Free Understanding Quadrilaterals 02 Practice Test - 8th Grade 

A line segment connecting any two non-consecutive vertices of a polygon is known as diagonal.A polygon with an exterior diagonal is known as concave polygon.

Only in these two options, there exists an exterior diagonal. So, these are concave polygons. But II is a concave pentagon. Only IV is a concave quadrilateral. Moreover, a concave polygon has one or more interior angle greater than 180°.

It is given that AB = AD and BC = CD.
Since, two pairs of adjacent sides are equal, the ABCD is a kite.

According to the properties of a kite: 

$AC\perp BD$ 
[Diagonals of kite are perpendicular.]


In $\Delta ABD,$
$\angle ADB$ = $\angle ABD$  $[\because BA=DA$ and angles opposite to equal sides are equal$]$

But, $\angle DAB$ $\ne$ $\angle DCB$

Rhombus has unequal diagonals which are perpendicular bisectors.

Square has equal diagonals which are perpendicular bisectors.

Rectangle has equal diagonals which are perpendicular bisectors .

Kite has unequal diagonals in which the longer one bisects the other.

Parallelogram has unequal diagonals unless it is a square or a rectangle. The diagonals bisect each other.

$\angle BCD+\angle BCH={180}^{\circ }$  [ Linear pair]
$\angle BCD$ = ${180}^{\circ }$ - ${35}^{\circ }$ = ${145}^{\circ }$ 
In quadrilateral BEDC, $\angle BED$ + $\angle EDC$ + $\angle DCB$ + $\angle CBE$ = ${360}^{\circ }$
$\angle CBE$ = ${360}^{\circ }$ - (${105}^{\circ }$ + ${50}^{\circ }$ + ${145}^{\circ }$) = ${60}^{\circ }$
So, $\angle ABF$ = $\angle CBE$ = ${60}^{\circ }$(Vertically opposite angles)

$\angle DCB+\angle ABC={180}^{\circ }$
     [Adjacent angles of a parallelogram are supplementary]

$\Rightarrow 2t+7t={180}^{\circ }$
 [$\because \angle DCB$ and $\angle ABC$ are in ratio 2:7]

$\Rightarrow t={20}^{\circ }$

$\therefore \angle DCB=2t=2\times {20}^{\circ }={40}^{\circ }$

and $\angle ABC=7t=7\times {20}^{\circ }={140}^{\circ }$

Now, $\angle DAB=\angle DCB={40}^{\circ }$
     [Opposite angles of a parallelogram are equal]
 
$\angle DAB+\angle ADB+\angle ABD={180}^{\circ }$
            [Sum of the angles of a triangle]

$\Rightarrow {40}^{\circ }+\angle ADB+\angle ABD={180}^{\circ }$

 $\Rightarrow \angle ADB+\angle ABD={140}^{\circ }$

Sum of the exterior angles of a polygon (irrespective of the number of sides) is ${360}^{\circ }$.

In a nonagon, the measure of all exterior angles will be the same i.e. ${360}^{\circ }$.

Also, nonagon has 9 sides.

So, measure of each exterior angle
= $\frac{{360}^{\circ }}{9}={40}^{\circ }$

Rectangles are also parallelograms. They are shown in color "blue" in the below figures. Hence, there are 7 parallelograms in the figure. 

A polygon is called a regular polygon if it is equiangular and equilateral. i.e., All the angles are equal and measure of all the sides are equal.
Square is quadrilateral with all the angles equal to ${90}^{\circ }$ and all the sides of equal length.
$⟹$ Square is a regular quadrilateral.
Hence, the given statement is true.

 

$\Delta ABD$ and $\Delta CDB$

AB = CD (Opposite sides of a parallelogram)

AD = CB (Opposite sides of a parallelogram)

BD = DB (Common)

So, $\Delta ABD$ $\cong$ $\Delta CDB$      (By SSS congruence condition)

Since, the opposite sides of a rhombus have the same length  it is also a parallelogram. So, a rhombus has all the properties of a parallelogram and also that of a kite. A quadrilateral with exactly two pairs of equal consecutive sides is a kite. This condition is also satisfied by Rhombus. So, all rhombuses are kites but converse is always not true