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Circles Test - 34

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Circles Test - 34
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  • Question 1
    1 / -0
    Four alternative answers for the following question is given. Choose the correct alternative.
    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
    Given, $$ABCD$$ is a cyclic quadrilateral.
    Also, $$2\angle A=3\angle C \, \, \, \, \,.......(1)$$.

    Now,
    $$\angle A+\angle C=180^{o}$$ ...(Opposite angles of cyclic quadrilateral are supplementary)
    $$\Rightarrow \dfrac{3}{2}\angle C+\angle C=180^{o} $$ [From $$(1)$$]
    $$\Rightarrow \dfrac{5}{2} \angle C=180^{o}$$
    $$\Rightarrow \angle C=\dfrac{2\times180^{o}}{5}=72^{o}$$.

    Thus, the measure of $$\angle C$$ is $$72^{o}$$
    Hence, option $$C$$ is correct.
  • Question 2
    1 / -0
    If the length of the chord increase its perpendicular distance from the centre
    Solution

    $$AB$$ is a chord and its distance from centre $$O$$ is $$OP_1$$
    $$CD$$ is a chord and its distance from centre $$O$$ is $$OP_2$$
    $$CD>AB\Rightarrow OP_2<OP_1$$
    As the length of the chord increases then the distance of the chord from the center decreases.

  • Question 3
    1 / -0
    In the figure, if $$A, B, C$$ and $$D$$ are vertices of cyclic quadrilateral, then $$\angle x$$ will be:

    Solution
    We know, $$ \angle CBA + 70^{0} = 180^{0}$$ (linear pair)
    Then, $$\angle CBA = 180^{0} – 70^{0} = 110^{0}$$.

    $$\because ABCD$$ is cyclic quadrilateral.
    We know, the sum of opposite angles of a cyclic quadrilateral is $$180^{o}$$.
    Thus, $$\angle CBA + \angle CDA = 180^{0}$$
    $$\implies$$ $$110^{0} + \angle CDA = 180^{0}$$
    $$\implies$$ $$\angle CDA = 180^{0} – 110^{0}$$ $$= 70^{0}$$.

    But $$\angle x + \angle CDA = 180^{0}$$ (linear pair)
    $$\implies$$$$\angle x + 70^{0} = 180^{0}$$
    $$\implies$$$$\angle x = 180^{0} – 70^{0}$$ $$= 110^{0}$$.

    Thus, option $$C$$ is correct.
  • Question 4
    1 / -0
    Side $$AB$$ of a quadrilateral is a diameter of its circumcircle and $$\angle ADC = 140^{0}$$, then $$\angle BAC$$ equals:

    Solution
    Here, $$ABCD$$ is a cyclic quadrilateral.
    $$\therefore \angle D + \angle B = 180^{0}$$ ....(Opposite angles of cyclic quadrilateral are supplementary)
    $$\implies$$ $$\angle B = 180° – \angle D$$
    $$\implies$$ $$\angle B = 180° – 140°$$
    $$\implies$$ $$\angle B = 40°$$.

    $$\angle ACB$$ is an  angle in a semicircle.
    $$\therefore \angle ACB = 90°$$.

    Now in $$\Delta ABC$$,
    $$\angle ABC + \angle ACB + \angle BAC = 180^{0}$$ ....(Angle sum property of a triangle)
    $$\implies$$ $$40° + 90° + \angle BAC = 180°$$
    $$\implies$$ $$\angle BAC= 180° – (40° + 90°)$$
    $$\implies$$ $$\angle BAC= 180° – 130°$$
    $$\implies$$ $$\angle BAC = 50°$$.

    Thus, option $$B$$ is correct.
  • Question 5
    1 / -0
    Chords equidistant to each other from center of circle are
    Solution

  • Question 6
    1 / -0
    In fig, $$ABCD$$ is a cyclic quadrilateral. if $$\angle D = 120^{0}$$, then $$\angle CBE$$ will be :

    Solution
    We know that opposite angles of cyclic quadrilateral are supplementary.

    Then, $$\angle ADC+\angle ABC=180^o$$
    $$\implies$$ $$120^o+\angle ABC=180^o$$
    $$\implies$$ $$\angle ABC=180^o-120^o=60^o$$.

