Understanding Quadrilaterals Test

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Understanding Quadrilaterals Test - 26

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Question 1
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In the adjoining figure, $$ABCD$$ is a parallelogram in which $$\angle BAO=35^{o}$$ and $$\angle COD=95^{o}, \angle DAC=50^{o}$$
So, the measure of $$\angle ABO$$ is_____________

A
$$35^{o}$$

B
$$70^{o}$$

C
$$50^{o}$$

D
$$60^{o}$$

Solution
Question 2
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If the quadrilateral $$ABCD$$ is a parallelogram. What is the value of $$x$$?

A
$$45^{\circ}$$

B
$$30^{\circ}$$

C
$$36^{\circ}$$

D
$$37^{\circ}$$

Solution

In parallelogram $$ABCD,$$

We know that the opposite angles of a parallelogram are equal.

Hence $$\angle A=\angle C $$

$$\Rightarrow 3{y} = 3x + {3}$$ $$...(1)$$

and $$\angle B=\angle D$$

$$\therefore\ \angle D = 2y - {5} $$ $$...(2)$$

We know that the sum of angles of a parallelogram is $$360^\circ$$

$$\Rightarrow (3y)+(3x+3)+(2y-{5})+(2y-{5})={360}$$

$$\Rightarrow 3y+3y+4y-10={360}$$

$$\Rightarrow 10y-{10}={360}$$

$$\Rightarrow 10y={360}+{10}$$

$$\Rightarrow 10y={370}$$

$$\Rightarrow y=\dfrac{370}{10} = {37^\circ}$$

Substitute $$y=37^\circ$$ in $$(1)$$

$$\Rightarrow 3\times {(37^\circ)} = 3x + {3^\circ}$$

$$\Rightarrow 111=3x+3$$

$$\Rightarrow 3x=111-3$$

$$\Rightarrow 3x=108$$

$$\Rightarrow x = \dfrac{{108}}{3} = 36^\circ$$
Question 3
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In parallelogram $$ABCD$$, the bisectors of $$\angle A$$ and $$\angle B$$ intersect at $$M$$.If $$\angle A=80^{o}$$, then $$\angle AMB=$$.........

A
$$40^{o}$$

B
$$50^{o}$$

C
$$80^{o}$$

D
$$90^{o}$$

Solution

Given in parallelogram $$ABCD, \; AM$$ and $$BM$$ are bisectors of $$\angle{A}$$ and $$\angle{B}$$ respectively.
As we know that sum of adjacent angles in a parallelogram is $$180°$$.
$$\therefore \angle{A} + \angle{B} = 180°$$
$$\Rightarrow 2 \angle{BAM} + 2 \angle{ABM} = 180°$$
$$\Rightarrow \angle{BAM} + \angle{ABM} = \cfrac{180°}{2} = 90° ..... \left( 1 \right)$$
Now, in $$\triangle{AMB}$$,
$$\angle{BAM} + \angle{AMB} + \angle{ABM} = 180°$$
$$\Rightarrow \angle{AMB} + 90° = 180°$$
$$\Rightarrow \angle{AMB} = 180° - 90° = 90°$$
Question 4
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Measures of opposite angles of parallelogram are $$(3x-2) ^{\circ} $$ and $$(50-x) ^{\circ} $$. Find the measure of its other angles.

A
$$143$$

B
$$153$$

C
$$163$$

D
$$173$$

Solution

Here, $$ABCD$$ is a parallelogram
Let $$\angle A=(3x-2)^o$$ and $$C=(50-x)^o$$
We know that in parallelogram opposite angles are equal.
$$\therefore$$ $$\angle A=\angle C$$
$$\Rightarrow$$ $$3x-2=50-x$$
$$\Rightarrow$$ $$4x=52$$
$$\Rightarrow$$ $$x=13$$
$$\Rightarrow$$ $$\angle A=(3x-2)^o=(3\times 13-2)^o=37^o$$
In parallelogram sum of adjacent angles are supplementary.
$$\therefore$$ $$\angle A+\angle B=180^o$$
$$\Rightarrow$$ $$37^o+\angle B=180^o$$
$$\therefore$$ $$\angle B=143^o$$
Hence other two equal opposite angles are $$143^o$$
Question 5
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In a parallelogram, the sum of adjacent angles is:

A
$$120^{\circ}$$

B
$$150^{\circ}$$

C
$$90^{\circ}$$

D
$$180^{\circ}$$
Question 6
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The diagonals do not necessarily intersect at right angles in a

A
parallelogram

B
rectangle

C
rhombus

D
kite

Solution

The diagonals do not necessarily intersect at right angles in a parallelogram. But opposite sides and opposite angles should be equal.
Question 7
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Two adjacent angles of a parallelogram are $$\left(2x+25\right)^{o}$$ and $$\left(3x-5\right)^{o}.$$ The value of $$x$$ is

A
$$28$$

B
$$32$$

C
$$36$$

D
$$42$$

Solution

We know that sum of adjucent angled of a parallelogram is $$180^o$$
$$(2x+25)^o+(3x-5)^o=180^o$$
$$5x+20=180^o$$
$$5x=180^o - 20^o$$
$$5x=160^o$$
$$x=160^o/5=32^o$$
Question 8
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In a parallelogram PQRS , if $$ \angle\, P = 60^{\circ} $$, then other three angles are:

A
$$ 45^{\circ} , 135^{\circ} , 120^{\circ} $$

B
$$ 60^{\circ} , 120^{\circ} , 120^{\circ} $$

C
$$ 60^{\circ} , 135^{\circ} , 135^{\circ} $$

D
$$ 45^{\circ} , 135^{\circ} , 135^{\circ} $$

Solution

Given: $$ \angle\, P = 60^{\circ} $$
$$ \therefore \, \angle\, R = 60^{\circ} $$ $$\therefore $$ opposite angles are equals

Similarly,
$$\angle Q = \angle S$$

Now,
$$\angle P+\angle Q = 180^o$$ {supplementary angles because these interior angles lie on the same side of the transversal}

Therefore,
$$ \angle\, Q = \angle S = 180^o - 60^{\circ} $$
$$ \angle\, Q = \angle S = 120^o $$
Question 9
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If two adjacent angles of a parallelogram are in the ratio $$ 2 : 3 $$ , then the measure of angles are:

A
$$ 72^{\circ} , 108^{\circ} $$

B
$$ 36^{\circ} , 54^{\circ} $$

C
$$ 80^{\circ} , 120^{\circ} $$

D
$$ 96^{\circ} , 144^{\circ} $$

Solution

Let the angles be $$ 2x , 3x $$
$$ 2 x + 3 x = 180 $$
$$ \Rightarrow \, 5x = 180 $$
$$ \Rightarrow \, 5x = 180 $$
So , the angles are
$$ 2 \times 36 = 72 $$
$$ 3 \times 36 = 108 $$
Question 10
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The angle between the two altitudes of a parallelogram through the same vertex of an obtuse angle of the parallelogram is $$ 30^{\circ} $$. The measure of the obtuse angle is

A
$$ 100^{\circ} $$

B
$$ 150^{\circ} $$

C
$$ 105^{\circ} $$

D
$$ 120^{\circ} $$

Solution

ABCD is a parallelogram.
From the question it is given that, ∠EBF =$$ 30^o$$
Sum of interior angles of a quadrilateral = $$360^o$$
Then, ∠EBF + ∠BED + ∠EDF + ∠DFB = $$360^o$$
∠EDF = $$360^o – (90^o + 90^o + 30^o) $$
∠EDF = $$150^o $$ which is an obtuse angle.