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Quadrilaterals Test - 21

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Quadrilaterals Test - 21
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  • Question 1
    1 / -0
    Three angles of a quadrilateral are $$75^{\circ}, 90^{\circ}$$ and $$75^{\circ}$$. The measure of fourth angle is?
    Solution
    We know, by angle sum property, the sum of angles of a quadrilateral is $$360^o$$.

    The given angles are, $$75^o, 90^o, 75^o$$.

    Let the fourth angle be $$x^o$$.
    Then, $$75^o+90^o+75^o+x^o=360^o$$.
    $$\Rightarrow$$ $$360^o - (75^o + 90^o + 75^o ) =x^{\circ}$$
    $$\Rightarrow$$ $$x^o=360^o - 240^o  =120^{\circ}$$.

    Therefore, the fourth angle is $$x^o=120^o$$.
    Hence, option $$D$$ is correct.
  • Question 2
    1 / -0
    One angle of a quadrilateral is of $$108^o$$ and the remaining three angles are equal. Find each of the three equal angles.
    Solution

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

    Given, one angle is $$108^o$$ and remaining three angles are equal.

    Let the remaining angles be $$x^o,y^o,z^o$$.

    Since they are equal, $$x^o+y^o+z^o=$$ $$x^o+x^o+x^o=3x^o$$.

    Then, $$108^o+x^o+y^o+z^o=360^o$$

    $$\Rightarrow$$ $$108^o+3x^o=360^o$$

    $$\Rightarrow$$ $$360^o - 108^o =3x^{\circ}$$

    $$\Rightarrow$$ $$252^o =3x^{\circ}$$

    $$\Rightarrow$$ $$x^o=\dfrac{252^o}{3}  =84^{\circ}$$.

    Therefore, each of the three remaining angle is $$x^o=84^o$$.

    Hence, option $$C$$ is correct.

  • Question 3
    1 / -0
    $$D$$ and $$E$$ are the mid-points of the sides $$AB$$ and $$AC$$ of $$ABC$$ and $$O$$ is any point on side $$BC. O$$ is joined to $$A$$. If $$P$$ and $$Q$$ are the mid-points of $$OB$$ and $$OC$$ respectively, then $$DEQP$$ is:
    Solution
    Given:
    $$ABC$$ is a triangle, $$D$$ & $$E$$ are the respective mid points of $$AB$$ &$$ AC$$.
    $$O$$ is any point on $$BC$$, $$P$$ & $$Q$$ are mid points of $$OB$$ & $$OC$$ respectively.

    To find:
    Type of quadrilateral of $$DEQP$$.

    Solution:
    In $$\Delta ABC, D \&E$$ are the respective mid points of $$AB$$ & $$AC.$$ 
    $$\therefore DE\parallel (BC or PQ)$$   ....(1)
    Again in $$\Delta ABO, P \&D$$ are the respective mid points of $$AB$$ & $$BO$$
    $$\therefore DP\parallel AO$$    ... (2) 
    Similarly, in $$\Delta ACO, EQ\parallel AO$$    ...(3)
    $$\therefore$$ From (2) & (3), we get 
    $$DP\parallel EQ$$    ... (4)
    $$\therefore$$ From (1) & (4), we have
    $$DE\parallel PQ$$ and $$DP\parallel EQ$$
    So, $$DEQP$$ is parallelogram.

    Hence, option D.

  • Question 4
    1 / -0
    Can all the angles of a quadrilateral be acute angles? 
    Solution
    Let us take a qudrilateral $$ABCD$$ with $$BD$$ as diagonal.
    In $$\Delta ABD$$ we have,
     $$\angle  ADB+\angle ABD+\angle DAB={ 180 }^{ O }$$    .......$$(i)$$
    And in $$\Delta DBC$$ we have
    $$ \angle  DCB+\angle CBD+\angle BDC={ 180 }^{ O }$$    ......$$(ii)$$
    Adding $$(i)$$ & $$(ii)$$, we get,
    $$ \angle  ADB+\angle ABD+\angle DAB+\angle  DCB+\angle CBD+\angle BDC={ 360 }^{ O }$$ 
    $$\Longrightarrow \angle A+\angle B+\angle C+\angle D={ 360 }^{ O }$$ 
    So, the sum of the angles of a quadrilateral is $${ 360 }^{ O }$$   .....$$(iii)$$.

