Quadrilaterals Test

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Quadrilaterals Test - 24

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Question 1
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Find $$\angle ABC$$.

A
$$92^o$$

B
$$104^o$$

C
$$120^o$$

D
$$106^o$$

Solution

We know, the sum of interior angles of a quadrilateral $$=360^o$$.
Given, the angles are:
$$ A=48^o+x,\quad B=4x,\quad C=3x\quad \& \quad D=4x.$$
Then,
$$48^o+x+4x+3x+4x=360^o\\ \Rightarrow 12x+48^o=360^o\\ \Rightarrow 12x=360^o-48^o\\ \Rightarrow 12x=312^o\\ \Rightarrow x=\dfrac { 312^o }{ 12 } =26^o.$$

$$\therefore \angle ABC=4x=4\times 26^o={ 104 }^{ o }.$$
Therefore, option $$B$$ is correct.
Question 2
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Given, in quadrilateral $$ABCD$$, $$\angle C = 64^{\circ}, \angle D = \angle C - 8^{\circ}, \angle A = 5(a + 2)^{\circ}$$ and $$\angle B = 2(2a+7)^{\circ}$$. Calculate $$\angle A$$.

A
$$120^o$$

B
$$130^o$$

C
$$140^o$$

D
$$150^o$$

Solution

Given$$\angle C={ 64 }^{ o }, \angle D= \angle C-{ 8 }^{ o }=64^o-8^o={ 56 }^{ o }, \angle A={ 5\left( a+2 \right) }^{ o }$$ and $$ \angle B={ 2\left( 2a+7 \right) }^{ o }.$$

We know, by angle sum property, the sum of angles of a quadrilateral is $$360^o$$.

Then, $$\angle A+\angle B+\angle C+\angle D=360^o$$
$$ \Rightarrow { 5\left( a+2 \right) }+{ 2\left( 2a+7 \right) }+64^o+56^o=360^o$$
$$ \Rightarrow 5a+10^o+4a+14^o+120^o=360^o$$
$$ \Rightarrow 9a=360^o-144^o$$
$$ \Rightarrow 9a=216^o$$
$$ \Rightarrow a=\dfrac { 216^o }{ 9 } =24^o$$.

Therefore,
$$\angle A={ 5\left( a+2 \right) }^{ o }$$ $$={ 5\left( 24^o+2 \right) }^{ o }$$ $$={ 5\times 26^o }=130^o$$.
Question 3
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Sum of all angles of a square is

A
$$90^o$$

B
$$240^o$$

C
$$120^o$$

D
$$360^o$$

Solution

$$\Rightarrow$$ $$\square ABCD$$ is a square.
$$\Rightarrow$$ All four sides and angles of square are equal.
$$\Rightarrow$$ All four angles are right angle in square.
$$\Rightarrow$$ Sum of all angles of a square $$ABCD=\angle A+\angle B+\angle C+\angle D$$
$$=90^o+90^o+90^o+90^o$$
$$=360^o$$
$$\therefore$$ Sum of angles of square is $$360^o$$
Question 4
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The angles of a quadrilateral are in the ratio 3 : 2 : 4 : 1. Find the angles. Assign a special name to the quadrilateral.

A
$$108^{\circ}, \, 82^{\circ},\, 144^{\circ},\, 36^{\circ}$$, Trapezium

B
$$108^{\circ}, \, 92^{\circ},\, 144^{\circ}, \, 36^{\circ}$$,Trapezium

C
$$108^{\circ}, \, 72^{\circ},\, 144^{\circ}, \, 36^{\circ}$$,Trapezium

D
$$108^{\circ}, \, 142^{\circ},\, 144^{\circ}, \, 36^{\circ}$$, Trapezium

Solution

$$The\quad ratio\quad of\quad angles\quad of\quad quadilateral=3:2:4:1\\ Let\quad the\quad angles\quad of\quad quadilateral\quad \\ =3x+2x+4x+x=360\\ 10x=360,x=36\\ \therefore \quad angles\quad of\quad quadilateral\quad are\\ 3x=3\times 36=108\\ 2x=2\times 36=72\\ 4x=4\times 36=144\\ In\quad the\quad adjoining\quad figure\\ \angle A+\angle B=108+72=180\\ ie\quad the\quad sum\quad of\quad interior\quad angles\quad on\quad the\quad same\quad side\quad of\quad transversal\\ AB=180\\ \therefore AD\parallel BC,\\ Hence\quad quadilateral\quad ABCD\quad is\quad a\quad trapezium$$
Question 5
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The angles $$P, Q, R$$ and $$S$$ of $$ PQRS$$ are in the ratio $$1:3:7:9$$, find the measure of each angle.

A
$$19^{\circ}, 57^{\circ}, 136^{\circ}, 171^{\circ}$$

B
$$18^{\circ}, 54^{\circ}, 126^{\circ}, 162^{\circ}$$

C
$$17^{\circ}, 66^{\circ}, 116^{\circ}, 152^{\circ}$$

D
$$12^{\circ}, 36^{\circ}, 108^{\circ}, 120^{\circ}$$

Solution

Given ratio of angles of quadrilateral $$PQRS$$ is $$1:3:7:9$$
Let the angles of quadrilateral $$PQRS$$ be $$x,3x,7x,9x$$, respectively.

We know, by angle sum property, the sum of angles of a quadrilateral is $$360^o$$.
$$\Rightarrow \angle P+\angle Q+\angle R+\angle S=360^o$$
$$\Rightarrow x+3x+7x+9x=360^o$$
$$\Rightarrow 20x=360^o$$
$$\Rightarrow x=\dfrac{360^o}{20}$$
$$\therefore x=18^o$$.

$$\therefore \angle P =x= 18^o$$,
$$ \angle Q =3x=3\times 18^o =54^o$$,
$$ \angle R =7x=7\times 18^o =126^o$$
and $$ \angle S =9x=9 \times 18^o =162^o$$.

