Quadrilaterals Test

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Quadrilaterals Test - 27

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Question 1
1 / -0

The diagram shows the construction of

A
a trapezium.

B
a parallelogram.

C
a rectangle.

D
a kite.

Solution

From the given signs in the construction we can comprehend that
the diagonals of the given quadrilateral bisect each other.
No other clue is given.
So, we can conclude that it is a construction of a parallelogram.
Ans- Option B.
Question 2
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If angles $$\angle A,\angle B,\angle C,\angle D$$ of a quadrilateral $$ABCD$$ are taken in order, then they are found to be in ratio $$3:7:6:4$$, then $$ABCD$$ is a:

A
rhombus

B
parallelogram

C
trapezium

D
kite

Solution

Given ratio of angles of quadrilateral $$ABCD$$ is $$3:7:6:4$$

Let the angles of quadrilateral $$ABCD$$ be $$3x,7x,6x,4x$$, respectively.

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

$$\Rightarrow 3x+7x+6x+4x=360^o$$

$$\Rightarrow 20x=360^o$$

$$\therefore x=18^o$$.

$$\therefore \angle A =3x=3 \times 18^o =54^o$$,

$$ \angle B =7x=7\times 18^o =126^o$$,

$$ \angle C =6x=6 \times 18^o =108^o$$

and $$ \angle D =4x=4 \times 18^o =72^o$$.

We know that in a trapezium we have two parallel lines which form two pairs of co-interiors angles, which are Supplementary.

Clearly $$\angle A+ \angle B=180^o$$ and $$\angle C+\angle D=180^o$$
Therefore, the given quadrilateral is a trapezium.

Hence, option $$C$$ is correct.
Question 3
1 / -0

The diagonals of a quadrilateral PQRS bisect each other at right angles. If $$PQ=5.5\:cm$$, then the perimeter of PQRS is

A
$$11$$ cm

B
$$22$$ cm

C
$$16.5$$ cm

D
$$55(1+\sqrt{2})$$ cm

Solution

Given that the diagonals of a quadrilateral bisect each other at right angles.
Since, the diagonal bisect each other, the quadrilateral is a parallelogram.
The diagonals of the quadrilateral bisect at right angles, hence it must be a rhombous in which all sides are equal.
Then, perimeter
$$=4S\\ = 4 \times 5.5\\ = 22\ cm$$
Question 4
1 / -0

The sum of the angles in a quadrilateral is equal to .......... .

A
$$2$$ right angles

B
$$3$$ right angles

C
$$4$$ right angles

D
$$360$$ right angles

Solution

Consider a quadrilateral $$ABCD$$.

Then, $$\angle A+\angle B+\angle C+\angle D=360^{\circ}$$.

To prove this, we join $$A$$ and $$C$$, i.e. we draw the diagonal $$AC$$.

In $$\triangle ABC$$,

$$\angle CAB+\angle ABC+\angle BCA=180^{\circ}$$ [Sum of all angles of a triangle is $$180^{\circ}$$].....$$(1)$$.

In $$\triangle ACD$$,

$$\angle CAD+\angle ADC+\angle DCA=180^{\circ}$$ [Sum of all angle of a triangle is $$180^{\circ}$$]....$$(2)$$.

Adding $$(1)$$ and $$(2)$$, we get,

$$\left(\angle CAB+\angle ABC+\angle BCA \right)+ \left(\angle CAD+\angle ADC+\angle DCA \right)=180^{\circ}+180^{\circ}$$

$$\implies$$ $$\angle ABC+\angle ADC+(CAB+CAD)+(BCA+DCA)=360^{\circ}$$

$$\implies$$ $$\angle ABC+\angle ADC+\angle BAD+\angle BCD=360^{\circ}$$

$$\implies$$ $$\therefore \angle A+\angle B+\angle C+\angle D=360^{\circ}$$.

That is, the sum of all angles of a quadrilateral is $$360^o=4\times90^o$$, i.e. $$4$$ right angles.

Therefore, option $$C$$ is correct.
Question 5
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In a quadrilateral $$PQRS$$, if $$\angle P\, =\, \angle R\, =\, 100^{\circ}$$ and $$\angle S\, =\, 75^{\circ}$$, then $$\angle Q = $$ ......... .

A
$$50^{\circ}$$

B
$$85^{\circ}$$

C
$$120^{\circ}$$

D
$$360^{\circ}$$

Solution

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

The given angles are $$\angle P=100^o, \angle R=100^o, \angle S=75^o, \angle Q$$.

Then, $$\angle P+\angle Q+\angle R+\angle S=360^o$$

$$\implies$$ $$100^o+100^o+75^o+\angle Q=360^o$$.

$$\Rightarrow$$ $$\angle Q=360^o - (100^o + 100^o + 75^o )$$

$$\Rightarrow$$ $$\angle Q=360^o - 275^o =85^{\circ}$$.

