Quadrilaterals Test

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Quadrilaterals Test - 23

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Weekly Quiz Competition

Question 1
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Two angles of a quadrilateral are $$68^{\circ}$$ and $$76^{\circ}$$. If the other two angles are in the ratio $$5 : 7$$, find the measure of largest of them.

A
$$126^{\circ}$$

B
$$90^{\circ}$$

C
$$63^{\circ}$$

D
$$100^{\circ}$$

Solution

We know, by angle sum property, the sum of interior angles of a quadrilateral=$${ 360 }^{ o }$$.
Given two angles are $${ 68 }^{ o }$$ & $${ 76 }^{ o }$$ and the other two angles are in the ratio of $$5:7$$.
Let the other two angles be $$5x^{ o }$$ & $$7x^{ o }$$.
Then, $$5x^{ o }+7{ x }^{ o }+68^o+76^o=360^o$$
$$ \Rightarrow 12x+144^o=360^o$$
$$ \Rightarrow 12x=360^o-144^o=216^o$$
$$\Rightarrow x=\dfrac { 216^o }{ 12 } =18^o$$.

So unknown angles are $$5x=5\times 18^o={ 90 }^{ o }$$
and $$7x=7\times 18^o={ 126 }^{ o }$$.

Therefore, the largest angle is $$126^o$$.
Hence, option $$A$$ is correct.
Question 2
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$$ P,Q, R $$ and $$ S $$ are the mid-points of sides $$ AB, BC, CD $$ and $$ DA $$ respectively of rhombus $$ ABCD $$. Quadrilateral $$ PQRS $$ is a rectangle. Under what condition will $$ PQRS $$ be a square ?

A
When $$ ABCD $$ is a square

B
When $$ ABCD $$ is a parallelogram

C
When $$ ABCD $$ is a rectangle

D
When $$ ABCD $$ is a square or a rectangle

Solution

Given: ABCD is a rhombus. P, Q, R, S are mid points of AB, BC, CD, DA respectively.
Join: AC and BD

In $$\triangle ABC$$
P is mid point of AB and Q is mid point of AC.
Thus, by mid point theorem, $$PQ \parallel AC$$ and $$PQ = \frac{1}{2} AC$$
Similarly, In $$\triangle ACD$$,
S is mid point of AD and R is mid point of CD
Thus, by mid point theorem, $$SR \parallel AC$$ and $$SR = \frac{1}{2} AC$$
Hence, $$PQ \parallel SR$$ and $$PQ = SR$$

Similarly, $$PS = QR$$ and $$PS \parallel QR$$
Thus, the opposite sides of PQRS are equal and parallel.
We know the diagonals of a rhombus bisect each other at right angles.
Now, since $$AC \perp BD$$ thus, $$PS \perp PQ$$ (Angle between two lines is same as the angle between their corresponding parallel lines)

Now, the opposite sides of PQRS are equal and parallel and the sides meet each other at right angles. Hence, PQRS is a rectangle.

If PQRS had to be a square, the diagonals must be equal and bisect at right angles. It is possible only if ABCD is a square.
Question 3
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In triangle $$ ABC $$, $$ M $$ is mid-point of $$ AB $$ and a straight line through $$ M $$ and parallel to $$ BC $$ cuts $$ AC $$ in $$ N $$. Find the lenghts of $$ AN $$ and $$ MN $$ if $$ BC= 7 $$ cm and $$ AC= 5 $$ cm.

A
$$ AN= 2.5 $$ cm and $$ MN= 3.5 $$ cm

B
$$ AN= 1.5 $$ cm and $$ MN= 3.5 $$ cm

C
$$ AN= 2.5 $$ cm and $$ MN= 4.5 $$ cm

D
none of the above

Solution

M is the mid point of AB and MN II BC. Thus, N is the mid point of AC and

$$MN = \dfrac{1}{2} BC$$ (Mid point theorem)

$$MN = \dfrac{1}{2} (7) $$

$$MN = 3.5 cm$$

Also, $$AN = \dfrac{1}{2} AC$$

$$AN = \dfrac{1}{2} (5)$$

$$AN = 2.5 cm$$
Question 4
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Two angles of a quadrilateral are $$89^{\circ}$$ and $$113^{\circ}$$. If the other two angles are equal, find the equal angles.

