Understanding Quadrilaterals Test

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Understanding Quadrilaterals Test - 28

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Weekly Quiz Competition

Question 1
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Which of the following statement(s) is/ are correct?

A
A parallelogram in which two adjacent angles are equal is a rectangle

B
A quadrilateral in which both pairs of opposite angles are equal is parallelogram

C
In a parallelogram the number of acute angles is two

D
All of these

Solution
Question 2
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Classify the following into open and closed curves

A
Open (b,p,r) Closed ( a,c,q)

B
Open (b,q ,c) Closed (a,p,r)

C
Open (p,q,a ) Closed (r,c,b)

D
Open (r,b,a) closed (p,q,c)
Question 3
1 / -0

Select the CORRECT match. (A) (B)

A
Two pairs of parallel sides - Trapezium

B
A rhombus with 4 right angles - Rectangle

C
Parallelogram with 4 right angles - Rectangle

D
Diagonals are perpendicular to one another -

Trapezium
Question 4
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$$p$$ is a point in the interior of $$\parallel gm$$ $$ABCD$$. If the area of $$\parallel gm$$ $$ABCD$$ is $$60\ {cm}^{2}$$.$$ar\left( \triangle ADP \right) +ar\left( \triangle BPC \right) =$$

A
$$15\ { cm }^{ 2 }$$

B
$$30\ { cm }^{ 2 }$$

C
$$45\ { cm }^{ 2 }$$

D
$$20\ { cm }^{ 2 }$$

Solution

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,
$$ar(\Delta ADP)+ar(\Delta BPC)$$

$$=\dfrac{1}{2}ar(AEFD)+\dfrac{1}{2}ar(EFCB)$$

$$=\dfrac{1}{2}ar(ABCD)$$.

$$=\dfrac{1}{2}\times 60\ cm^2=30\ cm^2$$.
Question 5
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In the given figure, if $$TS||QV, \, SV || TQ, \, \angle QRS = {115^ \circ}$$, then $$\angle TQR$$ is equal to

A
$${65^ \circ}$$

B
$${115^ \circ}$$

C
$${45^ \circ}$$

D
$${25^ \circ}$$

Solution

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
Question 6
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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:

A
$$4\ cm$$

B
$$6\ cm$$

C
$$9\ cm$$

D
$$12\ cm$$
Question 7
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$$ABCD$$ is a parallelogram. $$ { AP } $$ bisects $$\angle A$$ and $$ { CQ } $$ bisects $$\angle C.P$$ lies on $$ { CD } $$ and Q lies on $$ { AB }. $$ Then

A
$$ { AB } \parallel { CQ } $$

B
$$ { AP } \parallel { CQ } $$

C
$$ { AQ } \parallel { BC } $$

D
None of these

Solution
Question 8
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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

A
$${ 20 }^{ \circ }$$

B
$${ 10 }^{ \circ }$$

C
$${ 30 }^{ \circ }$$

D
$${ 40 }^{ \circ }$$
Question 9
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$$ABCD$$ 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 $$\dfrac {AQ}{AC}=$$

A
$$\dfrac {2000}{2016}$$

B
$$\dfrac {2015}{2016}$$

C
$$\dfrac {2018}{2016}$$

D
$$\dfrac {1}{2016}$$

Solution
Question 10
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The following figure shows a ABCD in which AB is quarlled to DC and AD=BC

A
$$\angle DAB=\angle CBA$$

B
$$\angle ADC=\angle BCD$$

C
$$ADC=BD$$

D
$$OA=OB AND OC=OD$$