Therefore, the unknown angle is $$\angle Q=85^o$$.

Hence, option $$B$$ is correct.
Question 6
1 / -0

Find the value of $$x$$.

A
$$60^o$$

B
$$120^o$$

C
$$90^o$$

D
$$140^o$$

Solution

Clearly, ext $$ \angle A=90^o $$ (given)
$$ \angle A=180^o-90^o=90^o $$ .....[Linear pair].
We know, by angle sum property, the sum of the measures of interior angles of a quadrilateral is $$ 360^o$$,
$$\implies$$ $$ 90^o+60^o+70^o+x=360^o $$
$$\implies$$ $$ 220^o+ x= 360^o$$
$$\implies$$ $$ x=360^o-220^o$$ $$= 140^o $$.
Therefore, option $$D$$ is correct.
Question 7
1 / -0

Find the value of $$x$$.

A
$$60^o$$

B
$$40^o$$

C
$$50^o$$

D
$$90^o$$

Solution

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

The given angles are $$\angle A=130^o, \angle B=120^o, \angle C=x, \angle D=50^o$$.

Then, $$\angle A+\angle B+\angle C+\angle D=360^o$$

$$\implies$$ $$130^o+120^o+x+50^o=360^o$$.

$$\implies$$ $$x=360^o - (130^o + 120^o + 50^o )$$

$$\implies$$ $$x=360^o - 300^o =60^{\circ}$$.

Therefore, the unknown angle is $$x=60^o$$.

Hence, option $$A$$ is correct.
Question 8
1 / -0

$$P$$ is the mid-point of side $$AB$$ to a parallelogram $$ABCD$$. A line through $$B$$ parallel to $$PD$$ meets $$DC$$ at $$Q$$ and $$AD$$ produced at $$R$$. Then $$BR$$ is equal to ______.

A
$$BQ$$

B
$$\dfrac {1}{2}BQ$$

C
$$2BQ$$

D
None of these

Solution

In $$\triangle ABR$$, $$ P $$ is the mid-point of $$AB$$ and $$DP || BR $$
$$\therefore D$$ is the mid-point of $$AR$$ .....(Converse of mid-point theorem).
Then, $$AR=2AD $$
$$\implies$$ $$AR=2BC$$ (Opposite sides of parallelogram are equal, i.e. $$AD=BC$$)
$$\implies$$ $$AD = BC$$.

Also, $$D$$ is the mid-point of $$AR$$ and $$DQ||AB$$ ....(Since, $$ABCD$$ is parallelogram)
Then, $$Q$$ is the mid-point of $$DC$$ ....(Converse of mid-point theorem)
$$\therefore BR = 2 BQ $$.

Hence, option $$C$$ is correct.
Question 9
1 / -0

Suppose the triangle ABC has an obtuse angle at C and let D be the midpoint of side AC. Suppose E is on BC such that the segment DE is parallel to AB. Consider the following three statements
i) E is the midpoint of BC
ii) The length of DE is half the length of AB
iii) DE bisects the altitude from C to AB

A
only (i) is true

B
only (i) and (ii) are true

C
only (i) and (iii) are true

D
all three are true
Question 10
1 / -0

The measures of angles of a quadrilateral in degrees are $$x, 3x - 40 , 2x$$ and $$4x +20$$. Find the measure of each angle.

A
The measures of the angles are $$55^0, 74^0, 62^0$$ and $$100^0$$

B
The measures of the angles are $$28^0, 66^0, 102^0$$ and $$165^0$$

C
The measures of the angles are $$38^0, 74^0, 76^0$$ and $$172^0$$

D
The measures of the angles are $$50^0, 72^0, 80^0$$ and $$150^0$$

Solution

Given, $$x , (3x-40) , 2x , (4x+20) $$.
We know, by angle sum property, the sum of the angles of a quadrilateral is $$360^0$$.
Therefore, $$x + (3x-40) + 2x+ (4x+20) = 360^o$$
$$\Rightarrow 10x - 20^o = 360^o$$
$$\Rightarrow 10x = 360^o + 20^o =380^o$$
$$\Rightarrow x =\displaystyle \cfrac{380^o}{10}= 38^o$$.

Now, $$x = 38^o$$,
$$2x = 2 \times 38^o = 76^o$$,
$$3x - 40^o = 3 \times 38^o - 40^o =74^o$$
and $$4x + 20^o = 4 \times 38^o + 20^o = 172^o$$.

Thus, the measures of the angles are $$38^o, 74^o, 76^o$$ and $$172^o$$.

Hence, option $$C$$ is correct.