A
$$69^{\circ}$$

B
$$89^{\circ}$$

C
$$158^{\circ}$$

D
$$79^{\circ}$$

Solution

$$\textbf{Step-1: Apply the concept of interior angles of a quadrilateral.}$$

$$\text{We know, by angle sum property, the sum of interior angles of a quadrilateral =}$$ $${ 360 }^{ o }$$.

$$\text{Given two angles are}$$ $${ 89 }^{ o }$$ & $${ 113 }^{ o }$$ $$\text{and the other two angles are equal.}$$

$$\text{Let the unknown angle be}$$ $${ x }^{ o }$$.

$$\text{So,}$$ $${ x }+{ x }+89^o+113^o=360^o$$

$$\Rightarrow 2x+202^o=360^o\\ \Rightarrow 2x=360^o-202^o=158^o\\ \Rightarrow x=\dfrac { 158^o }{ 2 } ={ 79^o }.$$

$$\text{So the other angles are}$$ $${ 79 }^{ o } $$ $$\text{each.}$$

$$\textbf{Hence, option D is correct.}$$
Question 5
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In the given figure:
$$\angle b = 2a + 15^{\circ}$$ and $$\angle c = 3a + 5^{\circ}$$, find the value of $$\angle b $$.

A
$$100^o$$

B
$$105^o$$

C
$$95^o$$

D
$$92^o$$

Solution

Given the angles are:
$${ a }, { 70 }^{ o }, b=2a+15^o$$ and $$ c=3a+5^o$$.

We know, by angle sum property, the sum of angles of a quadrilateral is $$360^o$$.
$$a+b+c+70^o=360^o$$
$$ \Rightarrow a+2a+15^o+3a+5^o+70^o=360^o$$
$$ \Rightarrow 6a+90^o=360^o$$
$$ \Rightarrow 6a=270^o$$
$$ \Rightarrow a={ 45 }^{ o }$$.

Therefore,
$$ \angle b=2a+15^o$$
$$=2(45^o)+15^o$$
$$=90^o+15^o$$
$$={ 105 }^{ o}$$

Hence, option $$B$$ is correct.
Question 6
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Three angles of a quadrilateral are equal. If the fourth angle is $$69^{\circ}$$, find the measure of equal angles.

A
$$86^o$$

B
$$97^o$$

C
$$68^o$$

D
None of the above

Solution

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

Given, three angles are equal. Let the measure of each of these angles be $$x$$.
Also, fourth angle $$=69^o$$.

Then, $$x+x+x+69^o=360^o$$

$$\Rightarrow 3x+69^o=360^o$$.

$$\Rightarrow$$ $$3x=360^o - 69^o $$

$$\Rightarrow$$ $$3x=291^o$$
$$\Rightarrow x=\dfrac{291^o}{3}\\=97^o$$.

Therefore, the equal angles are each $$=97^o$$.

Hence, option $$B$$ is correct.
Question 7
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Angles of a quadrilateral are $$(4x)^{\circ},\ 5(x + 2)^{\circ},\ (7x - 20)^{\circ}$$ and $$6(x + 3)^{\circ}$$. Find the value of $$x$$ (in degrees).

A
$$14$$

B
$$15$$

C
$$16$$

D
$$18$$

Solution

We know, the sum of interior angles of a quadrilateral $$={ 360 }^{ o }$$.
Given the angles are:
$$4x, 5(x+2), (7x-20)$$ and $$6(x+3).$$
Then,
$$ 4x+5\left( x+2 \right) +\left( 7x-20 \right) +6\left( x+3 \right) =360$$
$$ \Rightarrow 4x+5x+10+7x-20+6x+18=360\\ \Rightarrow 22x+8=360\\ \Rightarrow 22x=360-8=352\\ \Rightarrow x=\dfrac { 352 }{ 22 } =\dfrac { 176 }{ 11 } =16$$
Therefore, $$ x=16$$.
Hence, option $$C$$ is correct.
Question 8
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From the following figure, find $$x$$.

