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Quadrilaterals Test - 30

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Quadrilaterals Test - 30
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  • Question 1
    1 / -0
    A quadrilateral has three acute angles, each measuring $$75^o$$. Find the measure of the fourth angle.
    Solution
    Let the four angles be $$ \angle A , \angle B , \angle C$$ and $$\angle D$$ .
    Given $$ \angle A , \angle B , \angle C = 75^{o} $$ .
    We know, by angle sum property, the sum of angles of a quadrilateral is $$360^o$$.
    $$ \angle A + \angle B + \angle C + \angle D = 360^{o}  $$.
    Then, $$  \angle D $$ will be given by:
    $$\angle D = 360^o - \angle A - \angle B-\angle C $$ 
    $$ \Rightarrow \angle D =  360 ^o-75^o -75^o -75^o $$ 
    $$ \Rightarrow \angle D =  360 - 225^o$$
    $$ \Rightarrow \angle D =  135^o$$.
    The measure of fourth angle is $$ 135 ^{o}$$.

    Therefore, option $$B$$ is correct.

  • Question 2
    1 / -0
    $$X, Y$$ are mid-points of opposite sides $$AB$$ and $$DC$$ of a parallelogram $$ABCD. AY$$ and $$DX$$ are joined intersecting in P; $$CX$$ and $$BY$$ are joined intersecting in Q. What is $$PXQY$$?

    Solution
    $$ABCD$$ is the given parallelogram

    $$AB=CD$$ ..... (Opposite sides of parallelogram are equal)
    So, $$\dfrac{1}{2}AB=\dfrac{1}{2}CD$$

    $$AX=CY$$ ...... (as $$X$$ is the mid point of $$AB$$ and $$Y$$ is the mid point of $$CD$$)
    $$AX||CY$$ ......... (We know AB||CD, as $$ABCD$$ is the parallelogram)

    Quadrilateral $$AXCY$$ is a prallelogram, as a pair of opposite sides is equal and parallel.

    So, $$XC=YA$$ and $$XC\parallel YA$$
    $$XQ\parallel YP$$ ...... $$(i)$$ (As $$XQ$$ is part of line $$XC$$ and $$YP$$ is part of line $$YA$$)
    We have $$ABCD$$ as the given parallelogram
    Now, $$AB=CD$$ ....... (Opposite sides of parallelogram are equal)

    So, $$\dfrac{1}{2}AB=\dfrac{1}{2}CD$$

    $$DY=XB$$ ....... (As $$X$$ is the mid point of $$AB$$ and $$Y$$ is the mid point of $$CD$$)
    $$DY\parallel XB$$ ..... (We know $$AB \parallel CD$$, as $$ABCD$$ is the parallelogram)

    Quadrilateral $$DXBY$$ is a prallelogram, as a pair of opposite sides is equal and parallel.

    So, $$DX=YB$$ and $$DX||YB$$
    $$PX||YQ$$ ...... $$(ii)$$ (As $$XP$$ is part of line $$XD$$ and $$YQ$$ is part of line $$YB$$)

    From $$(i)$$ and $$(ii)$$ we get,
    $$PXQY$$ is also a parallelogram.
  • Question 3
    1 / -0
    In quadrilateral $$ABCD$$, if $$\angle A = 60^{\circ}$$ and $$\angle B : \angle C : \angle D = 2 : 3 : 7$$, then find $$\angle D$$.
    Solution
    In quadrilateral $$ABCD$$, $$\angle A$$ = $$60^\circ$$ and $$\angle B: \angle C: \angle D = 2 : 3 : 7  $$        [Given].

    Let $$\angle B, \angle C, \angle D$$ are $$2x^\circ$$ , $$3x^\circ$$ and $$7x^\circ$$.

    $$\Rightarrow$$  $$\angle A+ \angle B+ \angle C + \angle D $$=$$ 360^\circ$$    [Sum of all angles of quadrilateral is $$360$$$$^\circ$$]

    $$\Rightarrow$$  $$60^\circ + 2x^\circ+ 3x^\circ +7x^\circ$$ = $$360^\circ$$

    $$\Rightarrow$$  $$12x^\circ$$ = $$360^\circ - 60^\circ$$

    $$\Rightarrow$$  $$x^\circ$$ = $$\dfrac {300^\circ}{12}$$

    $$\Rightarrow$$  $$x^\circ$$ = $$25^\circ$$.

    Now, $$\angle D$$=$$7x^\circ$$.

    Substitute value of $$x$$ we get,
    $$\Rightarrow$$  $$\angle D$$=$$ 7\times 25^o$$

    $$\therefore$$  $$\angle D = 175^\circ$$.

    Hence, option $$A$$ is correct.
  • Question 4
    1 / -0
    The diagonals of a parallelogram $$ABCD$$ intersect at $$O$$. A line through O intersect AB at X and DC at Y. Identify the correct option

