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CBSE Questions for Class 8 Maths Understanding Quadrilaterals Quiz 11 - MCQExams.com
CBSE
Class 8 Maths
Understanding Quadrilaterals
Quiz 11
A parallelogram and a triangle stand on the same side of the base with the same base and on the same side of the base with the same height. If $${l}_{1}$$, $${l}_{2}$$ be the perimeters of the parallelogram and the rectangle respectively, then the which one of the following is correct?
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$${l}_{1}< {l}_{2}$$
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$${l}_{1}={l}_{2}$$
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$${l}_{1}> {l}_{2}$$ but $${l}_{1}\ne 2{l}_{2}$$
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$${l}_{1}=2{l}_{2}$$
A school was having 4 Hexagonal buildings joined to each other. They wanted to utilize the space between the 4 buildings to make a playground. The shape is that of a parallelogram. Can you find the measure of the angles as opposite angles are equal?
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$$60^o, 120^o$$
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$$20^o, 60^o$$
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$$80^o, 120^o$$
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$$40^o, 40^o$$
Explanation
$$x$$ and $$y$$ are exterior angles of the regular polygon.
$$\displaystyle \frac{360}{6} = 60$$ i.e value of $$x$$ angle
Opposite angles are $$60^o+60^o$$
$$360^o - 120^o (60^o + 60^o)$$
$$360^o = 60^o + 60^o + y + y$$
$$2y = 360^o - 120^o$$
$$y =$$ $$ \dfrac{240^o }{ 2}$$
$$y = 120^o$$
Therefore, option A is the correct answer.
Classify the following into open and closed curves
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Open (b,p,r) Closed ( a,c,q)
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Open (b,q ,c) Closed (a,p,r)
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Open (p,q,a ) Closed (r,c,b)
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Open (r,b,a) closed (p,q,c)
Which of the following statement(s) is/ are correct?
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A parallelogram in which two adjacent angles are equal is a rectangle
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A quadrilateral in which both pairs of opposite angles are equal is parallelogram
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In a parallelogram the number of acute angles is two
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All of these
A parallelogram with all equal sides are called _________.
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Rhombus
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Rectangle
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Polygon
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None of these
Select the CORRECT match. (A) (B)
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Two pairs of parallel sides - Trapezium
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A rhombus with 4 right angles - Rectangle
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Parallelogram with 4 right angles - Rectangle
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Diagonals are perpendicular to one another -
Trapezium
In the given figure, ABCD is a parallelogram, AL $$\perp$$ BC, AM$$\perp$$ CD, AL$$=4$$cm and AM$$=5$$cm. If BC$$=6.5$$cm, then find CD.
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$$5.2$$cm
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$$8.7$$cm
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$$6.5$$cm
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$$3.3$$cm
Which of the following closed plane figures is/are not polygons?
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Fill in the blanks.
Any drawing ( straight or non-Straight) done without lifting the pencil may be a_____P_____. A____Q____is the one that does not cross itself . A curve is said to be ___R__ if its ends are joined. A____S___is a simple closed curve made up of lines segments.
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P-Curve, Q-Open curve, R-Closed, S-Line
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P-Line, Q-Curve, R-Open, S-Line
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P-Curve, Q-Simple curve, R-Closed, S-Polygon
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P-Curved, Q-Closed curve, R-Open, S-Circle
In the given figure, if $$TS||QV, \, SV || TQ, \, \angle QRS = {115^ \circ}$$, then $$\angle TQR$$ is equal to
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$${65^ \circ}$$
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$${115^ \circ}$$
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$${45^ \circ}$$
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$${25^ \circ}$$
Explanation
Given, A parallelogram $$QTSV,\angle QRS=115^\circ$$
From parallelogram
$$\displaystyle \angle S +\angle V = 180^{\circ} $$
Because $$ TS \parallel QV $$
So, $$\displaystyle 90^{\circ}+\angle V = 180 $$
$$\displaystyle \angle V = 90^{\circ} $$
similarity $$\displaystyle \angle TQV+\angle V = 180^{\circ} $$
so, $$\displaystyle \angle TQV = 90^{\circ} $$
$$\displaystyle \angle TRQ = 180-\angle QRS = 180-115 = 65^{\circ} $$
Now we know that alternate angle will be equal as $$RQ$$ is transversal to $$ TS \parallel QV $$
so, $$\displaystyle \angle TRQ = \angle RQV = 65^{\circ} $$
But $$\displaystyle \angle TQV = 90 \Rightarrow \angle RQV+\angle TQR = 90^{\circ} $$
$$\displaystyle \angle TQR = 90-65 $$
$$\displaystyle \angle TQR = 25^{\circ} $$ option$$-$$D
The length of the diagonal $$BD$$ of the parallelogram $$ABCD$$ is $$18\ cm$$. If $$p$$ and $$Q$$ are the Centroid of the $$\triangle ABC$$ and $$\triangle ADC$$ respectively then length of the line segment $$PQ$$ is:
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$$4\ cm$$
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$$6\ cm$$
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$$9\ cm$$
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$$12\ cm$$
$$p$$ is a point in the interior of $$\parallel gm$$ $$ABCD$$. If the area of $$\parallel gm$$ $$ABCD$$ is $$60\ {cm}^{2}$$.$$ar\left( \triangle ADP \right) +ar\left( \triangle BPC \right) =$$
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$$15\ { cm }^{ 2 }$$
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$$30\ { cm }^{ 2 }$$
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$$45\ { cm }^{ 2 }$$
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$$20\ { cm }^{ 2 }$$
Explanation
In parallelogram ABCD, in quadrilateral AEFD,
In $$\Delta APD $$ and quadrilateral AEFD lie on same base between same parallels AD & EF.
