Quadrilaterals Test

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Quadrilaterals Test - 25

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Question 1
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In a quadrilateral $$ABCD$$, if $$\angle$$ $$A = 60$$ $$^o$$ and $$\angle$$ $$B :$$ $$\angle$$ $$C :$$ $$\angle$$ $$D = 2 : 3 : 7$$, then $$\angle$$ $$D =$$ ?

A
$$175$$ $$^o$$

B
$$135$$ $$^o$$

C
$$150$$ $$^o$$

D
None of these

Solution

We know, by angle sum property, the sum of all angles of a quadrilateral is $$ 360^o $$.
Given, $$\angle B:\angle C:\angle D=2:3:7$$.
Let the $$ \angle B = 2x , \angle C = 3x , \angle D= 7x $$ and $$\angle A = 60^o $$.
$$\Rightarrow 2x + 3x + 7x + 60^o = 360^o $$
$$\Rightarrow 12x = 300^o $$
$$\Rightarrow x = 25^o $$.

$$\therefore \angle D= 7x = 7 \times 25^o = 175^o $$.

Hence, option $$A$$ is correct.
Question 2
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The angles of a quadrilateral are in the ratio $$3 : 5 : 9 : 13$$. Find all the angles of the quadrilateral in degrees.

A

$$36, 60, 108, 156$$

B
$$36, 70, 108, 156$$

C
$$36, 60, 118, 156$$

D
none of the above

Solution

Let the four angles of the quadrilateral be $$3x, 5x, 9x$$ and $$13x$$

$$ 3x + 5x + 9x + 13x = { 360 }^{ \circ } $$

[sum of all angles of the quadrilateral is $$ { 360 }^{ \circ }$$]

$$ 30x = { 360 }^{ \circ } $$

$$ x = { 12 }^{ \circ } $$

Hence, the angles of the quadrilateral are $$ 3\times{ 12 }^{ \circ }={ 36 }^{ \circ },5\times{ 12 }^{ \circ }={ 60 }^{ \circ },9\times{ 12 }^{ \circ }={ 108 }^{ \circ }$$ and $$13\times{ 12 }^{ \circ }={ 156 }^{ \circ }$$
Question 3
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If three angles of a quadrilateral are of magnitudes $$80^\circ,\ 95^\circ,\ 120^\circ,$$ then what is the measure of the fourth angle?

A
$$80$$ $$^o$$

B
$$65$$ $$^o$$

C
$$75$$ $$^o$$

D
$$70$$ $$^o$$

Solution

The given angles are, $$80^\circ,\ 95^\circ,\ 120^\circ$$.

Let the fourth angle be $$x^\circ$$
Angle sum property states that the sum of angles of a quadrilateral is $$360^\circ$$.

Then, $$80^\circ+95^\circ+120^\circ+x^\circ=360^\circ$$

$$\Rightarrow 360^\circ - (80^\circ + 95^\circ + 120^\circ ) =x^{\circ}$$

$$\Rightarrow$$ $$x^\circ=360^\circ - 295^\circ$$
$$ =65^{\circ}$$

Therefore, the fourth angle is $$x^\circ=65^\circ$$

Hence, option $$B$$ is correct.
Question 4
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In the given figure, $$AB=\,AD$$; $$CB=\,CD$$, $$\angle A={ 42 }^{ 0 }$$ and $$\angle C={ 108 }^{ 0 }$$, Find $$\angle ABC$$

A
$${ 105 }^{ 0 }$$

B
$${ 150 }^{ 0 }$$

C
$${ 75 }^{ 0 }$$

D
$${ 110 }^{ 0 }$$

Solution

Given: $$AB=\,AD$$; $$CB=\,CD$$, $$\angle A={ 42 }^{ 0 }$$ and $$\angle C={ 108 }^{ 0 }$$
Join AC.
Now, in $$\triangle ADC$$ and $$\triangle ABC$$
$$AD = AB$$ (Given)
$$BC = CD$$ (Given)
$$AC = AC$$ (Common)
Thus, $$\triangle ADC \cong \triangle ABC$$ (SSS rule)
Hence, $$\angle CAB = \angle CAD = \frac{1}{2} \angle DAB = 21$$ (By CPCT)
$$\angle ACB = \angle ACD = \frac{1}{2} \angle DCB = 54$$ (By CPCT)

Now, In $$\triangle ABC$$
$$\angle ABC + \angle CAB + \angle ACB = 180$$ (Angles sum property)
$$\angle ABC + 21 + 54 = 180$$
$$\angle ABC = 105^{\circ}$$
Question 5
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In the adjoining figure $$ABC$$ is a triangle in which $$D$$ and $$E$$ are the mid points $$AB$$ and $$AC$$ respectively if $$BC = 4.6\ \text{cm}$$ then $$DE =$$ _______

