Understanding Quadrilaterals Test

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Understanding Quadrilaterals Test - 12

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Question 1
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The diagonals $$AC$$ and $$BD$$ of a parallelogram $$ABCD$$ intersect each other at the point $$O$$.
If $$\angle DAC = 32^o$$ and $$\angle AOB = 70^o$$, then $$\angle DBC$$ is equal to?

A
$$24^o$$

B
$$86^o$$

C
$$38^o$$

D
$$32^o$$

Solution

In given figure,
Quadrilateral ABCD is a parallelogram.
So, AD $$\left|\right|$$ BC
$$\therefore$$ $$\angle$$DAC = $$\angle$$ACB --- ( Alternate angle)
$$\therefore$$ $$\angle$$ACB = 32$$^\circ$$
$$\angle$$AOB + $$\angle$$BOC = 180$$^\circ$$ --- (straight angle)
$$\Rightarrow$$70$$^\circ$$ + $$\angle$$BOC = 180$$^\circ$$
$$\therefore$$ $$\angle$$BOC = 110$$^\circ$$
In $$\triangle$$BOC,
$$\angle$$OBC + $$\angle$$BOC + $$\angle$$OCB = 180$$^\circ$$
$$\Rightarrow$$$$\angle$$OBC + 110$$^\circ$$ + 32$$^\circ$$ = 180$$^\circ$$
$$\Rightarrow$$ $$\angle$$OBC = 38$$^\circ$$
$$\therefore$$ $$\angle$$DBC = 38$$^\circ$$
Question 2
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In Fig.8.2, $$ABCD$$ and $$AEFG$$ are two parallelograms. If $$\angle C = 55^o$$, determine $$\angle F$$.

A
$$27.5^o$$

B
$$45^o$$

C
$$55^o$$

D
$$65^o$$

Solution

$$ ABCD$$ and $$ AEFG$$ are two parallelograms sharing two sides $$AD$$ (or $$AG$$) and $$AE$$ (or $$AE$$) and the included angle.
$$\angle C={ 55 }^{ o }$$.
To find out- $$ \angle F$$
Solution-
We have $$ \angle F=\angle A$$ ......(opposite angles of a parallelogram).........(i)
Again $$ \angle A=\angle C$$ ......(opposite angles of a parallelogram)...........(ii)
$$ \therefore \angle F=\angle C$$ ......[From (i) and (ii)]
And $$ \angle C={ 55 }^{ o }$$ .......(given)
$$ \Rightarrow \angle F={ 55 }^{ o }$$
Question 3
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Examine the table. (Each figure is divided into triangles and the sum of the angles deduced from that)
What can you say about the angle sum of a convex polygon with the number of sides (n) = 7?

A
$$700^{0}$$

B
$$800^{0}$$

C
$$900^{0}$$

D
$$1000^{0}$$

Solution

$$ After\quad analysing\quad the\quad given\quad table\quad it\quad is\quad observed\quad that\quad when\\ (i)\quad numbre\quad of\quad sides\quad of\quad the\quad polygon=3\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \longrightarrow sum\quad of\quad the\quad internal\quad angles={ 180 }^{ o }=\left( 3-2 \right) \times { 180 }^{ o }\\ (ii)\quad numbre\quad of\quad sides\quad of\quad the\quad polygon=4\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \longrightarrow sum\quad of\quad the\quad internal\quad angles={ 360 }^{ o }=\left( 4-2 \right) \times { 180 }^{ o }\\ (iii)\quad numbre\quad of\quad sides\quad of\quad the\quad polygon=5\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \longrightarrow sum\quad of\quad the\quad internal\quad angles={ 540 }^{ o }=\left( 5-2 \right) \times { 180 }^{ o }\\ (iv)\quad numbre\quad of\quad sides\quad of\quad the\quad polygon=6\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \longrightarrow sum\quad of\quad the\quad internal\quad angles={ 720 }^{ o }=\left( 6-2 \right) \times { 180 }^{ o }\\ \therefore \quad By\quad induction,\quad it\quad can\quad be\quad concluded\quad that\\ When\quad the\quad numbre\quad of\quad sides\quad of\quad the\quad polygon=n\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \longrightarrow sum\quad of\quad the\quad internal\quad angles=\left( n-2 \right) \times { 180 }^{ o }\\ \therefore \quad For\quad a\quad 7\quad sided\quad polygon\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \rightarrow sum\quad of\quad the\quad internal\quad angles=\left( 7-2 \right) \times { 180 }^{ o }={ 900 }^{ o }\\ \\ \\ \\ \\ \\ \\ $$
Question 4
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How many sides does a regular polygon have if each of its interior angles is $$165^{0}$$?

