Understanding Quadrilaterals Test

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Understanding Quadrilaterals Test - 13

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Weekly Quiz Competition

Question 1
1 / -0

ABCD is a parallelogram in which $$\angle DAC = 40^{\circ} ; \angle BAC = 30^{\circ}; \angle DOC = 105^{\circ}$$ ; then $$\angle CDO$$ equals :

A
$$75^{\circ}$$

B
$$70^{\circ}$$

C
$$40^{\circ}$$

D
$$45^{\circ}$$

Solution

$$\angle AOB = \angle COD = 105^{\circ}$$ (Vertically Opposite Angles)
In triangle $$AOB,$$
$$\angle BAO + \angle AOB + \angle OBA = 180^{\circ} $$
$$30^{\circ} + 105^{\circ} + \angle OBA = 180^{\circ} $$
$$\angle OBA = 45^{\circ}$$

$$AB$$ and $$CD$$ are parallel
$$\therefore \angle OBA = \angle CDO = 45^{\circ}$$ (Alternate Interior angles)
Question 2
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If PQRS is a parallelogram, then $$\angle{Q}- \angle{S}$$ is equal to :

A
$$90^{\circ}$$

B
$$120^{\circ}$$

C
$$180^{\circ}$$

D
$$0^{\circ}$$

Solution

Given that PQRS is parallelogram,
Opposite angles in a parallelogram are equal.
i.e, $$\angle P=\angle R$$ and $$\angle Q=\angle S$$
$$\therefore$$ If $$\angle Q=\angle S$$ then $$\angle Q-\angle S=0°$$
Hence, the answer is $$0°.$$
Question 3
1 / -0

The diagonals of a parallelogram PQRS intersect at O. If $$ \angle QOR= 90^{\circ}\ and\ \angle QSR=50^{\circ}, \ then\ \angle ORS$$ is :

A
$$90^{\circ}$$

B
$$40^{\circ}$$

C
$$70^{\circ}$$

D
$$50^{\circ}$$

Solution

Given: $$ \angle QOR= 90^{\circ}\ and\ \angle QSR=50^{\circ}$$
Now, $$\angle SOR = 180^{\circ} - \angle QOR = 90^{\circ}$$ (Since angles forming a linear pair are complementary)
Consider $$\Delta SOR$$,
$$\angle SOR + \angle OSR + \angle ORS = 180^{\circ}$$ (angle sum property of triangles)
$$\therefore \angle ORS = 180^{\circ} - 90^{\circ} - 50^{\circ} = 40^{\circ}$$
Question 4
1 / -0

In the given figure, ABCD is a parallelogram. If $$\angle B = 100^{\circ}$$, then ($$\angle A + \angle C$$) is equal to :

A
$$360^{\circ}$$

B
$$200^{\circ}$$

C
$$180^{\circ}$$

D
$$160^{\circ}$$

Solution

Given ABCD is a parallelogram.
The sum of co-interior angles is always $$180°$$.
Given $$\angle B =100°$$
and, $$\angle B +\angle C=180°$$
$$\Rightarrow \angle C = 80°$$
The opposite angles of a parallelogram are equal
i.e, $$\angle A= \angle C=80°$$
$$\therefore$$ Sum $$\angle A+ \angle C=80°+80°=160°$$
Hence, the answer is $$160°.$$
Question 5
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Consider the following statements:
(1) The diagonals of a parallelogram are equal.
(2) The diagonals of a square are perpendicular
to each other.
(3) If the diagonals of a quadrilateral intersect at
right angles, it is not necessarily a rhombus.
(4) Every quadrilateral is either a trapezium or a
parallelogram or a kite.
Which of the above statements is/are correct?

