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CBSE Questions for Class 8 Maths Understanding Quadrilaterals Quiz 4 - MCQExams.com
CBSE
Class 8 Maths
Understanding Quadrilaterals
Quiz 4
All the angles of a quadrilateral are equal. What special name is given to this quadrilateral?
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Rectangle
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Trapezium
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Rhombus
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Parallelogram
Explanation
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.
In the above figure, it is given that $$BDEF$$ and $$FDCE$$ are parallelograms. Can you say that $$BD = CD$$? Why or why not?
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Yes
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No
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May be
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Cannot be determined
Explanation
Given:-
$$ABC$$ is a triangle. $$D,\;E\;F$$ are points on $$BC,\;CA.\;AB$$ taken in order. $$BDEF$$ anf $$FDCE$$ are parallelograms.
To Prove:-
$$BD=CD$$
Proof:-
In parallelogram $$BDEF$$,
$$BD=EF\quad\quad\quad\dots(i)$$[Opposite sides of a parallelogram are equal.]
In parallelogram $$FDCE$$,
$$CD=EF\quad\quad\quad\dots(ii)$$[Opposite sides of a parallelogram are equal.]
From $$(i)$$ and $$(ii)$$,
$$BD=CD$$
Hence, proved.
The adjacent figure $$\Box HOPE$$ is a parallelogram. Find the angle measures $$x, y$$ and $$z$$. State the properties you use to find them.
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$$ x = 110^{o}, y = 40^{o}, z = 30^{o}$$
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$$ x = 110^{o}, y = 30^{o}, z = 30^{o}$$
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$$ x = 110^{o}, y = 10^{o}, z = 30^{o}$$
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$$ x = 110^{o}, y = 20^{o}, z = 30^{o}$$
Explanation
Given-
$$HOPE$$ is a parallelogram with diagonal $$HP$$.
$$\angle PHE={ 40 }^{ o }$$
The external angle by extending $$HO$$ is $${ 70 }^{ o }$$
To find out- $$\angle HEP=x$$
$$ \angle HPO=y=?$$ and $$ \angle PHO=z=?$$
Solution
-
The external $$\angle HOP={ 70 }^{ o }$$
$$ \therefore \angle HOP={ { 180 }^{ o }-70 }^{ o }$$ ....(linear pair)
$$={ 110 }^{ o }$$
Now $$\angle HOP=x$$ ....(opposite angle of a parallelogram)
$$ \therefore x={ 110 }^{ o }$$
Again $$HE\parallel OP$$ .....(opposite sides of a parallelogram)
$$ \therefore y=\angle HPE$$ ....(alternate angles) or $$ y={ 40 }^{ o }.$$
To find out $$z$$, we have
$$ z={ 180 }^{ o }-\angle HOP-y={ 180 }^{ o }-{ 110 }^{ o }-{ 40 }^{ o }$$ ....(angle sum property of triangles)
$$\Rightarrow z={ 30 }^{ o }$$
Ans- $$ x={ 110 }^{ o }, y={ 40 }^{ o }, z={ 30 }^{ o } $$
Given a parallelogram ABCD, if $$\angle A = 60^{\circ}$$ then $$\angle D$$ is equal to :
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$$110^{\circ}$$
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$$140^{\circ}$$
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$$120^{\circ}$$
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$$130^{\circ}$$
Explanation
$$\angle A=60^o,$$ therefore the opposite angle $$C=60^o$$
$$\angle B=\angle D$$ [Rule of parallelogram] as they are opposite angles.
