Quadrilaterals Test

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Quadrilaterals Test - 32

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Weekly Quiz Competition

Question 1
1 / -0

If the angle of a quadrilateral are $${ x }^{o },(x-{ 10) }^{ o },(x+{ 30 })^{o }$$ and $$2{ x }^{ o }$$, then the greatest angle is:

A
$${ 136 }^{ 0 }$$

B
$${ 180 }^{ 0 }$$

C
$${ 68 }^{ 0 }$$

D
$${ 148 }^{ 0 }$$

Solution

The given angles of a quadrilateral are $${ x }^{ o },(x-{ 10) }^{ o },(x+{ 30 })^{ o }$$ and $$2{ x }^{ o }.$$
We know, that sum of four interior angles of a quadrilateral is $$360^o$$.
$$\therefore$$ $$x+(x-10)+(x+30)+2x=360^o$$
$$\Rightarrow$$ $$5x+20^o=360^o$$
$$\Rightarrow$$ $$5x=340^o$$
$$\Rightarrow$$ $$x=\dfrac{340^o}{5}$$
$$\Rightarrow$$ $$x=68^o$$.

Now, measure of an angles:
$$x^o=68^o$$,
$$(x-10)^o=(68-10)^o=58^o$$,
$$(x+30)^o=(68+30)^o=98^o$$
and $$2x^o=2(68^o)=136^o$$.
$$\therefore$$ The greatest angle is $$136^o$$.

Hence, option $$A$$ is correct.
Question 2
1 / -0

The three angles of a quadrilateral are $${ 80 }^{ \circ }$$,$${ 70 }^{ \circ }$$and$${ 120 }^{ \circ }$$. The fourth angle is:

A
$${ 110 }^{ \circ }$$

B
$${ 100 }^{ \circ }$$

C
$${ 90 }^{ \circ }$$

D
$${ 80 }^{ \circ }$$

Solution

Three angles of quadrilateral are :

$$ \theta _{1} = 80^o$$, $$ \theta _{2} = 70^o$$, $$ \theta _{3} = 120^o$$.

To find $$ \theta _{4} :$$

We know, by angle sum property, total angle of quadrilateral $$ = 360^o$$

Then, $$ \theta _{1} + \theta _{2} + \theta _{3} + \theta _{4} = 360^o$$

$$\implies$$ $$ 80^o + 70^o + 120^o + \theta _{4} = 360^o$$

$$\implies$$ $$ 270^o + \theta _{4} = 360^o$$

$$\implies$$ $$ \theta _{4} = 360^o - 270^o$$

$$ \therefore \theta _{4} = 90^o$$.

Therefore, the fourth angle is $$ 90^o $$.

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

Which of the following statements is false?

A
A quadrilateral has four sides and four vertices

B
A quadrilateral has four angles

C
A quadrilateral has four diagonals

D
A quadrilateral has two diagonals

Solution

$$\textbf{Step -1: Characteristics of quadrilateral,}$$
$$\text{We know from the characteristics of quadrilateral that it has four sides and}$$
$$\text{ four vertices and four angles}$$
$$\text{So, option A, B are TRUE.}$$
$$\textbf{Step -2: Checking number of diagonals.}$$
$$\text{Number of sides of a quadrilateral}=4$$
$$\text{Then, the number of diagonals}=\dfrac{4(4-3)}{2}=2$$ $$[\because\textbf{number of diagonals for side n}\mathbf{=\dfrac{n(n-3)}{2}}]$$
$$\text{So, option C is FALSE and D is TRUE.}$$
$$\textbf{Hence, Option C is correct.}$$
Question 4
1 / -0

The angles of a quadrilateral are in the ratio $$ 1 : 2 : 3 : 4 $$ . Then the smallest angle is:

A
$$ 72^{\circ} $$

B
$$ 144^{\circ} $$

C
$$ 36^{\circ} $$

D
$$ 18^{\circ} $$

Solution

Given ratio of angles of quadrilateral $$ABCD$$ is $$1:2:3:4$$

Let the angles of quadrilateral $$ABCD$$ be $$1x,2x,3x,4x$$, respectively.

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

$$\Rightarrow 1x+2x+3x+4x=360^o$$

$$\Rightarrow 10x=360^o$$

$$\therefore x=36^o$$.

$$\therefore \angle A =1x=1 \times 36^o =36^o$$,

$$ \angle B =2x=2\times 36^o =72^o$$,

$$ \angle C =3x=3 \times 36^o =108^o$$

and $$ \angle D =4x=4\times 36^o =144^o$$.

$$\therefore$$ The smallest angle $$=36^o$$.

Hence, option $$C$$ is correct.
Question 5
1 / -0

What is the maximum number of obtuse angles that a quadrilateral can have ?

A
$$ 1 $$

B
$$ 2 $$

C
$$ 3 $$

D
$$ 4 $$

Solution

We know that, the sum of all the $$4$$ interior angles of a quadrilateral is $$360^{\circ}$$
$$\implies \theta_1+\theta_2+\theta_3+\theta_4=360^{\circ}$$
An angle $$\theta$$ is obtuse if $$90^{\circ}< \theta < 180^{\circ}$$
If all the angles of a quadrilateral are obtuse
$$90^{\circ}< \theta_1 < 180^{\circ}$$ -----(1)
$$90^{\circ}< \theta_2 < 180^{\circ}$$ -----(2)
$$90^{\circ}< \theta_3 < 180^{\circ}$$ -----(3)
$$90^{\circ}< \theta_4 < 180^{\circ}$$ -----(4)
Adding (1), (2), (3) and (4)
$$\implies 360^{\circ}<\theta_1+\theta_2+\theta_3+\theta_4<720^{\circ}$$
Therefore all the $$4$$ angles can not be obtuse
If three angles are obtuse say each of $$100^{\circ}$$, then the fourth angle will be $$=360^{\circ}-300^{\circ}=60^{\circ}$$
Hence, utmost $$3$$ angles can be obtuse
Question 6
1 / -0