A
Number of sides$$ = 24$$

B
Number of sides$$ = 30$$

C
Number of sides$$ = 36$$

D
Number of sides$$ = 42$$

Solution

$$-Let\quad the\quad given\quad polygon\quad have\quad n\quad sides.\\ \therefore \quad the\quad sum\quad of\quad the\quad interior\quad angles=\left( n-2 \right) { 180 }^{ o }\quad \& \quad the\quad number\quad of\\ angles\quad is\quad n.\\ Again\quad the\quad polygon\quad is\quad a\quad regular\quad one.\\ So\quad the\quad interior\quad angles\quad are\quad equal\quad to\quad each\quad other.\\ Let\quad each\quad interior\quad angle\quad be\quad \theta .\quad $$

$$ \therefore \quad \theta =\dfrac { \left( n-2 \right) { 180 }^{ o } }{ n } \Longrightarrow n=\dfrac { { 360 }^{ o } }{ { 180 }^{ o }-\theta } .$$

$$Given\quad that\quad \theta =165^{ o }$$

$$ \therefore \quad n=\dfrac { { 360 }^{ o } }{ { 180 }^{ o }-165^{ o } } =24$$

$$ Ans-\quad 24\quad sides. $$
Question 5
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The adjacent figure $$\Box HOPE$$ is a parallelogram. Find the angle measures $$x, y$$ and $$z$$. State the properties you use to find them.

A
$$ x = 110^{o}, y = 40^{o}, z = 30^{o}$$

B
$$ x = 110^{o}, y = 30^{o}, z = 30^{o}$$

C
$$ x = 110^{o}, y = 10^{o}, z = 30^{o}$$

D
$$ x = 110^{o}, y = 20^{o}, z = 30^{o}$$

Solution

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- $$ x={ 110 }^{ o }, y={ 40 }^{ o }, z={ 30 }^{ o } $$
Question 6
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Given a parallelogram ABCD, if $$\angle A = 60^{\circ}$$ then $$\angle D$$ is equal to :

A
$$110^{\circ}$$

B
$$140^{\circ}$$

C
$$120^{\circ}$$

D
$$130^{\circ}$$

Solution

$$\angle A=60^o,$$ 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 $$120^o.$$
Question 7
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In figure, PQRS is a parallelogram in which $$\angle PSR=125^{\circ} , \angle RQT$$ is equal to

A
$$75^{\circ}$$

B
$$65^{\circ}$$

C
$$55^{\circ}$$

D
$$125^{\circ}$$

Solution

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 $$55°.$$
Question 8
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In figure, ABCD is a parallelogram. If $$ \angle DAB = 60^{\circ}\ and\ \angle DBC = 80^{\circ},\angle CDB $$ is :

A
$$40^{\circ}$$

B
$$80^{\circ}$$

C
$$60^{\circ}$$

D
$$20^{\circ}$$

Solution

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 $$40^o.$$
Question 9
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Two consecutive angles of a parallelogram are in the ratio $$1 : 3$$, then the smaller angle is :

A
$$50^{\circ}$$

B
$$90^{\circ}$$

C
$$60^{\circ}$$

D
$$45^{\circ}$$

Solution

$$\textbf{Step -1: Formulating the sum of angles}$$
$$\text{We know that, sum of two consecutive angles of a parallelogram = 180}^\circ$$
$$\text{Given angles are in the ratio 1 : 3}$$
$$\therefore \text{The angles are }x\text{ and }3x.$$
$$\therefore x + 3x = 180^\circ$$
$$\Rightarrow 4x = 180^\circ$$
$$\Rightarrow x = \dfrac{180}{4}$$
$$\therefore x = 45^\circ$$
$$\therefore \text{The smaller angle is 45}^\circ$$
$$\textbf{Hence, The smaller angle is 45}^\circ.$$ $$\textbf{Hence, option D is correct.}$$
Question 10
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In parallelogram ABCD, if $$ \angle A = 2x + 15^{\circ}, \angle B = 3x - 25^{\circ}, $$ then value of $$x$$ is :

A
$$91^{\circ}$$

B
$$89^{\circ}$$

C
$$34^{\circ}$$

D
$$38^{\circ}$$

Solution

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 $$38°.$$