CBSE Questions for Class 9 Maths Quadrilaterals Quiz 6 - MCQExams.com

$$\textbf{Step-1: Comparing with trapezium}$$ 

                $$\text{The definition of a square does not comply with the}$$

                $$\text{definition of a trapezoid. The definition of a trapezoid}$$

                $$\text{is a quadrilateral (a closed plane figure with 4 sides) with exactly one pair of parallel sides.}$$

$$\textbf{Step-2: Comparing with other option}$$ 

                $$\text{On the other hand, a square is a very special kind of}$$

                $$\text{quadrilateral; A square is also a parallelogram because}$$

                $$\text{it has two sets of parallel sides and four right angles. A}$$

                $$\text{square is also a parallelogram because opposite sides}$$

                $$\text{are parallel. A square is always a rhombus. A rhombus}$$

                $$\text{is a quadrilateral with four congruent sides. If the rhombus has 4}$$

                $$\text{right angles it may also be called a square. So every square is a rhombus.}$$ 

$$\textbf{Hence , Square is not a (C) trapezium}$$

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

Given two angles are equal.

Then $$ \angle A =\angle B=x $$.

Also, $$\angle C = 70^{o} $$ and $$\angle D = 80^{o} $$.

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

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

$$x + x + 70^o + 80^o = 360^{o}  $$

$$2x+ 150^o = 360^{o}  $$

$$ \Rightarrow  2x =  360 ^o-150^o  $$ 

$$ \Rightarrow 2x =  210^o$$

$$ \Rightarrow x = 105^o$$.

Hence, the equal angles are $$ =  105^o$$ each

$$\textbf{Step -1: Find the correct option.}$$

                

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

The given angles are, $$70^o, 60^o, 90^o$$.

Let the fourth angle be $$x^o$$.

Then, $$70^o+60^o+90^o+x^o=360^o$$.

$$\Rightarrow$$ $$360^o - (70^o + 60^o + 90^o ) =x^{\circ}$$

$$\Rightarrow$$ $$x^o=360^o - 220^o  =140^{\circ}$$.

Therefore, the fourth angle is $$x^o=140^o$$.

Hence, option $$D$$ is correct.

Given ratio of angles of a quadrilateral $$ABCD$$ is $$3:5:9:13$$

Let the angles of the quadrilateral $$ABCD$$ be $$3x,5x,9x$$ and $$13x$$, respectively.

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

$$\Rightarrow 3x+5x+9x+13x=360^o$$

$$\Rightarrow 30x=360^o$$

$$\therefore x=12^o$$.

$$\therefore \angle A =3x=3 \times 12^o =36^o$$,

$$ \angle B =5x=5\times 12^o =60^o$$,

$$ \angle C =9x=9 \times 12^o =108^o$$

and $$ \angle D =13x=13 \times 12^o =156^o$$.

Hence, option $$A$$ is correct.