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 7
1 / -0

The three angles of a quadrilateral are $$80^o,70^o$$ and $$120^o$$. Then the fourth angle is:

A
$$110^0$$

B
$$100^0$$

C
$$90^0$$

D
$$80^0$$

Solution

Given the three angles of a quadrilateral are $$80^o,70^o$$ and $$120^o$$.
Let the fourth angle be $$x$$.
We know, by angle sum property, the sum of all the angles of a quadrilateral is $$360^0$$.
$$\implies$$ $$80^o+70^o+120^o+x=360^o$$
$$\implies$$ $$270^o+x=360^o$$
$$\implies$$ $$x=360^o-270^o$$
$$\implies$$ $$x=90^o$$.

$$\therefore$$ The fourth angle is $$90^o$$.
Hence, option $$C$$ is correct.
Question 8
1 / -0

Area of a quadrilateral ABCD is $$20 cm^2$$ and perpendiculars on BD from opposite vertices are 1 cm and 1.5 cm. The length of BD is

A
4 cm

B
15 cm

C
16 cm

D
18 cm

Solution

Area of the quadrilateral given=
$$ \dfrac{1}{2} (sum\, of\, altitudes)\times corresponding\, diagonal $$
$$ \rightarrow $$ $$ 20=\dfrac{1}{2}(1+1.5)\times BD $$
$$ \rightarrow $$ $$ \dfrac{1} {2}(2.5) \times BD = 20 cm ^{2} $$
$$ \rightarrow BD = 20 \times\dfrac{2}{2.5} $$ = $$ \dfrac{40}{2.5} = 16 cm$$
Question 9
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Three angles of quadrilateral are $$ 47^{\circ}, 102^{\circ} $$ and $$ 111^{\circ} $$, then the fourth angle is equal to:

A
$$ 102^{\circ} $$

B
$$ 100^{\circ} $$

C
$$ 360^{\circ} $$

D
$$ 260^{\circ} $$

Solution

Let the four angles be $$ \angle A , \angle B , \angle C$$ and $$\angle D$$ .

Given $$ \angle A=47^o , \angle B =102^o, \angle C = 111^{o} $$ .

We know, by angle sum property, the sum of angles of a quadrilateral is $$360^o$$.
$$ \angle A + \angle B + \angle C + \angle D = 360^{o} $$.

Then, $$ \angle D $$ will be given by:

$$\angle D = 360^o - \angle A - \angle B-\angle C $$

$$ \Rightarrow \angle D = 360 ^o-47^o -102^o -111^o $$

$$ \Rightarrow \angle D = 360 - 260^o$$
$$ \Rightarrow \angle D = 100^o$$.

The measure of fourth angle is $$ 100 ^{o}$$.

Therefore, option $$B$$ is correct.
Question 10
1 / -0

In a quadrilateral $$PQRS, \angle P, \angle Q, \angle R$$ and $$\angle S$$ are interior angles. If $$\angle P: \angle Q: \angle R: \angle S = 1:2:3: 4$$, then which angle is equal to $$144^\circ$$.

A
$$\angle P$$

B
$$\angle Q$$

C
$$\angle R$$

D
$$\angle S$$

Solution

Given ratio of angles of quadrilateral $$PQRS$$ is $$1:2:3:4$$
Let the angles of quadrilateral $$PQRS$$ be $$1x,2x,3x,4x$$, respectively.

We know, by angle sum property, the sum of angles of a quadrilateral is $$360^o$$.
$$\Rightarrow 1x+2x+3x+4x=360^o$$
$$\Rightarrow 10x=360^o$$
$$\therefore x=36^o$$.

$$\therefore \angle P =1x=1 \times 36^o =36^o$$,
$$ \angle Q =2x=2\times 36^o =72^o$$,
$$ \angle R =3x=3 \times 36^o =108^o$$
and $$ \angle S =4x=4\times 36^o =144^o$$.

$$\therefore$$ $$\angle S=144^o$$.

Hence, option $$D$$ is correct.

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Quadrilaterals Test - 32

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Quadrilaterals Test - 32

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Question 1

$${ 136 }^{ 0 }$$

$${ 180 }^{ 0 }$$

$${ 68 }^{ 0 }$$

$${ 148 }^{ 0 }$$

Solution

Question 2

$${ 110 }^{ \circ }$$

$${ 100 }^{ \circ }$$

$${ 90 }^{ \circ }$$

$${ 80 }^{ \circ }$$

Solution

Question 3

A quadrilateral has four sides and four vertices

A quadrilateral has four angles

A quadrilateral has four diagonals

A quadrilateral has two diagonals

Solution

Question 4

$$ 72^{\circ} $$

$$ 144^{\circ} $$

$$ 36^{\circ} $$

$$ 18^{\circ} $$

Solution

Question 5

$$ 1 $$

$$ 2 $$

$$ 3 $$

$$ 4 $$

Solution

Question 6

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

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

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

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

Solution

Question 7

$$110^0$$

$$100^0$$

$$90^0$$

$$80^0$$

Solution

Question 8

4 cm

15 cm

16 cm

18 cm

Solution

Question 9

$$ 102^{\circ} $$

$$ 100^{\circ} $$

$$ 360^{\circ} $$

$$ 260^{\circ} $$

Solution

Question 10

$$\angle P$$

$$\angle Q$$

$$\angle R$$

$$\angle S$$

Solution

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