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Circles Test - 32

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Circles Test - 32
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  • Question 1
    1 / -0
    In the given figure, $$\Delta$$ABC is an isosceles triangle with AB$$=$$AC and $$\angle$$ABC$$=50^o$$. Find $$\angle$$BDC and $$\angle$$BEC respectively.

    Solution
    $$\triangle ABC$$ is an isosceles triangle.
     $$\therefore \angle B=\angle C=50^o$$
    $$\angle BAC=180-50-50=80^o$$

    $$\angle BAC=\angle BDC=80^o$$ (angles on circle by the same chord on the same segment are equal)
    $$\angle BEC$$ is supplementary angle of $$\angle BAC$$
    $$\therefore \angle BEC=180-\angle BAC\\=180-80=100^o$$
  • Question 2
    1 / -0
    In the given figure, $$ABCD$$ is a cyclic quadrilateral whose side $$AB$$ is a diameter of the circle passing through $$A, B, C$$ and $$D$$. If $$\angle$$ADC$$=130^o$$, find $$\angle$$BAC.

    Solution
    $$\angle ACB=90^o$$ (angle subtended by the diameter of circle)
     and $$\angle ABC $$ is the supplementary angle of $$\angle ADC$$...(opposite angles of cyclic quadrilateral)
     $$\therefore \angle ABC=180-\angle ADC=180-130=50^o$$.

    Now, by angle sum property,
    $$\angle BAC=180-(\angle ACB+\angle ABC)\\=180-50-90=40^o$$.

    Hence, option $$D$$ is correct.
  • Question 3
    1 / -0
    A circle (with centre at O) is touching two intersecting lines AX and BY. The two points of contact A and B subtend an angle of $$65^{\circ}$$ at any point C on the circumference of the circle. If P is the point of intersection of the two lines, then the measure of $$\angle $$ APB is
    Solution

    $$\angle AOB = 2\angle ACB$$    (Angle subtended by an arc at the centre is double the angle subtended by it at the circumference)
    $$\implies \angle AOB=2 \times 65= 130^{\circ}$$

    In quadrilateral $$AOBP$$
    $$\angle OAP + \angle OBP + \angle AOB + \angle APB=360^{\circ}$$
    $$\implies 90^{\circ} + 90^{\circ} + \angle APB + 130^{\circ} = 360^{\circ}$$
    $$\implies \angle APB = 360^{\circ} - 310^{\circ}= 50^{\circ}$$ 

  • Question 4
    1 / -0
    In the figure given below, if $$\angle AOP \, = \, 75^\circ$$ and $$\angle AOB \, = \, 120^\circ$$, then what is $$\angle AQP$$ ?

    Solution
    $$2\angle AQP \, =  \angle AOP$$ (Angle subtended by an arc at the centre is double the angle subtended by the same arc at any other point on the circle ).
    $$\implies \angle AQP=\dfrac{1}{2}\angle AOP$$
    $$= \cfrac{75^\circ}{2} \, = \, 37.5^{\circ}$$
  • Question 5
    1 / -0
    In the adjacent figure, if $$\angle AOC = 110^0$$, then the value of $$\angle D\ and \angle B$$ respectively.

    Solution
    Given, $$\angle AOC = 110^0$$
    Also, we know $$\angle AOC = 2 \angle ADC$$  (angle subtended by an arc at the center is double the angle subtended by it at any point on the circumference)
    $$\therefore \angle ADC = 55^0$$
    Also, $$\angle B + \angle D = 180^0$$  (ABCD is a cyclic quadrilateral and opposite angles of it are supplementary)
    $$\implies \angle B = 125^0$$
  • Question 6
    1 / -0
    $$BD$$ is a chord parallel to the diameter $$AC$$ of a circle. A point $$B$$ is on the perimeter of the circle such that angle $$CBE={ 63 }^{ o }$$. The angle $$DCE$$ is equal to:
    Solution
    $$\angle ACE=\angle ABE=27°\\ (\because \angle ABE=27°\quad as\quad \angle ABC=90°)$$
    $$\angle ACE=\angle CED$$   (Alternate angles are equal) $$(\because AC||ED)$$
    BEDC is a cyclic quadrilateral
    $$\therefore \angle EBC+\angle EDC=180°\\ \therefore \angle EDC=117°\\ \therefore \angle ECD=180°-(117°+27°)=36°$$
  • Question 7
    1 / -0
    In a cyclic quadrilateral $$ ABCD$$, twice the measure of $$\angle A $$ is thrice the measure of $$\angle C$$, find the measure of $$\angle C$$.
    Solution
    The sum of the opposite angles of a cyclic quadrilateral is supplementary.

