Quadrilaterals Test

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Quadrilaterals Test - 19

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Weekly Quiz Competition

Question 1
1 / -0

Find the value of $$x + y + z + w$$

A
$$180^o$$

B
$$360^o$$

C
$$240^o$$

D
none of the above

Solution

Fourth angle of quadrilateral = $$360^o - (60^o + 80^o + 120^o)= 100^o$$
Angle x = $$180^o - 120^o = 60^o $$
Angle y = $$180^o - 80^o = 100^o $$
Angle z = $$180^o - 60^o = 120^o $$
Angle w = $$180^o - 100^o = 80^o $$
x+y+z+w = $$ 60^o + 100^o + 120^o + 80^o = 360^o $$
Question 2
1 / -0

If the diagonals of a quadrilateral bisect one another at right angles, then the quadrilateral is a:

A
rhombus

B
trapezium

C
rectangle

D
parallelogram

Solution

If the diagonals of a quadrilateral bisect one another at right angles then the quadrilateral is a rhombus.
Also, if $$ABCD$$ is a quadrilateral and diagonals $$AC$$ and $$BD$$ bisect each other at at right angles,
then in $$ \triangle AOB$$ & $$\triangle AOD$$,
$$DO=OB$$.....($$O$$ is mid point)
$$AO=AO$$.....(COMMON SIDE)
$$\angle AOB\quad =\angle AOD$$ .....(Right angle)
Therefore, $$\triangle AOB \cong \triangle AOD$$ ...(By $$RHS$$ congruency criterion).
Then, $$ AB=AD$$ (By $$CPCT$$ rule).

Similarly $$AB=BC=CD=AD$$ can be proved which means that quadrilateral is a rhombus.
Hence, option $$A$$ is correct.
Question 3
1 / -0

The angles of a quadrilateral are $$\displaystyle x^{\circ},\left ( x+5 \right )^{\circ},\left ( x+10 \right )^{\circ}\: and\: \left ( x+25 \right )^{\circ}$$ then value of x is

A
$$\displaystyle 50^{\circ}$$

B
$$\displaystyle 80^{\circ}$$

C
$$\displaystyle 60^{\circ}$$

D
$$\displaystyle 70^{\circ}$$

Solution

We have,
$$\displaystyle x^{\circ}+\left ( x+5 \right )^{\circ}+\left ( x+10 \right )^{\circ}+\left ( x+25 \right )^{\circ}=360^{\circ}$$
$$\displaystyle 4x+40^{\circ}=360^{\circ}$$
$$\displaystyle 4x=320^{\circ}$$
$$\displaystyle x=80^{\circ}.$$
Question 4
1 / -0

If in a quadrilateral ABCD, $$\angle A = 120^o, \angle B = 60^o, \angle C = 100^o$$ then $$\angle D =$$ ...............

A
80$$^o$$

B
90$$^o$$

C
180$$^o$$

D
360$$^o$$

Solution

$$\angle A + \angle B + \angle C + \angle D = 360^o$$
$$120^o + 60^o + 110^o + \angle D = 360^o$$
$$280^o + \angle D = 360^o$$
$$\angle D = 360^o - 280^o = 80^o$$
Question 5
1 / -0

In a quadrilateral $$ABCD$$, $$\displaystyle \angle A +\angle C = 180^{\circ}$$ then $$\displaystyle \angle B + \angle D$$ =

A
$$\displaystyle 360^{\circ}$$

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

C
$$\displaystyle 180^{\circ}$$

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

Solution

We have
$$\displaystyle \angle A + \angle B + \angle C + \angle D = 360^{\circ}$$
$$\Rightarrow \displaystyle \left ( \angle A +\angle C \right ) + \left ( \angle B + \angle D \right )= 360^{\circ}$$
$$\Rightarrow \displaystyle 180^{\circ}+ \left ( \angle B + \angle D \right )=360^{\circ}$$
$$\therefore \displaystyle \angle B +\angle D=180^{\circ}$$
Question 6
1 / -0

If in a quadrilateral the diagonals bisect each other, then it is a_____.