$$\therefore ar(\Delta ADP)=\dfrac{1}{2}ar(\Delta EFD)$$
In $$\Delta BPC$$ and quadrilateral EFCB lie on same base between same parallel EF & BC.
$$\therefore ar(\Delta BPC)=\dfrac{1}{2}ar(EFCB)$$
Now,
$$ar(\Delta ADP)+ar(\Delta BPC)$$
$$=\dfrac{1}{2}ar(AEFD)+\dfrac{1}{2}ar(EFCB)$$
$$=\dfrac{1}{2}ar(ABCD)$$.
$$=\dfrac{1}{2}\times 60\ cm^2=30\ cm^2$$.
$$ABCD$$ is a parallelogram $$P$$ is a point of $$AD$$ such that $$\dfrac {AP}{AD}=\dfrac {1}{2015}.Q$$ is the point of intersection of $$AC$$ and $$BP$$. Then $$\dfrac {AQ}{AC}=$$
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$$\dfrac {2000}{2016}$$
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$$\dfrac {2015}{2016}$$
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$$\dfrac {2018}{2016}$$
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$$\dfrac {1}{2016}$$
The following figure shows a ABCD in which AB is quarlled to DC and AD=BC
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$$\angle DAB=\angle CBA$$
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$$\angle ADC=\angle BCD$$
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$$ADC=BD$$
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$$OA=OB AND OC=OD$$
$$ABCD$$ is a parallelogram. $$ { AP } $$ bisects $$\angle A$$ and $$ { CQ } $$ bisects $$\angle C.P$$ lies on $$ { CD } $$ and Q lies on $$ { AB }. $$ Then
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$$ { AB } \parallel { CQ } $$
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$$ { AP } \parallel { CQ } $$
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$$ { AQ } \parallel { BC } $$
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None of these
A parallelogram, the lengths of whose sides are 11 cm and 13 cm, has one diagonal 20 cm, long the length of another diagonal is
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12 cm
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20 cm
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66 cm
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25 cm
The two adjacent sides of a parallelogram are 4 cm and 9 cm. The ratio of its altitude is
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21 : 22
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4 : 9
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16 : 81
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11 : 37
In the given figure, ABCD is a parallelogram, $$\angle ADE={ 50 }^{ \circ }$$ and $$\angle ACE={ \angle BED=90 }^{ \circ }.$$ The value of $$\angle EAC+{ \angle ABC }-2\angle DAC$$ is
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$${ 20 }^{ \circ }$$
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$${ 10 }^{ \circ }$$
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$${ 30 }^{ \circ }$$
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$${ 40 }^{ \circ }$$
In the given figure, ABCD is a parallelogram such that AB||CD and AD||BC with opposite angles equal. If DA=DX, then find the values of $$\angle{m}$$ and $$\angle{n}$$.
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$$m=24^\circ.n=34^\circ$$.
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$$m=24^\circ.n=24^\circ$$.
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$$m=34^\circ.n=24^\circ$$.
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$$m=34^\circ.n=34^\circ$$.
In the given figure, ABCD is a parallelogram. If AB = 12 cm, AE = 7.5 cm, CF = 15 cm then AD =
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3 cm
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6 cm
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8 cm
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10.5 cm
ABCD is a parallelogram, AB=14 cm, BC=18 cm and AC=16 cm, then the length of the diagonal is_
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24cm
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36cm
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28cm
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32cm
The altitude of a parallelogram is twice of the length of the base and its area is 900 $$c{m^2}$$ . The length of base and altitude respectively are
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$$15\sqrt {2\,} \,cm,\,\,\,30\sqrt 2 cm$$
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$$25\sqrt 2 cm,\,\,\,\,\,35\sqrt 2 cm$$
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$$45\sqrt 2 cm,\,\,\,\,\,\,55\sqrt 2 cm$$
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$$65\sqrt 2 cm,\,\,\,\,\,75\sqrt 2 cm$$
In the given figure , $$ABCD$$ and $$BDCE$$ are parallelograms with common base $$DC$$ . If $$BC$$ $$ \perp $$ $$BD$$ , then $$\angle BEC$$ =
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$$ 60^{\circ} $$
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$$ 30^{\circ} $$
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$$ 150^{\circ} $$
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$$ 120^{\circ} $$
Explanation
$$\angle BAD=30^{\circ}$$
$$\angle BCD=30^{\circ}$$ [Opposite angles of parallelogram equal]
Now,, in $$\Delta CBD$$,
$$\angle DBC+\angle BCD+\angle CDB=180^{\circ}$$
$$\Rightarrow 90^{\circ}+30^{\circ}+\angle CDB=180^{\circ}$$
$$\Rightarrow CDB=60^{\circ}$$
$$\therefore \angle BEC=60^{\circ}$$ [Opposite angles are equal]
(A) $$60^{\circ}$$
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