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Circles Test - 31

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Circles Test - 31
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  • Question 1
    1 / -0
    Quadrilateral $$ABCD$$ and $$ECBA$$ are inscribed in a circle. If $$\angle B={ 125 }^{ \circ },$$ find the measure of $$\angle E.$$

    Solution

    In a cyclic quadrilateral, we know, opposite angles are supplementary.


    Since $$AECB$$ is a cyclic quadrilateral,

    $$\angle E + \angle B = 180^\circ$$

                $$\angle E = 180^\circ - \angle B$$

                       $$ = 180^\circ – 125^\circ$$

                       $$= 55^\circ$$


    Hence, option $$A$$ is correct.

  • Question 2
    1 / -0
    Given that $$BC=CD$$, what is the value of $$x$$?

    Solution

    In $$\triangle DBC$$

    Given that $$CD = CB$$,

    $$\therefore \angle CDB = \angle CBD$$.

    Also, $$\angle DCB + \angle DBC + \angle CDB = 180^o$$...[Angle sum property]

     $$2\angle CDB = 180^o – 50^o $$

     $$\angle CDB = 65^o$$.

    In the first circle, $$AEBD$$ is a cyclic quadrilateral.

    Then, opposite angles are supplementary.

     $$\angle EAB + \angle EDB = 180^o$$

     $$\angle EAB + 180^o - \angle CDB = 180^o$$

     $$\angle EAB = \angle CDB = 65^o$$.

    Hence, option $$D$$ is correct.

  • Question 3
    1 / -0
    Angle $$ANB =$$ Angle $$AOB =$$ Angle $$AMB = ?$$

    Solution
    We know that all angles inscribed in a semicircle are right angles.

    So, $$\angle ANB =$$ $$\angle AOB =$$ $$\angle AMB =$$ $$90^\circ$$
  • Question 4
    1 / -0
    In the given figure, $$ST$$ is a tangent to the circle with centre $$O$$, at $$S$$.
    Find the value of $$x$$.

  • Question 5
    1 / -0
    In the given figure, if $$\angle AOB = 125^{\circ}$$ find the measure of $$\angle COD$$

    Solution
    Given $$\angle AOB=125^o \angle AOD=90^o$$ and $$\angle BOC=90^o, \angle O=360^o$$
    $$360=\angle AOB+\angle DOA+\angle COD+\angle COB$$
    $$\Rightarrow 360=125+90+\angle COD+90$$
    $$\therefore \angle COD = 55^o$$
      
  • Question 6
    1 / -0
    If one angle of cyclic quadrilateral is $$70^o$$, then the angle opposite to it is:
    Solution

    In the cyclic quadrilateral,

    angles $$A + C = 180^{\circ}$$, and angles $$B + D = 180^{\circ}$$......(opposite angles of acyclic quadrilaterals are supplimentry).

    So, if one of the angle is  $$70^{\circ}$$.

    Then, angle opposite to it is $$= 180^{\circ} - 70^{\circ}=110^o$$.

    Hence, option $$B$$ is correct.

  • Question 7
    1 / -0
    In a cyclic quadrilateral $$ABCD$$, $$\angle A=5x, \angle C=4x$$, the value of $$x$$ is:
    Solution

    Opposite angles of a cyclic quadrilateral are supplementary

    Then, $$\angle A + \angle C = 5x+4x = 180^{\circ}$$

    $$\implies$$ $$9x = 180 ^{\circ}$$

    $$\implies$$ $$x= 20^{\circ}$$.

    Hence, option $$B$$ is correct.

  • Question 8
    1 / -0
    Draw a circle with centre O and radius $$2.5$$cm. Draw a chord $$AB$$ passing through its centre. The $$\angle AOB$$ is:
    Solution

    Chord AB passing through the centre of a circle .
    Therefore it is the diameter of a circle and diameter makes $$180^{\circ}$$ at the centre of a circle
    So , $$\angle{AOB}=180^{\circ}$$

    option A is the answer.

  • Question 9
    1 / -0
    In the given figure, $$O$$ is the centre of the circle, Reflex $$\angle AOB=260^\circ$$. $$C$$ is a point on the minor arc $$AB$$ of the circle. Find $$\angle ACB$$.

    Solution
    We know, the angle at the center of a circle is twice the angle at the circumference subtended by the same arc. Hence,

    $$\angle ACB=\dfrac{1}{2}\times reflex(\angle AOB)$$

                  $$=\dfrac{260}{2}$$

                  $$=130^\circ$$
  • Question 10
    1 / -0
    The sum of pairs of opposite angles of a cyclic quadrilateral is:
    Solution
    Sum of opposite angles of a cyclic quadrilateral is $$ 180 ^{o} $$ 
    Let $$ABCD$$ is cyclic quadrilateral.
    To prove that  $$\angle A+\angle C=180°$$  and $$\angle B+\angle D=180°$$
    Construction: Join $$OB$$ and $$OD$$.
    We know that  $$\angle BOD=2\angle BAD$$ ...(The angle at the centre is twice the angle at the circumference )  
    $$\implies$$ $$\angle BAD=\dfrac { 1 }{ 2 } \angle BOD$$.

    Similarly, we can write $$\angle BCD=\frac { 1 }{ 2 } \angle BOD$$.

    Now, add above two equations, 
    $$\angle BAD+\angle BCD=\dfrac { 1 }{ 2 } \angle BOD+\dfrac { 1 }{ 2 } \angle DOB$$
    $$\implies$$ $$\angle BAD+\angle BCD=\dfrac { 1 }{ 2 } (\angle BOD+\angle DOB)$$   ...(complete angle)
    $$\implies$$$$\angle BAD+\angle BCD=\dfrac { 1 }{ 2 } \times 360°=180°$$.

    So we get, $$\angle A+\angle C=180°$$  
    And similarly $$\angle B+\angle D=180°$$.

    Hence,option $$C$$ is correct.

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