CBSE Questions for Class 8 Maths Understanding Quadrilaterals Quiz 1 - MCQExams.com

Polygons that have at least one of their diagonals in their exterior are called concave polygons.

$$Rectangle$$ is the only parallelogram which like a parallelogram has opposite sides parallel and all angles $$equal$$.
Therefore, C is the correct answer.

Adjacent angles are supplementary i.e $$180^o$$.
Opposite angles are equal, so they can be only adjacent angles.
Therefore, C is the correct answer.

A polygon that has one or more interior angles greater than 180 known as concave polygon.

All the diagonals are in the interiors of the polygons because each angle is less than $$180^o$$. So, It has no reflex angle.
Therefore, C is the correct answer.

Polygons that have no portions of their diagonals in their exterior are called convex polygons.

 The diagonals of all the parallelogram does not intersect at right angles.Square and rhombus are the parallelogram in which diagonals intersect at right angles.
Hence the above statement is false

Option(A) is not true. A trapezium has one pair of parallel sides. 
A parallelogram has two pairs of parallel sides. So a parallelogram is also a trapezium, but the trapezium cant be a parallelogram.  
Option  (B) is not true as the opposite sides of a parallelogram are equal and parallel.
Option D is the correct answer.

There are six general and important properties of parallelograms to know:

There are six important properties of parallelograms to know:

Opposite sides are congruent (AB = DC).

Opposite angels are congruent (D = B).

Consecutive angles are supplementary (A + D = 180°).

If one angle is right, then all angles are right.

The diagonals of a parallelogram bisect each other.

Each diagonal of a parallelogram separates it into two congruent triangles.

These properties does not say anything about angle between diagonals . So diagonals $$may$$ $$be$$ perpendicular $$may$$ $$be$$ $$not$$.

Special type of parallelogram with perpendicular diagonals are square and rhombus.

So correct answer is option is C 

A parallelogram is a closed figure having 4 sides and 4 angles. 

In a convex polygon, none of the internal angles is a reflex angle. Instead, all the internal angles are less than $$180^{\circ}$$. 

The given figure is a regular pentagon since it has $$5$$ sides and all are equal 

In a concave polygon, at least one of the internal angles is a reflex angle i.e greater than 180.

Let number of polygon = n (which must be an integer)
Sum of all interior angles is $$ n \times 138^o $$
Sum of all interior angles of a polygon is $$ 180^o (n-2) $$
$$ => n \times 138^o = 180^o (n-2) $$
$$ => 42^o n = 360^o $$ 
$$=> n = \dfrac{60}{7} $$ 
Their is no possibility to have a regular polygon whose each interior angle is $$ 138^o  $$

$$ After\quad analysing\quad the\quad given\quad table\quad it\quad is\quad observed\quad that\quad when\\ (i)\quad numbre\quad of\quad sides\quad of\quad the\quad polygon=3\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \longrightarrow sum\quad of\quad the\quad internal\quad angles={ 180 }^{ o }=\left( 3-2 \right) \times { 180 }^{ o }\\ (ii)\quad numbre\quad of\quad sides\quad of\quad the\quad polygon=4\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \longrightarrow sum\quad of\quad the\quad internal\quad angles={ 360 }^{ o }=\left( 4-2 \right) \times { 180 }^{ o }\\ (iii)\quad numbre\quad of\quad sides\quad of\quad the\quad polygon=5\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \longrightarrow sum\quad of\quad the\quad internal\quad angles={ 540 }^{ o }=\left( 5-2 \right) \times { 180 }^{ o }\\ (iv)\quad numbre\quad of\quad sides\quad of\quad the\quad polygon=6\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \longrightarrow sum\quad of\quad the\quad internal\quad angles={ 720 }^{ o }=\left( 6-2 \right) \times { 180 }^{ o }\\ \therefore \quad By\quad induction,\quad it\quad can\quad be\quad concluded\quad that\\ When\quad the\quad numbre\quad of\quad sides\quad of\quad the\quad polygon=n\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \longrightarrow sum\quad of\quad the\quad internal\quad angles=\left( n-2 \right) \times { 180 }^{ o }\\ \therefore \quad For\quad a\quad 8\quad sided\quad polygon\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad sum\quad of\quad the\quad internal\quad angles=\left( 8-2 \right) \times { 180 }^{ o }={ 1080 }^{ o }\\ Ans-(i)\quad If\quad the\quad numbre\quad of\quad sides\quad of\quad a\quad polygon=n\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \longrightarrow sum\quad of\quad the\quad internal\quad angles=\left( n-2 \right) \times { 180 }^{ o }\\ \quad \quad \quad \quad (ii){ 1080 }^{ o }\\ \\ \\ \\ \\ \\  $$

