\begin {align} {BAC}=30^\circ. \end {align}\) \(\begin {align}{AB = BC.}\end {align}\)

What is unknown?

Value of \(\begin {align} \angle {BCD} \text { and } \angle {ECD} \end {align}\)

Reasoning:

  • A quadrilateral \(\begin {align} {ABCD}\end {align}\) is called cyclic if all the four vertices of it lie on a circle.
  • The sum of either pair of opposite angles of a cyclic quadrilateral is \(180^\circ.\)
  • Sum of angles in a triangle is \(180^\circ.\)
  • Angles in the same segment are equal.

Steps:

Based on the data given, draw the figure.

In the triangles \({ABD} \) and \({BCD,} \) \(\begin {align} \angle {CAD}=\angle {CBD}=70^{\circ}. \end {align}\) (Angles in the same segment)

So \(\begin {align} \angle {BAD}=30^{\circ}+70^{\circ}=100^{\circ} \end {align}\)

Since \(\begin {align} {ABCD }\end {align}\)is a cyclic quadrilateral, the sum of either pair of opposite angles of cyclic quadrilateral is \(180^{\circ}. \)

\[\begin{align} \angle {BAD}+ \angle {BCD}&=180^{\circ} \\ \angle {BCD} &=180^{\circ}-100^{\circ} \\&=80^{\circ} \end{align}\]

Also given \(\begin{align} {AB = BC.}\end{align}\)

So, \(\begin {align} \angle {BCA}=\angle {BAC}=30^{\circ} \end {align}\) (Base angles of isosceles triangle are equal)

\[\begin{align} ∴ \quad \angle {ECD} &=\angle {BCD}-\angle {BCA} \\ &=80^{\circ}-30^{\circ} \\ &=50^{\circ} \end{align}\]

Chapter 10 Ex.10.5 Question 7

If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.

Solution

Video Solution

What is known?

Diagonals of a cyclic quadrilateral are diameters of a circle passing through the vertices.

What is unknown?

To prove lines joining the vertices is a rectangle.

Reasoning:

  • The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
  • The sum of either pair of opposite angles of a cyclic quadrilateral is \(180^{\circ}.\)
  • Diameter is a chord.

Steps:

Let \({DB}\) be the diameter of the circle which is also a chord.

Then  \(\begin {align} \angle {BOD}=180^{\circ} \end {align}\)

We know that, the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

\[\begin{align} \angle {BAD} &=\frac{1}{2} \times \angle {BOD} \\ &=90^{\circ} \end{align}\]

Similarly,  \(\begin {align} \angle {BCD}=90^{\circ} \end {align}\)

Now considering \(\text{AC}\) as the diameter of the circle, we get \(\begin {align} \angle {AOC}=180^{\circ} \end {align}\)

We know that, the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

\[\begin{align} \angle {ABC} &=\frac{1}{2} \times \angle {AOC} \\ &=90^{\circ} \end{align}\]

Similarly, \(\begin {align} \angle {ADC}=90^{\circ} \end {align}\)

As you can see, all the angles at the corners are \(90^\circ \)we can say that the shape joining the vertices is a rectangle.

This problem can also be solved by using the property of cyclic quadrilaterals.

Chapter 10 Ex.10.5 Question 8

If the non-parallel sides of a trapezium are equal, prove that it is cyclic.

Solution

Video Solution

What is known?

Non-parallel sides of trapezium are equal.

What is unknown?

To prove that trapezium is cyclic.

Reasoning:

If the sum of a pair of opposite angles of a quadrilateral is \(180^{\circ},\) the quadrilateral iscyclic. Using Right angled-Hypotenuse-Side (RHS) criteria and Corresponding parts of congruent triangles (CPCT) we prove the statement.

Steps:

Draw a trapezium \( {ABCD} \) 

with \(  {AB} \| {CD} \)

\(AD\) and \(BC\) are the non-parallel sides which are equal. \({AD = BC.}\)

Draw \(\begin {align} {AM} \perp {CD} \end {align}\)

and \(\begin {align} {BN} \perp {CD} \end {align}\)

Consider \(\Delta {AMD}\) and \(\Delta \text{BNC.}\)

\(\begin{align} {AD} &={BC} \qquad \; (\text { Given }) \\\angle{AMD}&=\angle {BNC} \quad \left(90^{\circ}\right) \end{align}\) 

\(AM =BN \) (Perpendicular distance between two parallel lines is same)

By RHS congruence,

\[\begin {align}\Delta {AMD} \cong \Delta {BNC} \end {align}\]

Using CPCT , 

\[\begin {align}  \angle {ADC}=\angle {BCD} \ldots . .(1) \end {align}\]

\(\angle {BAD} \) and \(\angle {ADC} \) are on the same side of transversal \(\angle \text{AD}. \)

\[\begin{align}\angle {BAD}+\angle {ADC}&=180^{\circ} \\ \angle {BAD}+\angle {BCD}&=180^{\circ} \text{[Using}(1)]\end{align}\]

This equation proves that the opposite angles are supplementary.

Hence, \({ABCD} \) is a cyclic quadrilateral.

Chapter 10 Ex.10.5 Question 9

Two circles intersect at two points \({B}\) and \({C.}\) Through \({B,}\) two line segments \({ABD}\) and \({PBQ}\) are drawn to intersect the circles at \(A, D, P\) and \(Q\) respectively. Prove that \(\begin {align} \angle {ACP}=\angle {QCD.} \end {align}\)

 

Solution

Video Solution

What is given ?

Two circles intersect at two points

What is unknown?

Proof of \( \angle {ACP}=\angle {QCD} \)

Reasoning:

\(\angle {ACP} \)  and  \(\angle {ABP} \) lie on the same segment.