    Solution
    $$\Rightarrow$$  In given figure ABCD is a parallelogram.
    $$\therefore$$    AB$$\parallel$$DC
    $$\Rightarrow$$  Also AC is a transversal of AB$$\parallel$$DC.
    $$\therefore$$  $$\angle1=\angle2$$             [Alternate interior angles]
    $$\Rightarrow$$  Now in $$\triangle$$AXO and $$\triangle$$CYO, we have
    $$\Rightarrow$$  $$\angle1=\angle2$$         [Alternate interior angles]
    $$\Rightarrow$$   $$\angle3=\angle4$$       [Vertically opposite angles]
    $$\Rightarrow$$  $$CO=OA$$            [Diagonals bisect each other]
    $$\therefore$$    $$\triangle AXO \cong \triangle CYO$$      [ASA Criteria]
    $$\therefore$$   $$OX=OY$$      [CPCT]
  • Question 5
    1 / -0
    In a quadrilateral $$ABCD$$, the angles $$\angle A, \angle B, \angle C$$ and $$\angle D$$ are in the ratio $$2 : 3 : 4 : 6$$. Find the measure of each angle of the quadrilateral.
    Solution
    Let the angles of quadrilateral $$ABCD$$ are $$2x,3x,4x$$ and $$6x$$.
    We know, by angle sum property, the sum of angles of quadrilateral is $$360^o$$.
    $$\therefore$$  $$\angle A+\angle B+\angle C+\angle D=360^o$$
    $$\Rightarrow$$  $$2x+3x+4x+6x=360^o$$
    $$\Rightarrow$$  $$15x=360^o$$
    $$\therefore$$  $$x=24^o$$.

    $$\Rightarrow$$  $$\angle A=2x=2\times 24^o=48^o$$
    $$\Rightarrow$$  $$\angle B=3x=3\times 24^o=72^o$$
    $$\Rightarrow$$  $$\angle C=4x=4\times 24^o=96^o$$
    $$\Rightarrow$$  $$\angle D=6x=6\times 24^o=144^o$$.

    Therefore, option $$A$$ is correct.
  • Question 6
    1 / -0
    Which of the following group is true for quadrilateral $$ ABCD$$?
    (1) $$ABCD$$ is a Rhombus(a) $$\overline { AC } $$ and $$\overline { BD } $$ bisect each other.
    (2) $$ABCD$$ is a Parallelogram(b) $$\overline { AC } $$ and $$\overline { BD } $$ bisect each other at an right angle.
    (3) $$ABCD$$ is Rectangle(c) $$\overline { AC } $$ and $$\overline { BD } $$ are congruent and bisect each other at an right angle.
    (4) $$ABCD$$ is Square(d) $$\overline { AC } $$ and $$\overline { BD } $$ are congruent and bisect each other.
    Solution
    Diagonals of rhombus $$ABCD$$ intersect at right angles. $$1 \rightarrow b$$

    Diagonals of parallelogram $$ABCD$$ bisect each other. $$2\rightarrow a$$

    Diagonals of rectangle $$ABCD$$ are equal/congruent and bisect each other, $$3\rightarrow d$$

    Diagonals of square $$ABCD$$ bisect at right angles and are equal/congruent. $$4\rightarrow c$$
  • Question 7
    1 / -0
    The angles of a quadrilateral are in the ratio $$3:5:9:13$$. Then angles of the quadrilateral respectively are ___________.
    Solution

    Given ratio of angles of a quadrilateral $$ABCD$$ is $$3:5:9:13$$

    Let the angles of the quadrilateral $$ABCD$$ be $$3x,5x,9x$$ and $$13x$$, respectively.

     

    We know, by angle sum property, the sum of angles of a quadrilateral is $$360^o$$.

    $$\Rightarrow 3x+5x+9x+13x=360^o$$

    $$\Rightarrow 30x=360^o$$

    $$\therefore x=12^o$$.

     

    $$\therefore \angle A =3x=3 \times 12^o =36^o$$,

    $$ \angle B =5x=5\times 12^o =60^o$$,

    $$ \angle C =9x=9 \times 12^o =108^o$$

    and $$ \angle D =13x=13 \times 12^o =156^o$$.

     

    Hence, option $$A$$ is correct.

  • Question 8
    1 / -0
    In a quadrilateral, the angles are $$x^{\circ}, (x + 10)^{\circ}, (x + 20)^{\circ}, (x + 30)^{\circ}$$. Find the value of $$x$$.
    Solution
    The given quadrilateral angles are $$x^o,(x+10)^o,(x+20)^o$$ and $$(x+30)^o$$.
    We know, by angle sum property, the sum of angles of quadrilateral is $$360^o$$.
    $$\therefore$$  $$x^o+(x+10)^o+(x+20)^o+(x+30)^o=360^o$$
    $$\Rightarrow$$  $$4x^o+60^o=360^o$$
    $$\Rightarrow$$  $$4x^o=300^o$$
    $$\Rightarrow$$  $$x^o=75^o$$.
    $$\therefore$$  The value of $$x$$ is $$75^o$$.
    Hence, option $$D$$ is correct.
  • Question 9
    1 / -0
    The sum of pairs of opposite angles of a cyclic quadrilateral is:
    Solution
    Sum of pair of opposite angles of a cyclinc quadrilateral is $${ 180 }^{ o }$$
  • Question 10
    1 / -0
    Match the following.
    (1)Rectangle(p) A quadrilateral having its opposite sides equal and parallel.
    (2)Square(q)A parallelogram having its opposite sides equal and each of the angle is a right angle.
    (3)Parallelogram(r) A parallelogram having all sides equal and each of the angle is a right angle.
    (4)Rhombus(s)A quadrilateral in which a pair of opposite sides are parallel.
    (5)Trapezium(t)A parallelogram having all sides equal.
    Solution
    $$(1)$$ Rectangle - A parallelogram having its opposite sides equal and each of the angle is a right angle.

    $$(2)$$ Square - A parallelogram having its opposite sides equal and each of the angle is a right angle.

    $$(3)$$ Parallelogram -  A quadrilateral having its opposite sides equal and parallel.

    $$(4)$$ Rhombus - A parallelogram having all sides equal.

    $$(5)$$ Trapezium - A quadrilateral in which a pair of opposite sides are parallel.

    $$\therefore$$   $$1\rightarrow (q),2\rightarrow (r),2\rightarrow (p), 4\rightarrow (t),5\rightarrow (s)$$
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