A
$$18^o$$

B
$$26^o$$

C
$$24^o$$

D
$$36^o$$

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, option $$B$$ is correct.
Question 9
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Find $$\angle ACD$$.

A
$$56^o$$

B
$$28^o$$

C
$$45^o$$

D
$$36^o$$

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,
$$48^o+x+4x+3x+4x=360^o\\ \Rightarrow 12x+48^o=360^o\\ \Rightarrow 12x=360^o-48^o\\ \Rightarrow 12x=312^o\\ \Rightarrow x=\dfrac { 312^o }{ 12 } =26^o.$$

Then,
$$ \angle D=4x=4\times 26^o={ 104 }^{ o }$$.

Now in $$\triangle ACD$$,
$$ \angle ACD+\angle DAC+\angle D=180^o$$ ....[Angle sum property]
$$ \Rightarrow \angle ACD+48^o+104^o=180^o$$
$$ \Rightarrow \angle ACD=180^o-152^o$$
$$\Rightarrow \angle ACD={ 28^o }$$.

Therefore, option $$B$$ is correct.
Question 10
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Angles of a quadrilateral are $$(4x)^{\circ}, 5(x + 2)^{\circ}, (7x - 20)^{\circ}$$ and $$6(x + 3)^{\circ}$$. Find the smallest angle of the quadrilateral.

A
$$90^{\circ}$$

B
$$64^{\circ}$$

C
$$114^{\circ}$$

D
$$78^{\circ}$$

Solution

We know, the sum of interior angles of a quadrilateral $$={ 360 }^{ o }$$.
Given the angles are:
$$4x, 5(x+2), (7x-20)$$ and $$6(x+3).$$
Then,
$$ 4x+5\left( x+2 \right) +\left( 7x-20 \right) +6\left( x+3 \right) =360$$
$$ \Rightarrow 4x+5x+10+7x-20+6x+18=360\\ \Rightarrow 22x+8=360\\ \Rightarrow 22x=360-8=352\\ \Rightarrow x=\dfrac { 352 }{ 22 } =\dfrac { 176 }{ 11 } =16$$
Therefore, $$ x=16$$.

Each angle is as follows:
$$4x=4\times 16={ 64 }^{ o } $$,
$$ 5(x+2)=5\left( 16+2 \right) =5\times 18={ 90 }^{ o },$$
$$ (7x-20)=7\times 16-20=112-20=92^{ o } $$
and $$ 6(x+3)=6\left( 16+3 \right) =6\times 19=114^{ o }$$.

Therefore, the smallest angle is $$64^o$$.
Hence, option $$B$$ is correct.

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Re-Attempt Weekly Quiz Competition

Quadrilaterals Test - 23

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Quadrilaterals Test - 23

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SHARING IS CARING

Question 1

$$126^{\circ}$$

$$90^{\circ}$$

$$63^{\circ}$$

$$100^{\circ}$$

Solution

Question 2

When $$ ABCD $$ is a square

When $$ ABCD $$ is a parallelogram

When $$ ABCD $$ is a rectangle

When $$ ABCD $$ is a square or a rectangle

Solution

Question 3

$$ AN= 2.5 $$ cm and $$ MN= 3.5 $$ cm

$$ AN= 1.5 $$ cm and $$ MN= 3.5 $$ cm

$$ AN= 2.5 $$ cm and $$ MN= 4.5 $$ cm

none of the above

Solution

Question 4

$$69^{\circ}$$

$$89^{\circ}$$

$$158^{\circ}$$

$$79^{\circ}$$

Solution

Question 5

$$100^o$$

$$105^o$$

$$95^o$$

$$92^o$$

Solution

Question 6

$$86^o$$

$$97^o$$

$$68^o$$

None of the above

Solution

Question 7

$$14$$

$$15$$

$$16$$

$$18$$

Solution

Question 8

$$18^o$$

$$26^o$$

$$24^o$$

$$36^o$$

Solution

Question 9

$$56^o$$

$$28^o$$

$$45^o$$

$$36^o$$

Solution

Question 10

$$90^{\circ}$$

$$64^{\circ}$$

$$114^{\circ}$$

$$78^{\circ}$$

Solution

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