CBSE Questions for Class 8 Maths Understanding Quadrilaterals Quiz 5 - MCQExams.com

The sum of the interior angles of a polygon is four times the sum of its exterior angles.
The sum of the exterior angles of a polygon is always equal to $$360^o$$.
The sum of the interior angles of polygon = $$180 (n-2)$$
=> $$180 (n-2) = 4 \times 360$$
=> $$n -2 = 8$$ 
=> $$n =10$$ 
Number of sides in the polygon = $$10$$

Each interior angle of regular polygon of $$n$$ sides is given by $$\dfrac{180^o(n-2)}{n}$$
According to question

$$
Given\quad ABCD\quad is\quad a\quad parallelogram,\quad AD\parallel BC\quad \& \quad AB\parallel DC,\quad \angle CBE=\angle EBA=\angle 1,\quad \angle DAE=\angle EAB=\angle 2\\ Construction-\quad Draw\quad EF\quad from\quad E\quad such\quad that\quad EF=BC\quad \& \quad EF\parallel BC.\\ So\quad BCEF\quad is\quad a\quad parallelogram.\\ Since\quad EF=BC\quad \& \quad EF\parallel BC\quad so\quad EF=AD\quad \& \quad EF\parallel AD.\\ Now\quad \angle CBE=\angle BEF=\angle 1\quad \left\{ Alternate\quad angles\quad when\quad EF\parallel BC,\quad BE\quad is\quad the\quad transversal \right\} \\ So\quad \angle BEF=\angle EBF=\angle 1\\ \Rightarrow \triangle EFB\quad is\quad a\quad isosceles\quad triangle\quad \left\{ Since\quad base\quad angles\quad are\quad equal \right\} \\ \therefore \quad in\quad \triangle EFB,\quad EF=FB\Rightarrow FB=BC---\left( 1 \right) \\ Now\quad \angle FEA=\angle EAD=\angle 2\quad \left\{ Alternate\quad angles\quad when\quad EF\parallel AD,\quad EA\quad is\quad the\quad transversal \right\} \\ So\quad \angle FEA=\angle EAF=\angle 2\\ \Rightarrow \triangle EAB\quad is\quad a\quad isosceles\quad triangle\quad \left\{ Since\quad base\quad angles\quad are\quad equal \right\} \\ \therefore \quad in\quad \triangle EAB,\quad EF=AF\Rightarrow AF=BC---\left( 2 \right) \\ Adding\quad \left( 1 \right) \quad \& \quad \left( 2 \right) ,\quad we\quad get\\ FB+AF=BC+BC\Rightarrow AB=2BC
$$

Given is a parallelogram PQRS whose diagonals intersect at M.

$$\textbf{Step -1: Find each exterior angle of the given polygon.}$$

                

                $$\text{Let the number of sides of the polygon be }n.$$

                $$\text{Let }AB\text{ and }DC\text{ are the alternate sides of the polygon produced and meet at a right angle at }P.$$

                $$\text{So, }\angle PBC\text{ and }\angle PCB\text{ are exterior angles of the polygon.}$$

                $$\text{As the given polygon is regular so, all of its interior angles are equal, i.e. }\angle ABC=\angle DCB$$

                $$\therefore\angle PBC=180^\circ-\angle ABC$$               $$\textbf{(Linear pair)}$$

                $$\text{and }\angle PCB=180^\circ-\angle DCB$$          $$\textbf{(Linear pair)}$$

                $$\Rightarrow \angle PCB=180^\circ-\angle ABC$$

                $$\text{So, }\angle PBC=\angle PCB$$

                $$\text{In }\triangle PCB,$$

                $$\angle PBC+\angle PCB+90^\circ=180^\circ$$

                $$\Rightarrow 2\angle PBC=90^\circ$$

                $$\Rightarrow \angle PBC=45^\circ$$

                $$\text{Thus, each exterior angle of the polygon}=45^\circ.$$

$$\textbf{Step -2: Find the number of sides of the polygon.}$$

                $$\mathbf{\because\text{Each exterior angle of a regular polygon}=\dfrac{360^\circ}{\text{Number of sides of the polygon}}.}$$

