Understanding Quadrilaterals Test

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Understanding Quadrilaterals Test - 15

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Question 1
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PQRS is a parallelogram whose diagonals intersect at M.
If $$\angle PMS = 54^{\circ}$$, $$\angle QSR = 25^{\circ}$$ and $$\angle SQR = 30^{\circ}$$: find $$\angle RPS $$

A
$$86^o$$

B
$$96^o$$

C
$$92^o$$

D
$$100^o$$

Solution

Given is a parallelogram PQRS whose diagonals intersect at M.
Also given $$\angle PMS={ 54 }^{ o }, \angle QSR={ 25 }^{ o }$$ and $$\angle SQR={ 30 }^{ o }$$
We know that in a parallelogram opposite sides are parallel.
So PQ$$\parallel$$ SR & if SQ is considered as the transversal then
$$ \angle PSM=\angle RQM={ 30 }^{ o }$$ [ Alternate Angles ]
Now in $$\triangle$$ PMS,
$$\angle MPS+\angle PSM+PMS={ 180 }^{ o }$$
$$\Rightarrow \angle MPS+{ 30 }^{ o }+{ 54 }^{ o }={ 180 }^{ o }$$
$$ \Rightarrow \angle MPS+{ 84 }^{ o }={ 180 }^{ o }$$
$$\Rightarrow \angle MPS={ 180 }^{ o }-{ 84 }^{ o }$$
$$\Rightarrow \angle MPS={ 96 }^{ o }$$
Now $$ \angle RPS=\angle MPS={ 96 }^{ o }$$
Question 2
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PQRS is a parallelogram whose diagonals intersect at M.
If $$\angle PMS = 54^{\circ}$$, $$\angle QSR = 25^{\circ}$$ and $$\angle SQR = 30^{\circ}$$: find $$\angle PRS$$

A
$$39^o$$

B
$$29^o$$

C
$$18^o$$

D
$$24^o$$

Solution

Given is a parallelogram PQRS whose diagonals intersect at M.
Also given $$\angle PMS={ 54 }^{ o }$$, $$\angle QSR={ 25 }^{ o }$$ and$$\angle SQR={ 30 }^{ o }$$
We know that in a parallelogram opposite sides are parallel.
So PQ$$\parallel $$SR & if SQ is considered as the transversal then
$$ \angle PSM=\angle RQM={ 30 }^{ o } $$[ Alternate Angles ]
Also $$\angle PSR=\angle PSQ+\angle QSR={ 30 }^{ o }+{ 25 }^{ o }={ 55 }^{ o }$$
Now in
$$\triangle PMS,\\ \angle MPS+\angle PSM+PMS={ 180 }^{ o }\\ \Rightarrow \angle MPS+{ 30 }^{ o }+{ 54 }^{ o }={ 180 }^{ o }\\ \Rightarrow \angle MPS+{ 84 }^{ o }={ 180 }^{ o }\\ \Rightarrow \angle MPS={ 180 }^{ o }-{ 84 }^{ o }\\ \Rightarrow \angle MPS={ 96 }^{ o }$$
Now $$\angle RPS=\angle MPS={ 96 }^{ o }$$
Now in
$$\triangle PSR,\\ \angle PSR+\angle PRS+RPS={ 180 }^{ o }\\ \Rightarrow \angle PRS+{ 96 }^{ o }+{ 55 }^{ o }={ 180 }^{ o }\\ \Rightarrow \angle PRS+{ 151 }^{ o }={ 180 }^{ o }\\ \Rightarrow \angle PRS={ 180 }^{ o }-{ 151 }^{ o }\\ \Rightarrow \angle PRS={ 29 }^{ o }\\
$$
Question 3
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In parallelogram PQRS $$\angle Q = (4x-5)^{\circ}$$ and $$\angle S = (3x+10)^{\circ}$$. Calculate $$\angle Q$$

A
$$45^o$$

B
$$50^o$$

C
$$55^o$$

D
$$60^o$$

Solution

Opposite angles of a parallelogram are equal.
Therefore,
$$\angle Q=\angle S$$
$$(4x-5)^{\circ}=(3x+10)^{\circ}$$
$$x=15^{\circ}$$
$$\angle Q=(4*15-5)^{\circ}=55^{\circ}$$
Question 4
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Find the number of sides in a regular polygon, if its each interior angle is $$160^{\circ}$$.