A
$$9.2\ \text{cm}$$

B
$$2.3\ \text{cm}$$

C
$$5.0\ \text{cm}$$

D
$$6.4\ \text{cm}$$

Solution

According to mid point theorem the line segment joining the mid-points of two sides of a triangle is parallel to the third side and equal to half the third side.
Here $$D$$ and $$E$$ are the mid points of side $$AB$$ aand $$AC$$
$$\Rightarrow DE||BC$$ and $$BC=2DE$$
$$BC=4.6\ \text{cm}$$
$$2DE=4.6\ \text{cm}$$
$$DE=2.3\ \text{cm}$$
So option $$B$$ is correct.
Question 6
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In a quadrilateral the angles are in the ratio $$3:4:5:6$$. Then the difference between the greatest and the smallest angle is?

A
$${108}^{o}$$

B
$${60}^{o}$$

C
$${80}^{o}$$

D
$${90}^{o}$$

Solution

Given ratio of angles of quadrilateral $$ABCD$$ is $$3:4:5:6$$

Let the angles of quadrilateral $$ABCD$$ be $$3x,4x,5x,6x$$ respectively.

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

$$\Rightarrow 3x+4x+5x+6x=360^o$$
$$\Rightarrow 18x=360^o$$
$$\therefore x=20^o$$.

$$\therefore \angle A =3x=3 \times 20^o =60^o$$,

$$ \angle B =4x=4\times 20^o =80^o$$,

$$ \angle C =5x=5 \times 20^o =100^o$$

and $$ \angle D =6x=6 \times 20^o =120^o$$.

$$\therefore$$ The greatest angle is $$=120^o$$
and the smallest angle is $$=60^o$$.

Therefore, the difference between the greatest and the smallest angle is $$=120^o-60^o=60^o$$.

Hence, option $$B$$ is correct.
Question 7
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Look at the pairs of shapes Which is a pair of rectangles?

A

B

C

D

Solution

Option B shows rectangle.
Hence, the answer is Option B.
Question 8
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Which of the following statement(s) is/are true?

A
In a trapezium the diagonals bisect each other

B
In a rectangle diagonals intersect at right angles

C
The diagonals of a rhombus are equal

D
None of these

Solution

No option is correct because
1. Diagonals of trapezium do not bisect each other
2. Diagonal of rectangle do not intersect each other at 90 degree.
3. Diagonals of rhombus are unequal.
So option D is correct answer
Question 9
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The sides of a quadrilateral are extended to make the angles as shown in the figure. Find the value of $$x.$$

A
$${80}^{o}$$

B
$$60^o$$

C
$$40^o$$

D
$$130^o$$

Solution

The sum of the exterior angles of a quadrilateral is equal to $${360}^{o}.$$ So,
$$\Rightarrow$$ $$x+{75}^{o}+{115}^{o}+{90}^{o}={360}^{o}$$
$$\Rightarrow$$ $$x+280^o={360}^{o}$$
$$\Rightarrow$$ $$x={360}^{o}-{280}^{o}$$
$$\Rightarrow x={80}^{o}$$
Question 10
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If the bisectors of the angles $$A$$ and $$B$$ of a quadrilateral $$ABCD$$ meet at $$O$$, then $$\angle AOB$$ is equal to:

A
$$\angle C+\angle D$$

B
$$\cfrac{1}{2}(\angle C+\angle D)$$

C
$$\cfrac{1}{2}\angle C+\cfrac{1}{3}\angle D$$

D
$$\cfrac{1}{3}\angle C+\cfrac{1}{2}\angle D$$

Solution

In quad $$ABCD$$

$$\angle A+\angle B+\angle C+\angle D={360}^{o}$$

$$\Rightarrow$$ $$\angle A+\angle B={360}^{o}-(\angle C+\angle D)$$

$$\Rightarrow$$ $$\cfrac{\angle A}{2}+\cfrac{\angle B}{2}={180}^{o}-\cfrac{1}{2}(\angle C+\angle D)$$

$$\Rightarrow$$ $$\angle OAB+\angle OBA={180}^{o}-\cfrac{1}{2}(\angle C+\angle D)$$

($$\therefore$$ $$OA$$ bisects $$\angle A$$ and $$OB$$ bisects $$\angle B$$)

In $$\triangle AOB$$, $$\angle AOB={180}^{o}-(\angle OAB+\angle OBA)$$

$$={180}^{o}-[({180}^{o}-\cfrac{1}{2}(\angle C+\angle D)]$$

$$=\cfrac{1}{2}(\angle C+\angle D)$$