A
Only (2)

B
Only (3)

C
Both (2) and (3)

D
(1), (2) and (3)
Question 6
1 / -0

Two adjacent angles of a parallelogram are $$(2x + 30)^{\circ}$$ and $$(3x + 30)^{\circ}$$. The value of $$x$$ is :

A
$$30^{\circ}$$

B
$$60^{\circ}$$

C
$$24^{\circ}$$

D
$$36^{\circ}$$

Solution

We know that the sum of the adjacent angles of parallelogram is equal to $$180^o$$
$$\therefore 2x+30^o+3x+30^o=180°$$
$$\Rightarrow 5x=120^o$$
$$\therefore x=\dfrac{120^o}{5}=24°$$
Hence, the answer is $$24^o.$$
Question 7
1 / -0

If the sum of all interior angles of a convex polygon is $$1440^{\circ}$$, then the number of sides of the polygon is?

A
$$8$$

B
$$10$$

C
$$11$$

D
$$12$$

Solution

If n is the number of sides of the polygon, then
$$(2n - 4)\times 90^{\circ}= 1440^{\circ}$$
or $$ 2n = 20$$
or $$n = 10$$
Question 8
1 / -0

The measurement of each angle of a polygon is 160$$^{\circ}$$. The number of its sides is?

A
$$15$$

B
$$18$$

C
$$20$$

D
$$30$$

Solution

Each angle of polygon $$= 160^{o}$$
Let there be n sides, then sum of all the angles $$= 160n$$
Sum of all the angles of any polygon $$= (n-2)(180^{o})$$
Therefore, $$160 n = (n-2)180^{o}$$
$$160^{o}n= 180^{o}n - 360^{o}$$
$$20n = 360$$
$$n= 18$$
The polygon has $$18$$ sides.
Question 9
1 / -0

The sum of the measures of all the angles of a pentagon are :

A
$$135^{\circ}$$

B
$$540^{\circ}$$

C
$$720^{\circ}$$

D
none of these

Solution

To find the sum of the interior angles of a pentagon, divide it up into triangles.There are three triangles. Because the sum of the angles of each triangle is 180 degrees.
We get $$=180 \times 3\ =\ 540°$$
Question 10
1 / -0

Each interior angle of a regular polygon is $$144^0$$. Find the interior angle of a regular polygon which has double the number of sides as the first polygon.

A
$$100^0$$

B
$$160^0$$

C
$$36^0$$

D
$$162^0$$

Solution

Since each interior angle of the first polygon $$=144^o$$
Each exterior angle of the first polygon $$=180-144$$
$$=36^o$$
$$\therefore$$ The number of sides of the first polygon $$=\frac {360}{36}=10$$
$$\therefore$$ The number of sides of the second polygon $$=2\times 10=20$$
$$\therefore$$ Each exterior angle of the second polygon $$=\frac{360^o}{20}=18^o$$
$$\therefore$$ Each interior angle of the second polygon $$=180^o-18^o=162^o$$

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Re-Attempt Weekly Quiz Competition

Understanding Quadrilaterals Test - 13

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Understanding Quadrilaterals Test - 13

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Question 1

$$75^{\circ}$$

$$70^{\circ}$$

$$40^{\circ}$$

$$45^{\circ}$$

Solution

Question 2

$$90^{\circ}$$

$$120^{\circ}$$

$$180^{\circ}$$

$$0^{\circ}$$

Solution

Question 3

$$90^{\circ}$$

$$40^{\circ}$$

$$70^{\circ}$$

$$50^{\circ}$$

Solution

Question 4

$$360^{\circ}$$

$$200^{\circ}$$

$$180^{\circ}$$

$$160^{\circ}$$

Solution

Question 5

Only (2)

Only (3)

Both (2) and (3)

(1), (2) and (3)

Question 6

$$30^{\circ}$$

$$60^{\circ}$$

$$24^{\circ}$$

$$36^{\circ}$$

Solution

Question 7

$$8$$

$$10$$

$$11$$

$$12$$

Solution

Question 8

$$15$$

$$18$$

$$20$$

$$30$$

Solution

Question 9

$$135^{\circ}$$

$$540^{\circ}$$

$$720^{\circ}$$

none of these

Solution

Question 10

$$100^0$$

$$160^0$$

$$36^0$$

$$162^0$$

Solution

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