A
$$12$$

B
$$14$$

C
$$16$$

D
$$18$$

Solution

Each interior angle is $$ 160^o $$
Sum of all angles of any polygon is $$ 180^o (n-2) $$

Each angle of any polygon = $$ \dfrac{180^o (n-2)}{n} $$

$$ => 160^o = \dfrac{180^o (n-2)}{n} $$

$$ 20n = 360^o $$
$$ n = 18 $$
Number of sides of given polygon is $$18$$.
Question 5
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The perimeter of a parallelogram $$ABCD= 40$$ cm, $$AB= 3x$$ cm, $$BC= 2x$$ cm and $$CD= 2\left ( y\, +\, 1 \right )$$ cm. Find the values of $$x$$ and $$y$$.

A
$$x= 5$$ and $$y= 4$$

B
$$x= 4$$ and $$y= 5$$

C
$$x= 5$$ and $$y= 5$$

D
$$x= 4$$ and $$y= 4$$

Solution

In the parallelogram $$ABCD$$ side $$AB$$ and $$CD$$ are opposite to each other. Opposite sides of parallelogram are equal.
Thus,
$$AB = CD$$ and $$BC = AD $$

$$\Rightarrow 3x = 2 (y+1)$$
$$\Rightarrow 2x = AD $$

$$\Rightarrow (y+1) = 1.5x$$

Now, according to the given condition

$$AB + BC + CD + AD = 40 $$
$$\Rightarrow 3x + 2x + 2 (y+1) + 2x = 40 $$

$$ \Rightarrow 7x + 2 \times 1.5x$$ $$ = 40$$

$$\Rightarrow 7x + 3x = 40 $$

$$\Rightarrow x = 4$$

Simplify further,
$$(y+1) = 1.5x \\= 1.5 \times 4$$
$$ = 6$$

$$\Rightarrow y = 5 $$
Question 6
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In parallelogram $$ABCD$$, $$AP$$ and $$AQ$$ are perpendiculars from vertex of obtuse angle $$A$$ as shown in the figure. If $$\angle x\, :\, \angle y= 2\, :\, 1$$; find smallest angles of the parallelogram.(in degrees)

A
$$55^o$$

B
$$60^o$$

C
$$70^o$$

D
$$48^o$$

Solution

Given is a parallelogram ABCD. AP & AQ are perpendiculars drawn from vertex A.
So $$\angle AQD=\angle AQC=\angle APC=\angle PAB={ 90 }^{ o }$$
Also given $$\angle C=x $$ & $$\angle QAP=y$$ and $$x:y=2:1$$
Now in quadrilateral QCPA,
$$ \angle QAP+\angle APC+\angle PCQ+\angle CQA={ 360 }^{ o }$$
$$ \Rightarrow \angle QAP+\angle PCQ+{ 90 }^{ o }+{ 90 }^{ o }={ 360 }^{ o }$$
$$ \Rightarrow \angle QAP+\angle PCQ={ 180 }^{ o }$$
$$ \Rightarrow y+x={ 180 }^{ o }$$
Since $$x:y=2:1$$ so $$x=\dfrac { 2 }{ 3 } \times 180^o={ 120 }^{ o }$$
$$\Rightarrow \angle C={ 120 }^{ o }$$
So $$\angle A=\angle C={ 120 }^{ o }$$
Now $$\angle C+\angle B={ 180 }^{ o }$$ [ adjacent angles in parallelogram are supplementary ]
$$\Rightarrow \angle B=180^o-120^o={ 60 }^{ o }$$
So $$\angle B=\angle D={ 60 }^{ o }$$
Question 7
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In the given figure, $$E$$ is mid-point of $$AB$$ and $$DE$$ meets diagonal $$AC$$ at point $$F$$. If $$ABCD$$ is a parallelogram and area of $$\bigtriangleup ADF$$ is $$60 \: cm^{2}$$, then area of parallelogram $$ABCD$$ is