False
The diagonals of a parallelogram do always bisect each other. However, they only form right angles if the parallelogram is a rhombus or a square. 

Parallelogram . The diagonals (lines linking opposite corners) of parallelogram bisect each other. Each diagonal cuts the other into two equal parts.

Let one angle be $$\angle x$$
Thus, smallest angle is $$\dfrac{2}{3}\angle x$$
Now, $$\dfrac{2}{3}\angle x + \angle x =180$$

True.

  Rhombus have opposite sides  parallel and opposite angles are equal.Hence every rhombus is  a parallelogram..

$$
Given\quad that\quad ABCD\quad is\quad a\quad parallelogram.\\ We\quad know\quad that\quad in\quad a\quad parallelogram\quad opposite\quad angles\quad are\quad equal.\\ So\quad from\quad the\quad given\quad figure,\\ \angle B=\angle D\Rightarrow 6x+3y-8=7y\Rightarrow 6x-4y=8\Rightarrow x=\frac { 8+4y }{ 6 } \quad ----\left( 1 \right) \\ Also\quad we\quad know\quad that\quad adjacent\quad angles\quad in\quad a\quad parallelogram\quad are\quad supplementary.\\ So\quad \angle A+\angle D={ 180 }^{ o }\Rightarrow 4x+20+6x+3y-8={ 180 }^{ o }\Rightarrow 10x+3y=168\quad ----\left( 2 \right) \\ Using\quad 1\quad in\quad 2,\\ 10\left( \frac { 8+4y }{ 6 }  \right) +3y=168\Rightarrow 80+40y+18y=168\times 6\Rightarrow 58y=1008-80\Rightarrow 58y=928\Rightarrow y=\frac { 928 }{ 58 } \Rightarrow y={ 16 }^{ o }\\ Using\quad y=16\quad in\quad 1,\\ x=\frac { 8+4\times 16 }{ 6 } \Rightarrow x=\frac { 8+64 }{ 6 } \Rightarrow x=\frac { 72 }{ 6 } \Rightarrow x={ 12 }^{ o }\\ So\quad x={ 12 }^{ o }\quad \& \quad y={ 16 }^{ o }
$$

In a parallelogram, opposite angles are congruent(i.e. are of equal length).

Properties of rectangle are : 

If one of the angle measures more than  $$ {180}^{o} $$ in a quadrilateral, then that is known as a concave quadrilateral.

A curve which begins and ends at the same point is called a closed curve.

In simple closed curves, the shapes are closed by line-segments or by a curved line.

Triangle, quadrilateral, circle, etc., are examples of closed curves.

$$ABCD$$ is a parallelogram,
$$AD\parallel BC$$ and $$CD$$ is the transversal.
Thus, $$\angle C +\angle D=180^{o}$$ [Sum of the internal angles on the same side of the transversal is 180]

False,
If all angles of a quadrilateral are equal, it is a rectangle.
Given, $$ABCD$$ is a quadrilateral
$$\angle A=\angle B = \angle C=\angle D$$
Now, we know,
   $$\angle A+\angle B + \angle C+\angle D=360$$
$$=>\angle A=90$$
$$\angle A=\angle B = \angle C=\angle D=90$$
Thus, all angles are right angles.
Thus, $$ABCD$$ is a rectangle.