Similarly, \( \angle {DCQ} \)  and  \( \angle {DBQ} \) lie on the same segment.

Angles in the same segment of a circle are equal.

Steps:

We know that, angles in the same segment of a circle are equal.

So we get  \( \angle {ACP}=\angle {ABP} \)  and  \( \angle {QCD}=\angle {QBD} \)

Also, \(\begin {align} \angle {QBD}=\angle {ABP} \end {align}\)(Vertically opposite angles)

Therefore \( \angle {ACP}=\angle {QCD} \)

Chapter 10 Ex.10.5 Question 10

If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.

Solution

Video Solution

What is given ?

Two circles are drawn taking two sides of a triangle as diameters.

What is unknown?

To prove  that point of intersection of the \(2\) circles lie on the third side.

Reasoning:

Angle in a semicircle is a right angle. By using this fact we can show that \(BDC\) is a line which will lead to the proof that point of intersection lie on the third side.

Steps:

Since angle in a semicircle is a right angle, we get:

\( \begin{align} \angle  {ADB}&=90^{\circ} \text { and } \angle {ADC}=90^{\circ} \end{align}\)

\(\begin{align}  \angle {ADB}+\angle {ADC}&=90^{\circ}+90^{\circ} \\ \Rightarrow \;\; \angle {ADB}+\angle {ADC}&=180^{\circ} \end{align}\)

\(\Rightarrow \quad{BDC} \text { is a straight line. }\)

∴ \({D}\) lies on \({BC}\)

Hence, point of intersection of circles lie on the third side \(BC.\)

Chapter 10 Ex.10.5 Question 11

\({ABC}\) and \({ADC}\) are two right triangles with common hypotenuse \({AC.}\) Prove that \(\angle{CAD}= \angle {CBD}.\)

Solution

Video Solution

What is given ?

\({ABC}\) and \({ADC}\) are two right triangles with common hypotenuse \({AC.}\)

What is unknown?

Proof of \(\angle{CAD}= \angle {CBD}\)

Reasoning:

Sum of all angles in a triangle is \(180^\circ\)

If the sum of pair of opposite angles in a quadrilateral is \(180\) then it is cyclic quarilateral

Angles in the same segment of a circle are equal.

Steps:

Consider \(\Delta {ABC,}\)

\(\begin{align}&\angle {ABC}+  \angle {BCA}+\angle {CAB} =180^{\circ} \\&\text { (Angle sum property of a triangle) }\end{align}\)

\(\begin{align} 90^{\circ}+\angle {BCA}+\angle {CAB}&=180^{\circ} \\ \angle {BCA}+\angle {CAB}&=90^{\circ} \ldots(1)\end{align}\)

Consider \(\Delta {ADC,}\)

\(\begin{align}&\angle {CDA}+  \angle {ACD}+\angle {DAC} =180^{\circ} \\&\text { (Angle sum property of a triangle) }\end{align}\)

\(\begin {align} 90^{\circ}+\angle {ACD}+\angle {DAC}&=180^{\circ}\\\angle {ACD}+\angle {DAC}&=90^{\circ} \ldots(2) \end {align}\)

Adding Equations (\(1\)) and (\(2\)), we obtain

\( \begin {align} \begin{Bmatrix} \angle {BCA}+\angle {CAB}\\+\angle {ACD}+\angle {DAC} \end{Bmatrix}&=180^{\circ} \\ \begin{Bmatrix} (\angle {BCA}+\angle {ACD})+ \\ (\angle {CAB}+\angle {DAC}) \end{Bmatrix} &=180^{\circ} \end {align}\)

\( \begin {align}  \angle {BCD}+\angle {DAB}&=180^{\circ} \ldots(3) \end {align}\)

However, it is given that

\(\begin {align}\angle {B}+\angle {D} &=90^{\circ}+90^{\circ} \\ &=180^{\circ} \ldots(4) \end {align}\)

From Equations (\(3\)) and (\(4\)), it can be observed that the sum of the measures of opposite angles of quadrilateral \(\text{ABCD}\) is \(180^{\circ}.\) Therefore, it is a cyclic quadrilateral.

Consider chord \({CD.}\)

\(\begin {align}\angle {CAD} = \angle {CBD} \end {align}\) (Angles in the same segment)

Chapter 10 Ex.10.5 Question 12

Prove that cyclic parallelogram is a rectangle.

Solution

Video Solution

What is given ?

Cyclic quadrilateral is a parallelogram.

What is unknown?

Prove  that cyclic parallelogram is a rectangle.

Reasoning:

The sum of either pair of opposite angles of a cyclic quadrilateral is \(180^\circ.\) By using this fact we can show each angle of cyclic parallelogram as \(90^\circ\) which will prove the statement it is a rectangle.

Steps:

Let \({ABCD}\) be the cyclic parallelogram.

We know that opposite angles of a parallelogram are equal.

\[\begin {align}\angle {A}=\angle {C} \text { and } \angle {B}=\angle {D} \ldots .(1) \end {align}\]

We know that the sum of either pair of opposite angles of a cyclic quadrilateral is \(180^{\circ}.\)

\[\begin {align} \angle {A}+\angle {C}=180^{\circ} \ldots .(2) \end {align}\]

Substituting (\(1\)) in (\(2\)),

\[\begin{align}\angle {A}+\angle {C}&=180^{\circ} \\ \angle {A}+\angle {A}&=180^{\circ} \\ 2 \angle {A}&=180^{\circ} \\ \angle {A}&=90^{\circ}\end{align}\]

We know that if one of the interior angles of a parallelogram is \(90^{\circ},\) all the other angles will also be equal to \(90^{\circ}.\)

Since all the angles in the parallelogram is \(90^{\circ},\) we can say that parallelogram \({ABCD}\) is a rectangle.

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