Free Quadrilaterals 01 Practice Test - 9th Grade 

In a parallelogram, the adjacent sides and angles are not equal and no sides are perpendicular to other.

$\angle A+\angle B={130}^{\circ }+{50}^{\circ }={180}^{\circ }$
Here, $\angle A$ and $\angle B$ are co-interior angles and their sum is ${180}^{\circ }$. 
By the converse of co-interior angles theorem, AD || BC and AB is the transversal.

$\angle D+\angle A={50}^{\circ }+{130}^{\circ }={180}^{\circ }$
Here, $\angle D$ and $\angle A$ are co-interior angles and their sum is ${180}^{\circ }$. 
By the converse of co-interior angles theorem, AB || DC and AD is the transversal. 

In the given parallelogram, AD || BC and AB || DC
Consider the diagonal BD which acts as a transversal for BC and AD.
$\angle ADB=\angle DBC=x^{\circ }$ ( alternate angles)

Consider the diagonal BD which acts as a transversal for AB and BC.
$\angle ABD=\angle BDC=y^{\circ }$ ( alternate angles)

We can see that
$\angle B=\angle ABD+\angle DBC=x^{\circ }+y^{\circ }$
Also, $\angle D=\angle ADB+\angle BDC=x^{\circ }+y^{\circ }$
Thus, $\angle B=\angle D$.

Hence, opposite angles of a parallelogram are equal.

Let the fourth angle be $x^{\circ }$.

By angle sum property, we know that sum of interior angles of a quadrilateral is ${360}^{\circ }$.

$\Rightarrow {37}^{\circ }+{130}^{\circ }+{53}^{\circ }+x={360}^{\circ }$

$\Rightarrow x={360}^{\circ }-({37}^{\circ }+{130}^{\circ }+{53}^{\circ })={360}^{\circ }-{220}^{\circ }={140}^{\circ }$

The sum of internal angles of a quadrilateral is ${360}^{\circ }$. The sum of external angles of any polygon, irrespective of the number of sides is ${360}^{\circ }$. Hence, the given statement is true.

Given: In rectangle ABCD,
$\angle BAO=35°$

To find: Angle between the diagonals $\angle AODor\angle BOC$

In $\Delta$AOB
OA = OB
$\because$ Diagonals of a rectangle bisect each other
$\Rightarrow \angle OBA=\angle OAB={35}^{\circ }$ (As the angles opposite to equal sides are equal)

Applying exterior angle property in $\Delta$AOB
$\angle AOD=\angle OBA+\angle OAB$
$\angle AOD=35°+35°$
$\angle AOD=70°$

We know that,
the opposite angles of a parallelogram are equal.
ie. $\angle A=\angle Cand\angle B=\angle D...\left(i\right)$
If they are supplementary as well, then
$\angle A+\angle C={180}^{\circ }$
from (i),
$\angle A+\angle A={180}^{\circ }$
$2\angle A={180}^{\circ }$
$\angle A={90}^{\circ }$

Similarly,
$\angle C={90}^{\circ }$
It means that each angle will be equal to ${90}^{\circ }.$
A parallelogram whose internal angles are all ${90}^{\circ }$ is a rectangle. Thus, the figure will be a rectangle.

Here, the ratio is given to be a:b:b:a.
If KLMJ is a quadrilateral, the sides are in the ratio a:b:b:a.

Let KL = a, LM = b, MJ = b and JK = a.

Here, JK and LM are equal and LM and MJ are equal.

We know that, kite is a quadrilateral with two pairs of adjacent sides equal in length.

The data given implies that we have two disjoint pairs of adjacent sides which are equal i.e. there are two pairs of adjacent sides which are equal and the opposite sides are not equal. 

This is the very definition of a kite.

We know that the opposite angles of a parallelogram are equal.

So,
$(3x-10{)}^{\circ }=(x+80{)}^{\circ }\Rightarrow 3x-x={80}^{\circ }+{10}^{\circ }\Rightarrow 2x={90}^{\circ }\Rightarrow x={45}^{\circ }$

Also, the consecutive angles of a parallelogram are supplementary.

So,
$\angle B+\angle C={180}^{\circ }(3x-10{)}^{\circ }+(y+25{)}^{\circ }={180}^{\circ }(3\times {45}^{\circ })-{10}^{\circ }+y+{25}^{\circ }={180}^{\circ }{135}^{\circ }-{10}^{\circ }+y+{25}^{\circ }={180}^{\circ }\Rightarrow {125}^{\circ }+y+{25}^{\circ }={180}^{\circ }\Rightarrow y={180}^{\circ }–{125}^{\circ }-{25}^{\circ }\Rightarrow y={30}^{\circ }$

Let ABCD be a rectangle such as AB = CD and BC = DA. P,Q,R and S are the midpoints of the sides AB, BC, CD and DA respectively.

Let us join AC and BD 
In $\Delta$ABC,
P and Q are the mid-points of AB and BC respectively.
$\therefore$ PQ || AC and PQ = $\frac{1}{2}$AC (Midpoint theorem)...........(1)
Similarly in $\Delta$ ADC,
SR || AC and SR=$\frac{1}{2}$AC (Midpoint theorem)..........(2)
Clearly, PQ || SR and PQ = SR
Since, in quadrilateral PQRS, one pair of opposite side is equal and parallel to each other, it is parallelogram
$\therefore$ PS || QR and PS = QR (opposite sides of parallelogram).........(3)
In $\Delta$ BCD, Q and R are the mid-points of sides BC and CD respectively.
$\therefore$ QR || BD and QR=$\frac{1}{2}$BD (Midpoint theorem)..........(4)
However, the diagonals of a rectangle are equal.
$\therefore$ AC = BD ...........(5)
By using equation (1), (2), (3), (4), (5), we obtain
PQ = QR = SR = PS
Therefore, PQRS is a rhombus and hence, the given statement is true.
 

Since BDEF and FDCE are parallelograms,
BD = FE and DC = FE
Hence, we can see that BD is equal to DC and so, D is the mid-point of BC.