    So, in cyclic quadrilateral $$ABCD$$
    $$\angle A+\angle C=180^{\circ}$$ .............(1).

    It is given that
    $$2\angle A=3\angle C$$

    $$\implies$$ $$\angle A=\dfrac{3}{2}\angle C$$ ............ (2).

    From (1)  and  (2),

    $$\dfrac{3}{2}\angle C+\angle C=180^{\circ}$$

    $$\implies$$ $$\dfrac{5}{2}\angle C=180^{\circ}$$

    $$\implies$$ $$\angle C=\dfrac{180\times 2}{5}=72^{\circ}$$.

    Hence, option $$B$$ is correct.
  • Question 8
    1 / -0
    Quadilateral ABCD is cyclic. If $$ \angle B = 60^o$$, then $$\angle D = $$____.
    Solution
    In a cyclic quadrilateral, the sum of opposite angles $$={180}^{\circ}$$.
    Given, $$ABCD$$ is a cyclic quadrilateral.
    Then, $$\angle{B}$$ and $$\angle{D}$$ are opposite angles
    Hence $$\angle{B}+\angle{D}={180}^{\circ}$$
    $$\Rightarrow\,\angle{D}={180}^{\circ}-\angle{B}$$
    $$\Rightarrow\,\angle{D}={180}^{\circ}-{60}^{\circ}={120}^{\circ}$$
    Therefore, $$\angle{D}={120}^{\circ}$$.
    Thus, option $$C$$ is correct.
  • Question 9
    1 / -0
    In the given figure, AD=BC, $$\angle BAC={ 30 }^{ o }$$ and $$\angle CBD={ 70 }^{ o }$$. Find $$\angle BCD$$.

    Solution
    In figure, $$ABCD$$ is cyclic quadrilateral.
    Here, $$AC$$ and $$BD$$ are diagonals.
    Also, $$\angle BAC={30}^{\circ}$$ and $$\angle CBD={70}^{\circ }$$.

    Now,  $$\angle BAC=\angle BDC={30}^{\circ}$$.

    Then, $$\angle BAD=\angle BDC+\angle CAD$$
                             $$={30}^{\circ}+{70}^{\circ}={100}^{\circ}$$.

    Now, $$\angle BCD+\angle BAD={180}^{\circ}$$ ...(opposite angles are cyclic quadrilateral)
    $$\Rightarrow \angle BCD+{100}^{\circ }={180}^{\circ}$$
    $$\Rightarrow \angle BCD={180}^{\circ }-{100}^{\circ}$$
    $$\Rightarrow \angle BCD={80}^{\circ}$$.

    Therefore, option $$A$$ is correct.

  • Question 10
    1 / -0
    In the diagram above, $$O$$ is the centre of the circle. The value of $$x$$ is

    Solution
    Given $$O$$ is centre so $$OR=OS$$ (Radius)

    $$\angle ORS=\angle OSR$$  ($$OR=OS$$) So angle opposite to side will be equal

    Now $$\angle POR+\angle OSR=180^0$$ (cyclic quadrilateral)

    $$140+\angle OSR=180^0$$

    $$\angle OSR=40^0$$

    So $$\angle ORS=40^0$$

    In $$\Delta OSR$$

    $$x^0+\angle ORS+\angle OSR=180^0$$

    $$x^0+80^0=180^0$$

    $$\boxed{x^0=100^0}$$

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