A
rhombus

B
parallelogram

C
rectangle or square

D
all of the above

Solution

$$\Rightarrow$$ In rhombus opposite sides are parallel to each other and all sides are equal then, diagonals are bisect each other.
$$\Rightarrow$$ In parallelogram opposite sides are parallel to each other then diagonals are bisect each other.
$$\Rightarrow$$ And in rectangle and square also opposite sides are parallel and diagonals are bisect each other.
$$\Rightarrow$$ So, correct answer for these question is option D.
Question 7
1 / -0

A parallelogram with equal angles is called a ..............................

A
triangle

B
kite

C
rectangle

D
rhombus

Solution

$$Rectangle$$ is the only parallelogram which like a parallelogram has opposite sides parallel and all angles $$equal$$.
Therefore, C is the correct answer.
Question 8
1 / -0

Refer to the figure above and calculate the value of $$x$$, if one angle is a right angle.

A
$$54$$

B
$$56.25$$

C
$$60$$

D
$$67.5$$

E
$$72$$

Solution

We know, by angle sum property, the sum of angles of a quadrilateral is $$360^o$$.
$$\therefore 90^o+2x+x+x=360^o$$
$$\Rightarrow 4x=360^o-90^o$$
$$\Rightarrow 4x=270^o$$
$$\Rightarrow x=\dfrac{270^o}{4}=67.5^o$$.
Therefore, option $$D$$ is correct.
Question 9
1 / -0

Can the angles $$110^o, 80^o, 90^o$$ & $$105^o$$ be the angles of quadrilateral?

A
Yes

B
No

C
Cannot predict

D
None of the above

Solution

We know, by angle sum property, the sum of angles of a quadrilateral is $$360^o$$.
Here, the sum of angle will be $$= 110^o + 80^o + 90^o + 105^o = 385^o > 360^o$$.
Hence, the given angles cannot be the angles of a quadrilateral.
$$\therefore$$ The answer is No.
That is, option $$B$$ is correct.
Question 10
1 / -0

All the angles of a quadrilateral can be ___________.

A
right angles

B
acute angles

C
obtuse angles

D
Can't say

Solution

We know, by angle sum property, the sum of all angles of quadrilateral $$ = 360^o$$.
Then,
Sum of $$4$$ right angles $$= (90˚ + 90˚ + 90˚ + 90˚)$$ $$= 360˚$$.
Sum of $$4$$ acute angles $$= (<90˚ + <90˚ + <90˚ + <90˚)$$ $$< 360˚$$.
Sum of $$4$$ obtuse angles $$= (>90˚ + >90˚ + >90˚ + >90˚)$$ $$>360˚$$.
Hence, all the angles of a quadrilateral can be right angles.

Therefore, option $$A$$ is correct.

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Quadrilaterals Test - 19

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Quadrilaterals Test - 19

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Question 1

$$180^o$$

$$360^o$$

$$240^o$$

none of the above

Solution

Question 2

rhombus

trapezium

rectangle

parallelogram

Solution

Question 3

$$\displaystyle 50^{\circ}$$

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

$$\displaystyle 60^{\circ}$$

$$\displaystyle 70^{\circ}$$

Solution

Question 4

80$$^o$$

90$$^o$$

180$$^o$$

360$$^o$$

Solution

Question 5

$$\displaystyle 360^{\circ}$$

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

$$\displaystyle 180^{\circ}$$

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

Solution

Question 6

rhombus

parallelogram

rectangle or square

all of the above

Solution

Question 7

triangle

kite

rectangle

rhombus

Solution

Question 8

$$54$$

$$56.25$$

$$60$$

$$67.5$$

$$72$$

Solution

Question 9

Yes

No

Cannot predict

None of the above

Solution

Question 10

right angles

acute angles

obtuse angles

Can't say

Solution

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