Understanding Quadrilaterals Test - 6

$$ A\quad simple\quad closed\quad curve\quad encloses\quad a\quad single\quad area \quad and \quad do \quad not\quad have  \quad intersecting \quad lines.\\ Here\quad in\quad each\quad of\quad the\quad figures\quad 1,\quad 2,\quad 5,\quad 6,\quad 7\quad a\quad single\quad area\\ is\quad enclosed.\\ But\quad in\quad each\quad of\quad the\quad figures\quad 3,\quad 4,\quad 8\quad more\quad than\quad one\quad area\\ are\quad enclosed.\\ So\quad each\quad of\quad the\quad figures\quad 1,\quad 2,\quad 5,\quad 6,\quad 7\quad is\quad a\quad simple\quad curve.\\ Ans-\quad 1,\quad 2,\quad 5,\quad 6,\quad 7 $$

$$ 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 }\\ \\ \\ \\ \\ \\  $$

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 

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 

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

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

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

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.