Let $$\angle B$$ and $$\angle D$$ be $$x$$
$$\Rightarrow x+x+\angle A+\angle B=360^o$$ [ sum of all sides of a polygon ]
$$\Rightarrow 2x+60^o+60^o=360^o$$
$$\Rightarrow 2x=240^o$$
$$\Rightarrow x=120^o$$
$$\therefore \angle D=120^o$$
Hence, the answer is $$120^o.$$
Two consecutive angles of a parallelogram are in the ratio $$1 : 3$$, then the smaller angle is :
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$$50^{\circ}$$
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$$90^{\circ}$$
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$$60^{\circ}$$
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$$45^{\circ}$$
Explanation
$$\textbf{Step -1: Formulating the sum of angles}$$
$$\text{We know that, sum of two consecutive angles of a parallelogram = 180}^\circ$$
$$\text{Given angles are in the ratio 1 : 3}$$
$$\therefore \text{The angles are }x\text{ and }3x.$$
$$\therefore x + 3x = 180^\circ$$
$$\Rightarrow 4x = 180^\circ$$
$$\Rightarrow x = \dfrac{180}{4}$$
$$\therefore x = 45^\circ$$
$$\therefore \text{The smaller angle is 45}^\circ$$
$$\textbf{Hence, The smaller angle is 45}^\circ.$$ $$\textbf{Hence, option D is correct.}$$
In figure, ABCD is a parallelogram. If $$ \angle DAB = 60^{\circ}\ and\ \angle DBC = 80^{\circ},\angle CDB $$ is :
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$$40^{\circ}$$
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$$80^{\circ}$$
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$$60^{\circ}$$
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$$20^{\circ}$$
Explanation
Given $$ABCD$$ is a parallelogram and $$BD$$ is the diagonal.
$$AB\parallel CD$$
$$\Rightarrow \angle CBD=\angle ADB=80^o$$
We have $$\angle DAB=\angle BCD=60^o$$
Diagonal $$BD$$ forms a triangle $$CBD,$$
$$\Rightarrow \angle DCB+\angle CBD + \angle CDB =180^o$$ [ Sum of $$\angle$$ les of a triangle }
$$\Rightarrow 60^o+80^o+\angle CDB=180^o$$
$$\therefore \angle CBD=40^o$$ $$,$$ $$\angle ADB=80^o$$
Hence, the answer is $$40^o.$$
In figure, PQRS is a parallelogram in which $$\angle PSR=125^{\circ} , \angle RQT$$ is equal to
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$$75^{\circ}$$
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$$65^{\circ}$$
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$$55^{\circ}$$
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$$125^{\circ}$$
Explanation
Given PQRS is a parallelogram.
$$\Rightarrow \angle PSR=\angle PQR=125°$$ [ i.e, opposite angles ]
$$\therefore \angle RQT=180°-125°$$
$$=55°.$$
Hence, the answer is $$55°.$$
$$ABCD$$ is a parallelogram as shown in
figure. If $$AB = 2AD$$ and $$P$$ is mid-point of $$AB$$, then $$\angle{CPD}$$ is equal to
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$$90^{\circ}$$
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$$60^{\circ}$$
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$$45^{\circ}$$
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$$135^{\circ}$$
In parallelogram ABCD, if $$ \angle A = 2x + 15^{\circ}, \angle B = 3x - 25^{\circ}, $$ then value of $$x$$ is :
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$$91^{\circ}$$
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$$89^{\circ}$$
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$$34^{\circ}$$
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$$38^{\circ}$$
Explanation
Given the parallelogram ABCD.
In case of a parallelogram the sum of adjacent angles will be $$180°$$
i.e, $$\angle A +\angle B=180°$$
$$\Rightarrow 2x+15+3x-25^o=180°$$
$$\therefore 5x=190°$$
i.e, $$x=38°$$
Hence, the answer is $$38°.$$
ABCD is a parallelogram in which $$\angle DAC = 40^{\circ} ; \angle BAC = 30^{\circ}; \angle DOC = 105^{\circ}$$ ; then $$\angle CDO$$ equals :
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$$75^{\circ}$$
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$$70^{\circ}$$
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$$40^{\circ}$$
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$$45^{\circ}$$
Explanation
$$\angle AOB = \angle COD = 105^{\circ}$$ (Vertically Opposite Angles)
In triangle $$AOB,$$
$$\angle BAO + \angle AOB + \angle OBA = 180^{\circ} $$
$$30^{\circ} + 105^{\circ} + \angle OBA = 180^{\circ} $$
$$\angle OBA = 45^{\circ}$$
$$AB$$ and $$CD$$ are parallel
$$\therefore \angle OBA = \angle CDO = 45^{\circ}$$ (Alternate Interior angles)
If PQRS is a parallelogram, then $$\angle{Q}- \angle{S}$$ is equal to :
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$$90^{\circ}$$
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$$120^{\circ}$$
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$$180^{\circ}$$
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$$0^{\circ}$$
Explanation
Given that PQRS is parallelogram,
Opposite angles in a parallelogram are equal.