Hence, option $$B$$ is correct.
Question 6
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The angles of a quadrilaterals are in the ratio $$3 : 5 : 9 : 13$$. All the angles of the quadrilaterals are:

A
$$36^{\circ}, 60^{\circ}, 108^{\circ}, 156^{\circ}$$

B
$$30^{\circ}, 50^{\circ}, 90^{\circ}, 130^{\circ}$$

C
$$39^{\circ}, 65^{\circ}, 117^{\circ}, 169^{\circ}$$

D
$$30^{\circ}, 65^{\circ}, 100^{\circ}, 150^{\circ}$$

Solution

Given ratio of angles of quadrilateral $$PQRS$$ is $$3:5:9:13$$
Let the angles of quadrilateral $$PQRS$$ be $$3x,5x,9x,13x$$, respectively.

We know, by angle sum property, the sum of angles of a quadrilateral is $$360^o$$.
$$\Rightarrow \angle P+\angle Q+\angle R+\angle S=360^o$$
$$\Rightarrow 3x+5x+9x+13x=360^o$$
$$\Rightarrow 30x=360^o$$
$$\Rightarrow x=\dfrac{360^o}{30}$$
$$\therefore x=12^o$$.

$$\therefore \angle P =3x= 3\times 12^o=36^o$$,
$$ \angle Q =5x=5\times 12^o =60^o$$,
$$ \angle R =9x=9\times 12^o =108^o$$
and $$ \angle S =13x=13 \times 12^o =156^o$$.

Hence, option $$A$$ is correct.
Question 7
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In the figure, $$LM = LN,\ \angle PLN = 110^{\circ}$$. Then,the measure of $$\angle MLN$$ in degrees is____.

A
$$110^{\circ}$$

B
$$40^{\circ}$$

C
$$10^{\circ}$$

D
$$140^{\circ}$$

Solution

In quadrilateral $$PLQN$$,
$$\angle P + \angle Q + \angle PLN + \angle QNL = 360^o $$ ...(Sum of angles of quadrilateral)
$$90^o + 90^o + 110^o + \angle QNL = 360^o$$
$$\angle QNL = 70^{\circ}$$.

In $$\triangle LMN$$,
$$\angle QNL=\angle MNL=70^o$$
$$\angle LMN = \angle MNL$$ ...(Isosceles triangle property)
$$\therefore \angle MNL= \angle LMN = 70^{\circ}$$.

We know, sum of angles of the $$\triangle LMN$$,
$$\angle MLN + \angle LNM + \angle LMN = 180^o $$
$$\angle MLN + 70^o + 70^o = 180^o $$
$$\angle MLN = 180^o - 140^o $$
$$\angle MLN = 40^{\circ}$$.

Therefore, option $$B$$ is correct.
Question 8
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In the above figure, $$ABCD$$ is a parallelogram, $$AL \bot CD$$ and $$AM \bot BC$$. If $$AB = 12\space cm, AD = 8\space cm$$ and $$AL = 6\space cm$$, then $$AM = $$

A
$$15\space cm$$

B
$$9\space cm$$

C
$$10\space cm$$

D
None of these

Solution

$$ABCD$$ is a parallelogram, where $$AL\perp CD$$ and $$AM\perp BC$$
$$AB=12\,cm,\,AD=8\,cm$$ and $$AL=6\,cm$$
We know, opposite sides of a parallelogram are equal.
$$AB=CD$$ and $$AD=BC$$
$$\therefore$$ $$CD=12\,cm$$ and $$BC=8\,cm$$
$$\Rightarrow$$ Area of a parallelogram $$ABCD=AL\times CD$$
$$=6\times 12$$
$$=72\,cm^2$$ ------ ( 1 )
$$\Rightarrow$$ Area of a parallelogram $$ABCD=AM\times BC$$
$$=AM\times 8$$ ------- ( 2 )
Comparing ( 1 ) and ( 2 ), we get
$$\Rightarrow$$ $$AM\times 8=72$$
$$\Rightarrow$$ $$AM=9\,cm$$
Question 9
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Three angles of a quadrilateral are respectively equal to $$110^o$$, $$50^o$$ and $$40^o$$. Find the fourth angle.

A
$$160^o$$

B
$$140^o$$

C
$$120^o$$

D
$$100^o$$

Solution

Sum of all angles of a quadrilateral is $$360^0$$
Then fourth angle will be = $$360^0 - ( 110^0 + 50^0 + 40^0 )$$
= $$160^0$$

Hence, the fourth angle is $$160^0$$
Question 10
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Use the information given in figure to choose the correct value of $$x$$.

A
$$x = 86^{o}$$

B
$$x = 60^{o}$$

C
$$x = 68^{o}$$

D
$$x = 107^{o}$$

Solution

Since, $$EAB$$ is a straight line
$$\therefore \angle DAE + \angle DAB = 180^{\circ}$$ ....[Linear pair]
$$\Rightarrow 73^{\circ} + \angle DAB = 180^{\circ}$$.

Since, by angle sum property, the sum of the angles of quadrilateral $$ABCD$$ is $$360^o$$.
$$\angle ADC + \angle DCB + \angle CBA + \angle BAD = 360^o$$
$$\therefore 107^{\circ} + 105^{\circ} +x+80^{\circ} = 360^{\circ}$$
$$\Rightarrow 292^{\circ} + x = 360^{\circ}$$
$$ x = 360^{\circ} - 292^{\circ} = 68^{\circ}$$.

Therefore, option $$C$$ is correct.