A
$$360\ cm^{2}$$

B
$$460\ cm^{2}$$

C
$$450\ cm^{2}$$

D
None of These
Question 8
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Find the value of $$x , y$$

A
$$x = 12^{\circ}$$ and $$y = 16^{\circ}$$

B
$$x = 10^{\circ}$$ and $$y = 16^{\circ}$$

C
$$x = 15^{\circ}$$ and $$y = 16^{\circ}$$

D
$$x = 14^{\circ}$$ and $$y = 16^{\circ}$$

Solution

Given, ABCD is a parallelogram with $$\angle A=4x+20,\quad \angle B=7y,\quad \angle D=6x+3y-8$$

Opposite angles of a parallelogram are equal

$$\therefore \angle B=\angle D$$,

$$\quad 6x+3y-8=7y\\ or,6x+3y-8-7y=0\\ or,6x-4y-8=0\quad \\ or,6x-4y=8\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad (1)$$

Any two consecutive angles of a parallelogram are supplementary.
$$\therefore \angle A+\angle B=180\\ \quad \quad 4x+20+7y=180\\ or,4x+7y=180-20\\ or,4x+7y-160\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad (2)$$

From (1) and (2),

$$6x-4y=8...(i)$$ and $$4x+7y=160.....(ii)$$

Multiply eq.(i) by $$4$$ and multiply eq (ii) by $$6$$

$$24x-16y=32....(iii)$$ and $$24x+42y=960.....(iv)$$

Now Eq.$$(iv)$$ $$-$$ Eq.$$(iii)$$ we get,

$$x=12^{\circ} and\quad y=16^{\circ}$$
Question 9
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Directions For Questions

In the given figure, if area of triangle ADE is $$60\, cm^{2}$$;

...view full instructions

the area of $$\triangle ABE$$?

A
$$60\, cm^{2}$$

B
$$120\, cm^{2}$$

C
$$40\, cm^{2}$$

D
$$30\, cm^{2}$$

Solution

Given, $$Area \ of\ \Delta ADE = 60 cm^2 $$
From the figure,

$$AB \parallel CF$$ so, $$ AB \parallel DE $$
$$AD\parallel BE$$
So, $$ABED$$ is a parallelogram.
The diagonal of a parallelogram divides into two triangles of equal areas.
Therefore,
Area of $$\Delta ABE =$$ Area of $$\Delta ADE $$
=> Area of $$\Delta ABE = 60 \ cm^2 $$
Question 10
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In a quadrilateral ABCD, if AB || CD, $$\angle$$D $$= 2\angle$$B, AD = b and CD = a, then the side AB is of length:

A
$$\displaystyle \frac{a}{2}\, +\, 2b$$

B
$$a\, +\, 2b$$

C
$$2a\, -\, b$$

D
$$a\, +\, b$$

Solution

Given, $$AB\parallel CD$$, $$\angle D=2\angle B$$, $$AD=b$$, $$CD=a$$
Let $$\angle B=x$$ and draw a line segment, $$CM$$ equal and parallel to $$AD$$, as shown in the above figure.
Hence, $$AMCD$$ is a parallelogram. [Quadrilateral having one pair of opposite sides equal and parallel]

Now, $$AD=b$$
$$\therefore CM=AD=b$$
and $$CD=a$$
$$\therefore AM=CD=a$$

$$\therefore \angle D=\angle AMC$$ [Opposite angles of a parallelogram]
Hence, $$\angle AMC =2x\quad \quad [\angle D=2\angle B=2x]$$

Now, in $$\triangle MCB$$
$$2x=\angle MCB +\angle x\quad (\because\ $$ Exterior angle is equal to sum of opposite interior angle$$)$$
$$\therefore \ \angle MCB=x$$
Hence, $$ MB=MC \quad [$$As sides opposite to equal angles are equal$$]$$
$$\therefore \ MB=b$$
Now, $$AB=AM+MB$$

$$\therefore \ AB=a+b$$

Hence, option D is correct.