                $$\therefore 45^\circ=\dfrac{360^\circ}{n}$$

                $$\Rightarrow n=8$$

$$\textbf{Final Answer: Number of Sides are 8. The correct option is B.}$$

BDEF & FDCD are parallelogram
BF || DE, BF || DE, & FD || CE & CD || FE
$$\angle BDF = \angle EFD,$$ & $$\angle DFE = \angle DCE$$
$$\angle FBD = \angle FED$$ & $$\angle FED = \angle FED = \dfrac{1}{2} \angle FDC$$  (parallelogram purpose)
$$\triangle FDB$$ & $$\triangle EDC$$ are circular and conguent triangles.
$$\because BD = BC$$

$$
Given\quad ABCD\quad is\quad a\quad parallelogram;\quad \angle PBQ=\angle QBA=\angle 1\quad \& \quad \angle QAP=\angle PAB=\angle 2\\ From\quad the\quad figure,\quad \angle PBQ=\angle AQB=\angle 1\quad \left\{ AQ\parallel BP\quad \& \quad BQ\quad is\quad the\quad transversal \right\} \\ So\quad \angle AQB=\angle QBA=\angle 1\\ \Rightarrow \triangle BAQ\quad is\quad an\quad isosceles\quad triangle\quad \left\{ since\quad base\quad angles\quad are\quad equal. \right\} \\ \therefore \quad in\quad \triangle BAQ,\quad Side\quad AQ=Side\quad AB\quad ---\left( 1 \right) \\ Again\quad from\quad the\quad figure,\quad \angle QAP=\angle APB=\angle 2\quad \left\{ AQ\parallel BP\quad \& \quad AP\quad is\quad the\quad transversal \right\} \\ So\quad \angle PAB=\angle APB=\angle 2\\ \Rightarrow \triangle ABP\quad is\quad an\quad isosceles\quad triangle\quad \left\{ since\quad base\quad angles\quad are\quad equal. \right\} \\ \therefore \quad in\quad \triangle ABP,\quad Side\quad BP=Side\quad AB\quad ---\left( 2 \right) \\ From\quad \left( 1 \right) \quad \& \quad \left( 2 \right) ,\quad we\quad get\quad \\ AQ=BP
$$

$$AB = CD$$ and $$E$$ and $$F$$ are the mid points of $$AB$$ and $$CD$$ respectively. $$\therefore AE = DF $$ 

$$
Given\quad ABCD\quad is\quad a\quad parallelogram,\quad AD\parallel BC,\quad AB\parallel DC\quad \& \quad AD=DP=PC\\ So\quad AD=DP=PC=BC\\ Let\quad \angle DAP=\angle 2\quad \& \quad \angle CBP=\angle 1\\ In\quad isosceles\quad \triangle ADP,\quad \angle DAP=\angle DPA=\angle 2\\ In\quad isosceles\quad \triangle PCB,\quad \angle CPB=\angle CBP=\angle 1\\ Construction-\quad Draw\quad PQ\quad from\quad P\quad such\quad that\quad PQ=BC=AD\quad \& \quad PQ\parallel BC\parallel AD.\\ So\quad ADPQ\quad \& \quad BCPQ\quad are\quad both\quad parallelograms.\\ Now\quad DP\parallel AQ\quad \& \quad AP\quad is\quad the\quad transversal.\\ \angle DPA=\angle PAQ=\angle 2\quad \left\{ Alternate\quad angles\quad when\quad DP\parallel AQ\quad \& \quad AP\quad is\quad the\quad transversal \right\} \\ \Rightarrow \angle PAQ=\angle DAP\\ \Rightarrow AP\quad is\quad the\quad angle\quad bisector\quad of\quad \angle A
$$

Opposite angles of a parallelogram are equal
$$\therefore \angle SPQ = \angle SRQ = 70^o$$
In $$\triangle SQR, \angle RSQ + \angle SQR + \angle SRQ = 180^o$$
$$\angle RSQ = 180- 60 - 70$$
$$RSQ = 50^o$$

Ans : A) True.

