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Circles Test - 29

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Circles Test - 29
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  • Question 1
    1 / -0
    In given figure, BC is a diameter of the circle and $$\displaystyle \angle BAO=60^{\circ}\,$$. Then$$\,\angle ADC$$ is equal to

    Solution
    In $$\triangle AOB$$, 
    $$OA=OB$$
    $$=>\angle OAB=\angle OBA=60^{0}$$
    We know that the exterior angle of a triangle is equal to sum of the two opposite angles of the triangle.
    $$\angle AOC=\angle OAB+\angle OBA$$
    $$\angle AOC=60^{0}+60^{0}$$
    $$\angle AOC = 120^{0}$$
    We know that the angle subtended at the centre of the circle by an arch is twice the angle subtended at the circumference by the same arc.
    $$\therefore \angle ADC=\dfrac{1}{2}\angle AOC$$
                      $$=\dfrac{120}{2}$$
                      $$=60^{0}$$

  • Question 2
    1 / -0
    In given figure, if $$\displaystyle \angle OAB$$ = $$\displaystyle 40^{\circ},$$ then $$\displaystyle \angle ACB$$ is equal to:

    Solution
    In the circle $$OA=OB =$$ Radius
    So, $$\angle OAB=\angle OBA=40^°$$
    Therefore, $$\angle AOB=180^°-40^°-40^°=100^°$$             {Angle sum property}
    We know in a circle the angle subtended by an arc at the center is twice that of the angle made on the circle.
    Therefore, $$\angle ACB=\dfrac{\angle AOB}{2}=50^°$$
  • Question 3
    1 / -0
    In the figure above AQ = CT and $$\displaystyle \angle QAC=70^{\circ}$$ Find $$\displaystyle \angle ACT$$

    Solution
    Given that:
    $$AQ=CT, \angle QAC=70^\circ$$
    To find:
    $$\angle ACT=?$$
    Solution:
    If two opposite sides of a cyclic quadrilateral are equal then the other two opposite sides are parallel to each other .
    So, if $$AQ=CT$$ then $$QT\parallel AC$$
    $$\angle QAC+\angle QTC=180^\circ$$     (Sum of opposite angles of cyclic quadrilateral is $$180^\circ.$$) 
    or, $$\angle QTC=180^\circ-70^\circ=110^\circ$$
    Now, 
    $$\angle ACT+\angle QTC=180^\circ$$      (Sum of interior angles on the same side of the transversal line is $$180^\circ.$$)
    or, $$\angle ACT=180^\circ-110^\circ$$
    or, $$\angle ACT=70^\circ$$
    Hence, A is the correct option.
  • Question 4
    1 / -0
    In a cyclic trapezium ABCD, $$\angle{A}=100^\circ$$. Then, $$\angle{D}$$ is equal to:

    Solution

    Consider given trapezium $$ABCD$$.

    Given, $$\angle A={{100}^{\circ}}$$.

    Now,

    $$ \angle A=\angle B $$ ....[Since, $$DC$$ is the diameter of the circle]

    $$\implies$$ $$ \angle B={{100}^{\circ}} $$.


    Since, trapezium $$ABCD$$ is a cyclic quadrilateral,

    Sum of opposite angle of trapezium $$={{180}^{\circ}}$$....[Opposite angles of cyclic quadrilateral are supplementary].

    $$\implies \angle B+\angle D={{180}^{\circ}} $$

    $$ {{100}^{\circ}}+\angle D={{180}^{\circ}} $$

    $$ \angle D={{80}^{\circ}} $$.


    Hence, option $$A$$ is correct.

  • Question 5
    1 / -0
    For a $$\triangle PQR$$ with points $$P, Q$$ and $$R$$ lying on a circle, if  $$RP$$ is the diameter of the circle and $$PQ = 5\ cm$$ and $$QR = 12\ cm$$. Find radius of the circle. 
    Solution

    Because angle in a semicircle is always $$90^{\circ},$$ so

    So $$\angle RQP = 90^{\circ}$$ and $$RP$$ is the diameter

    Thus by pythagorean theorem we have

    $$RP^2 = RQ^2 + QP^2$$
              $$  = 12^2 + 5^2 = 169$$

    $$RP = 13 = d = 2r$$

    $$r = 6.5\ cm$$

  • Question 6
    1 / -0
    For a triangle ABC, with BC as the diameter of circle, if radius is 5 cm and AB = 8 cm. Find AC .
    Solution
    Given, $$\triangle ABC$$ is in a semicircle and BC is the diameter, therefore $$\angle BAC=90^o$$
    So, it is a right angled triangle, therefore
    $$BA^2+AC^2=BC^2$$
    $$\implies AC^2=10^2-8^2=36$$
    $$\implies AC=6$$
  • Question 7
    1 / -0
    Find the values of the x and y, where O denotes the centre of the circle.

  • Question 8
    1 / -0
    Find the values of $$x, y$$ and $$z$$, where $$O$$ denotes the centre of the circle.

    Solution
    Since the given quadrilateral is cyclic quadrilateral, we have,
    $$3x + x = 180^\circ$$
    $$\Rightarrow 4x =180^\circ$$
    $$\Rightarrow x=45^\circ$$.

    By the inscribed angle theorem, we have,
    $$y = 2\times x$$
    $$\Rightarrow y=90^\circ$$.

    We know that,
    $$y + z= 360^\circ$$
    $$\Rightarrow z=360^\circ- y$$
    $$\Rightarrow z= 360^\circ-90^\circ$$
    $$\Rightarrow z=270^\circ$$

    Hence, option $$B$$ is correct.
  • Question 9
    1 / -0
    Find the values of $$x$$ and $$y$$.

    Solution
    Since the given quadrilateral is a cyclic quadrilateral, we have,
    $$112^o + x^o = 180^o$$ ...[Opposite angles of cyclic quadrilateral are supplementary]
    $$\Rightarrow x = 68^o$$.

    Similarly,
    $$59^o + y^o = 180^o$$ ...[Opposite angles of cyclic quadrilateral are supplementary]
    $$\Rightarrow y = 121^o$$.

    Hence, option $$D$$ is correct.
  • Question 10
    1 / -0
    Find the values of $$y$$ and $$z$$, where $$O$$ denotes the centre of the circle.

    Solution
    Since the given quadrilateral is a cyclic quadrilateral,
    $$\angle y + 35^o =180^o$$
    $$\Rightarrow \angle y=180^o-35^o$$
    $$\Rightarrow \angle y=145^o$$.

    By the inscribed angle theorem, we have,
    $$\angle z=2\times \angle y$$
    $$\Rightarrow \angle z=2\times145^o$$
    $$\Rightarrow \angle z = 290^o$$.

    Hence, option $$C$$ is correct.
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