i.e, $$\angle P=\angle R$$ and $$\angle Q=\angle S$$
$$\therefore$$ If $$\angle Q=\angle S$$ then $$\angle Q-\angle S=0°$$
Hence, the answer is $$0°.$$
The diagonals of a parallelogram PQRS intersect at O. If $$ \angle QOR= 90^{\circ}\ and\ \angle QSR=50^{\circ}, \ then\ \angle ORS$$ is :
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$$90^{\circ}$$
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$$40^{\circ}$$
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$$70^{\circ}$$
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$$50^{\circ}$$
Explanation
Given:
$$ \angle QOR= 90^{\circ}\ and\ \angle QSR=50^{\circ}$$
Now, $$\angle SOR = 180^{\circ} - \angle QOR = 90^{\circ}$$ (Since angles forming a linear pair are complementary)
Consider $$\Delta SOR$$,
$$\angle SOR + \angle OSR + \angle ORS = 180^{\circ}$$ (angle sum property of triangles)
$$\therefore \angle ORS = 180^{\circ} - 90^{\circ} - 50^{\circ} = 40^{\circ}$$
Two adjacent angles of a parallelogram are $$(2x + 30)^{\circ}$$ and $$(3x + 30)^{\circ}$$. The value of $$x$$ is :
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$$30^{\circ}$$
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$$60^{\circ}$$
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$$24^{\circ}$$
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$$36^{\circ}$$
Explanation
We know that the sum of the adjacent angles of parallelogram is equal to $$180^o$$
$$\therefore 2x+30^o+3x+30^o=180°$$
$$\Rightarrow 5x=120^o$$
$$\therefore x=\dfrac{120^o}{5}=24°$$
Hence, the answer is $$24^o.$$
In the given figure, ABCD is a parallelogram. If $$\angle B = 100^{\circ}$$, then ($$\angle A + \angle C$$) is equal to :
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$$360^{\circ}$$
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$$200^{\circ}$$
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$$180^{\circ}$$
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$$160^{\circ}$$
Explanation
Given ABCD is a parallelogram.
The sum of co-interior angles is always $$180°$$.
Given $$\angle B =100°$$
and, $$\angle B +\angle C=180°$$
$$\Rightarrow \angle C = 80°$$
The opposite angles of a parallelogram are equal
i.e, $$\angle A= \angle C=80°$$
$$\therefore$$ Sum $$\angle A+ \angle C=80°+80°=160°$$
Hence, the answer is $$160°.$$
Consider the following statements:
(1) The diagonals of a parallelogram are equal.
(2) The diagonals of a square are perpendicular
to each other.
(3) If the diagonals of a quadrilateral intersect at
right angles, it is not necessarily a rhombus.
(4) Every quadrilateral is either a trapezium or a
parallelogram or a kite.
Which of the above statements is/are correct?
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Only (2)
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Only (3)
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Both (2) and (3)
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(1), (2) and (3)
If the sum of all interior angles of a convex polygon is $$1440^{\circ}$$, then the number of sides of the polygon is?
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$$8$$
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$$10$$
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$$11$$
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$$12$$
Explanation
If n is the number of sides of the polygon, then
$$(2n - 4)\times 90^{\circ}= 1440^{\circ}$$
or $$ 2n = 20$$
or $$n = 10$$
The measurement of each angle of a polygon is 160$$^{\circ}$$. The number of its sides is?
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$$15$$
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$$18$$
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$$20$$
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$$30$$
Explanation
Each angle of polygon $$= 160^{o}$$
Let there be n sides, then sum of all the angles $$= 160n$$
Sum of all the angles of any polygon $$= (n-2)(180^{o})$$
Therefore, $$160 n = (n-2)180^{o}$$
$$160^{o}n= 180^{o}n - 360^{o}$$
$$20n = 360$$
$$n= 18$$
The polygon has $$18$$ sides.