    Also, $$\angle ABC+ \angle CBE = 180^o$$ ...(Linear pair)
    $$\implies$$ $$\angle 60^o+ \angle CBE = 180^o$$
    $$\implies$$ $$\angle CBE=180^o-60^o=120^o$$

    Thus, option $$A$$ is correct.
  • Question 7
    1 / -0
    Sum of opposite angles in a cyclic quadrilateral is:
    Solution
    Given: $$ABCD$$ is a cyclic quadrilateral of a circle with centre $$O$$.

    To prove : Sum of opposite angles in a cyclic quadrilateral is $$180^0$$.

    Construction : Two diagonals $$AC$$ and $$BD$$ are drawn.

    Proof:
    $$\angle ADB=\angle ACB$$ [angles in the same segment of a circle]

               $$=\angle BAC=\angle BDC$$ 

    Again, $$\angle ADC=\angle ADB+\angle BDC$$ 

                $$=\angle ACB=\angle BAC$$ 

    $$\therefore \angle ADC+\angle ABC=$$ $$\angle ACB+\angle BAC+\angle ABC$$ 

    $$\therefore \angle ADC+\angle ABC=2$$ right angle [Sum of angles in a triangle is $$180^0$$]

    Similarly, $$\angle BAD+\angle BCD=2$$ right angle, i.e. $$180^0$$.

    Hence, option $$C$$ is correct.

  • Question 8
    1 / -0
    If in the given figure, $$\Delta ABC$$ is an equilateral triangle, then $$\angle BCD$$ will be:

    Solution
    In $$\Delta ABC, \angle A = 60^{0}$$...[Equilateral triangle].
    In cyclic quadrilateral $$ABCD$$,
    $$\angle A + \angle D = 180^{0}$$ ....[Opposite angles of cyclic quadrilateral are supplementary]
    $$\implies$$ $$\angle D = 180^{0} – \angle A$$ 
                    $$= 180^{0} – 60^{0}$$
    $$\implies$$ $$\angle BDC = 120^{0}$$.
    Thus, option $$B$$ is correct.
  • Question 9
    1 / -0
    In the given figure, $$ABCD$$ is a cyclic quadrilateral whose diagonals intersect each other at $$O$$. If $$\angle ACB = 50^{0}$$ and $$\angle ABC = 110^{0}$$, then $$\angle BDC$$ will be :

    Solution
    $$\because$$ Angles formed on same base $$AB$$,
    $$\therefore \angle ADB = \angle ACB$$
    $$\implies$$ $$\angle ADB = 50^{0}$$...[because $$\angle ACB = 50^{0}$$].

    We know that sum of opposite angles of cyclic quadrilateral is $$180^{0}$$
    $$\therefore \angle ADC + \angle ABC = 180^{0}$$
    $$\implies$$ $$\angle ADC + 110^{0} = 180^{0}$$....[because $$\angle ABC = 110^{0}]$$
    $$\implies$$ $$\angle ADC = 180^{0} – 110^{0} = 70^{0}$$.

    Now $$\angle ADC = \angle ADB + \angle BDC$$
    $$\implies$$ $$70^{0} = 50^{0} + \angle BDC$$
    $$\implies$$ $$70^{0} – 50^{0} = \angle BDC$$
    $$\implies$$ $$20^{0} = \angle BDC$$.

    Thus, $$\angle BDC = 20^{0}$$.
    Therefore, option $$D$$ is correct.
  • Question 10
    1 / -0
    In figure, $$D$$ is cyclic quadrilateral in which $$AC$$ and $$BD$$ are diagonals. If $$\angle DBC = 55^{0}$$ and $$\angle BAC = 45^{0}$$, then $$\angle BCD$$ will be :

    Solution
    Here, $$\angle CAD = \angle DBC = 55^{0}$$ (angles formed by same arc in the same segment are equal)
    We know, $$ \angle DAB = \angle CAD + \angle BAC$$
    $$=55^{0}+ 45^{0} = 100^{0}$$.
    But $$\angle DAB + \angle BCD = 180^{0}$$ (opposite angles of cyclic quadrilateral)
    $$\implies$$ $$\angle BCD = 180^{0} – 100^{0} = 80^{0}$$.
    Thus, option $$A$$ is correct.
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