    Now, if the angles are acute then $$\angle A<{ 90 }^{ O }, \angle B<{ 90 }^{ O }, \angle C<{ 90 }^{ O }, \angle D<{ 90 }^{ O }$$ 
    $$\Longrightarrow \angle A+\angle B+\angle C+\angle D<{ 90 }^{ O }+{ 90 }^{ O }+{ 90 }^{ O }+{ 90 }^{ O }$$ 
    $$\Longrightarrow \angle A+\angle B+\angle C+\angle D<{ 360 }^{ O }.$$ 
    But this does not comply with $$(iii)$$ .

    Therefore, all the angles of a quadrilateral cannot be acute angles.
    Hence, option $$B$$ is correct.

  • Question 5
    1 / -0
    State true or false:
    Every rhombus is a kite.
    Solution
    A rhombus is a quadrilateral for which all four sides and the opposite angles are equal. 
    A kite is a quadrilateral for which the adjacent sides and one pair of opposite angles are equal. Both these conditions of kite are satisfied by a rhombus.
    Hence, every rhombus is a kite.
  • Question 6
    1 / -0
    Can the angles $$110^o, 80^o, 70^o$$ and $$95^o$$ be the angles of a quadrilateral?
    Solution

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

    The given angles are $$110^o, 80^o, 70^o, 95^o$$.


    Then,

    $$110^o+80^o+70^o+95^o$$.

    $$=190^o + 165^o $$

    $$=355^o \ne 360^o$$.

    Therefore, the given angles are not the angles of a quadrilateral.

    Hence, option $$B$$ is correct.

  • Question 7
    1 / -0
    All the angles of a quadrilateral are equal. What special name is given to this quadrilateral?

    Solution
    We know that the sum of the angles of the quadrilateral is $$360^{o}$$
    Now, the quadrilateral in given question has all angles equal.
    $$ \therefore$$  Each angle $$={ 360 }^{ o }\div 4={ 90 }^{ o }$$
    So, this quadrilateral would be rectangle or a square.
  • Question 8
    1 / -0
    In a quadrilateral three angles are in the ratio $$3 : 3 : 1$$ and one of the angles is $$80^{\circ}$$, then other angles are :
    Solution
    In figure $$ABCD$$ is a quadrilateral.

    We have given ratio of three angle $$3 : 3 : 1$$.

    Let $$\angle A= 3x, \angle B=3x, \angle C=x$$ and  $$\angle D=80^\circ$$.

    We know, by angle sum property, the sum of all interior angles of quadrilateral is $$360^\circ$$]

    $$\Rightarrow$$  $$\angle A+\angle B+\angle C+\angle D=360^\circ$$ 

    $$\Rightarrow$$  $$3x+3x+x+80^\circ = 360^\circ$$

    $$\Rightarrow$$  $$ 7x =280^\circ$$     

    $$\therefore$$    $$x=40^\circ = \angle C$$.        ---- ( 1 )

    $$\Rightarrow$$   $$\angle A=3x=3\times 40^\circ = 120^\circ$$   -- ( 2 )

    $$\Rightarrow$$   $$\angle B=3x=3\times 40^\circ =120^\circ$$    -- ( 3 )

    From ( 1 ), ( 2 ) and ( 3 ),
    $$\Rightarrow$$  Other angles are $$120^\circ, 120^\circ, 40^\circ$$.

    Hence, option $$A$$ is correct.

  • Question 9
    1 / -0
    Three angles of a quadrilateral are $$60^{\circ}, 110^{\circ}$$ and $$86^{\circ}$$, the fourth angle of the quadrilateral is :
    Solution

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

    The given angles are, $$60^o, 110^o, 86^o$$.

    Let the fourth angle be $$x^o$$.

    Then, $$60^o+110^o+86^o+x^o=360^o$$.

    $$\Rightarrow$$ $$360^o - (60^o + 110^o + 86^o ) =x^{\circ}$$

    $$\Rightarrow$$ $$x^o=360^o - 256^o  =104^{\circ}$$.

    Therefore, the fourth angle is $$x^o=104^o$$.

    Hence, option $$A$$ is correct.

  • Question 10
    1 / -0
    State true or false:
    All parallelograms are trapeziums.
    Solution
    One pair of opposite sides of a trapezium are parallel.
    Opposite sides of parallelogram are equal and parallel.
    So, all parallelograms are trapeziums.
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