If one angle of the parallelogram is $$16^{\circ}$$ less than three times the smallest angle, then the largest angle of the parallelogram is
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$$131^{\circ}$$
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$$136^{\circ}$$
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$$112^{\circ}$$
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$$108^{\circ}$$
Explanation
Let $$x$$ and $$y$$ be the largest and smallest angles of parallelogram.
By hypothesis $$x = 3y - 16$$
Also in a parallelogram, we have
$$x + y = 180^{\circ}$$
$$\Rightarrow (3y - 16) + y = 180^{\circ}$$
$$\Rightarrow 4y = 196$$
$$\Rightarrow y = 49$$
Thus $$ x = 49\times 3 - 16$$
$$= 147 - 16 \\= 131^{\circ}$$
Each interior angle of a regular polygon is $$144^0$$. Find the interior angle of a regular polygon which has double the number of sides as the first polygon.
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$$100^0$$
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$$160^0$$
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$$36^0$$
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$$162^0$$
Explanation
Since each interior angle of the first polygon $$=144^o$$
Each exterior angle of the first polygon $$=180-144$$
$$=36^o$$
$$\therefore$$ The number of sides of the first polygon $$=\frac {360}{36}=10$$
$$\therefore$$ The number of sides of the second polygon $$=2\times 10=20$$
$$\therefore$$ Each exterior angle of the second polygon $$=\frac{360^o}{20}=18^o$$
$$\therefore$$ Each interior angle of the second polygon $$=180^o-18^o=162^o$$
If the sum of all the angles of a polygon except one angle is $$2220^\circ$$, then the number of sides of the polygon are:
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12
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13
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14
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15
Explanation
Sum of angles of polygon with $$n$$ sides $$= (n - 2) \times 180^{\circ}\dots(1)$$
The sum of the angles of a polygon should be a multiple of $$180^\circ$$.
But, $$2220^{\circ} = 180^{\circ}\times 12 + 60^{\circ}$$
$$2220^{\circ}$$ + $$120^{\circ}$$ = $$2340^{\circ}$$
$$( x - 2 ) = \dfrac{2340}{180}\dots$$ (From $$(1)$$)
$$x - 2 = 13$$
$$\Rightarrow x = 15$$
Therefore, the number of sides of the polygon is $$15$$.
If $$ABCD$$ is a parallelogram whose diagonals intersect at $$O$$ and $$BCD$$ is an equilateral triangle having each side of length $$6$$ cm, then the length of diagonal $$AC$$ is :
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$$3\sqrt{3}$$ cm
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$$6\sqrt{3}$$ cm
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$$3\sqrt{6}$$ cm
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$$12$$ cm
Explanation
$$ABCD$$ is a parallelogram
$$\therefore BC = AD = 6$$ cm
And $$AB = DC = 6$$ cm
Hence, $$ABCD$$ becomes a rhombus.
$$\therefore$$ Diagonals $$AC$$ and $$BD$$ bisect each other at right angle.
$$\therefore OD = \dfrac{1}{2} BD = 3$$ cm
From $$\Delta OCD$$,
$$OC^2 = CD^2 - OD^2$$ .... Pythagoras theorem
$$= 36 - 9$$
$$ = 27$$
$$\therefore OC = 3 \sqrt{3}$$
and $$AC = 2\cdot OC = 6 \sqrt{3}$$
Match List I with List II and select the correct answer using the codes given below the lists:
List I
(Regular plane figure)
List II
(Measure of interior angles)
I . Triangle
(A) $$30^{\circ}$$
II. Square
(B) $$60^{\circ}$$
III. Pentagon
(C) $$108^{\circ}$$
IV. Hexagon
(D) $$90^{\circ}$$
(E) $$120^{\circ}$$
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I-D, II-A, III-B, IV-E
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I-B, II-D, III-C, IV-E
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I-A, II-D, III-C, IV-B
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I-B, II-C, III-A, IV-D
Explanation
Each Angle (of a Regular Polygon) $$= \dfrac {(n-2) × 180°} { n}$$
Triangle =
$$(3-2)×180°/3=60°$$
Square=
$$(4-2)×180°/4=90°$$
Pentagon =
$$(5-2)×180°/5=108°$$
Hexagon =
$$(6-2)×180°/6=120°$$
$$\therefore$$ I-B, II-D, III-C, IV-E
A parallelogram with any pair of its consecutive sides equal is a
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Square
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Rectangle
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Rhombus
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Kite
State true or false:
It is possible to have a polygon whose sum of interior angles is
$$\displaystyle 320^{\circ}$$.
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True
0%
False
Explanation
A polygon whose sum of interior angles is $$ 320^o$$
Sum of interior angle of Polygon =$$ 180 (n -2) $$
=> $$180 (n -2) = 320 $$
=> $$n - 2 = \dfrac {16}{9} $$
=> $$n = \dfrac {25}{9} $$
Their is no possibility of a polygon whose sum of interior angles is $$ 320^o $$ because n must be an integer.
In parallelogram ABCD, $$\angle A = 3 \angle B$$
. In the same parallelogram, if AB $$= 5x-7$$ and $$CD = 3x + 1$$; find the length of CD.
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$$CD = 16$$ units
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$$CD = 11$$ units
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$$CD = 13$$ units
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$$CD = 15$$ units
Explanation
AB $$= $$ CD ( Opposite sides of parallelogram)
$$\implies 5x -7 = 3x + 1$$
$$\implies x = 4 $$
$$\therefore$$ AB $$=$$ CD$$ = 13 $$ units.
State true or false:
Is it possible to have a regular polygon whose exterior angle is $$40\%$$ of a right angle?
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True
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False
Explanation
$$40\%$$ of a right angle will be,
$$90^\circ \times \dfrac{{40}}{{100}} = 36^\circ$$
For a $$n$$-sided regular polygon all exterior angles are equal to $$\dfrac{360^\circ}{n}$$. So,
$$\begin{aligned}{}36^\circ& = \frac{{360^\circ}}{n}\\n &= \frac{{360^\circ}}{{36^\circ}}\\ &= 10\end{aligned}$$
We obtain an integral positive value of $$n$$ so, the given statement is true for a regular $$10$$-sided polygon.
In parallelogram PQRS $$\angle Q = (4x-5)^{\circ}$$ and $$\angle S = (3x+10)^{\circ}$$. Calculate $$\angle Q$$
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$$45^o$$
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$$50^o$$
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$$55^o$$
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$$60^o$$
Explanation
Opposite angles of a parallelogram are equal.
Therefore,
$$\angle Q=\angle S$$
$$(4x-5)^{\circ}=(3x+10)^{\circ}$$
$$x=15^{\circ}$$
$$\angle Q=(4*15-5)^{\circ}=55^{\circ}$$
Is it possible to have a regular polygon whose each interior angle is
$$170^{\circ}$$
State true or false:
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0%
True
0%
False
Explanation
Let number of polygon = n (which must be an integer)
Sum of all interior angles is $$ n \times 170^o $$
Sum of all interior angles of a polygon is $$ 180^o (n-2) $$
$$ => n \times 170^o = 180^o (n-2) $$
$$ => 10^o n = 360^o $$
$$=> n = 36^o $$
Yes, Their is a regular polygon whose each interior angle is $$ 170^o $$
Is it possible to have a polygon, whose sum of interior angles is
$$4500^{\circ}$$?
State true or false:
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0%
True
0%
False
Explanation
Sum of Interior angles of a polygon, having $$n$$ sides $$= 180^{\circ} (n-2) $$
Given sum of interior angles is $$ 4500^o $$
$$180^{\circ} (n-2) = 4500^{\circ} $$
$$ n - 2 = 25 $$
$$ n = 27 $$
Since $$n$$ is an integer so their exist a polygon whose sum of interior angles is $$ 4500^{\circ} $$
State true or false:
Is it possible to have a polygon whose sum of interior angles is
$$\displaystyle 540^{\circ}$$?
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0%
True
0%
False
Explanation
A polygon whose sum of interior angles is $$ 540^{\circ} $$
Sum of interior angle of Polygon $$=180 (n -2)$$
$$\implies 180 (n -2) = 540$$
$$\implies n - 2 = 3 $$
$$\implies n = 5 $$
Their exist a polygon whose sum of interior angles is $$ 540^{\circ} $$ and the polygon is pentagon.
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