ABCD is a parallelogram.
Draw a line PQ which is parallel to AD as well as BC.
$$ \angle DAP = \angle APQ $$ (Alternate interior angles are equal)......(1)
Similarly,
$$ \angle CBP = \angle BPQ $$  (Alternate interior angles are equal)......(2)  
Add equation (1) and (2) 
 $$ \angle DAP + \angle CBP = \angle APQ +  \angle BPQ $$
$$=> \angle DAP + \angle CBP = \angle APB $$

Given,
ABCD is a parallelogram P is a point on side AB and Q is a point on side BC 
Construction:Join diagonals AC and BD.
Proof:
$$CPD = BCD$$ [triangle with same base and between same set of parallel lines have equal area] (i)
Area $$BCD=\dfrac{1}{2}$$area of parallelogram ABCD.[daigonal of parallelogram bisects area in two equal parts] (ii)
From (i) and (ii)
Area$$CPD=\dfrac{1}{2}$$area ABCD (III)
Similarly,
Area $$AQD=Area ABD$$
                       =$$\dfrac{1}{2}$$area ABCD (IV)
From (iii) and (iv)
$$Area CPD  =Area AQD$$

Parallelogram $$ABCD$$  and $$\triangle BEC$$ have same base $$BC$$ and they are between same parallels $$BC$$ and $$AD$$.

To prove
$$Area\left( \triangle AQD \right) =Area\left( \triangle APD \right) +Area\left( \triangle CPB \right) $$
$$Area\left( \triangle APD \right) =Area\left( \triangle APC \right) $$ Area of triangle with same base AP  and between same set of parallel lines are equal.
$$Area\left( \triangle APD \right) =Area\left( \triangle CPB \right) $$
$$=Area\left( \triangle APC \right) =Area\left( \triangle CPB \right) $$
$$=Area\left( \triangle ABC \right) $$
$$=\frac { 1 }{ 2 } Area\quad of\quad parallelogram\quad ABCD$$( AC is diagonal and diagonals of parallelogram divides it into equal areas)
$$Area\left( \triangle AQD \right) =Area\left( \triangle APD \right) +Area\left( \triangle CPB \right) $$ Proved

Given, a polygon whose each interior angles is $$ 175^o $$
Sum of interior angles of a polygon is =  $$ 180^o (n-2) $$
Each interior angle of a polygon = $$ \dfrac {180^o (n-2)}{n} $$
$$ \dfrac {180^o (n-2)}{n}  = 175^o $$
$$  180^o n - 175^o n = 360^o $$
$$ n = \dfrac {360}{5} $$
$$ n = 72 $$
Since, n (number of sides) is an integer, therefore there exist a polygon whose each interior angles is $$ 175^o $$

Given, AM bisects angle A and DM bisects angle D of parallelogram ABCD.
AB is parallel to DC, so,
$$ \angle A + \angle D = 180^o $$ (Sum of interior angles on the same side of a transversal is supplementary.
AM bisects angle A and DM bisects angle D 
$$ \angle A = 2 \angle DAM   and    \angle D = 2 \angle ADM $$
Now, put it in above equality,
$$ 2 \angle DAM + 2 \angle ADM = 180^o $$
$$ \angle DAM  + \angle ADM = 90^o $$
In $$ \triangle AMD $$
$$  \angle DAM  + \angle ADM + \angle  AMD = 180^o $$
$$ 90^o +  \angle  AMD = 180^o $$
$$  \angle  AMD = 90^o $$

Given, ABCD is a parallelogram and $$AQ\bot CD\quad and\quad AP\bot BC$$
$$\therefore \angle AQC=\angle APC=90$$

AQCP is forming a quadrilateral.
As we know that sum of angles of quadrilateral is 360,
$$\therefore \angle QAP+\angle AQC+\angle QCP+\angle APC=360$$
$$\\ or,\angle QAP+90+\angle QCP+90=360\quad \quad \quad \quad \quad \quad \quad \quad (\because \angle AQC=\angle APC=90)\\ or,\angle QAP+\angle QCP=360-180\\ or,\angle QAP+\angle QCP=180\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad (1)$$
Given $$\angle QCP=x,\quad \angle QAP=y\\ x\quad :\quad y\quad =\quad 2\quad :\quad 1$$
Let angle be m. 
$$\therefore x=2m\quad and\quad y=m$$
from (1),
$$\quad \angle QAP+\angle QCP=180\\ or,m+2m=180\\ or,\quad 3m=180\\ or,\quad m=60$$
$$\angle QCP=2m\\ or,\angle C=2\times 60=120\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad   (2)$$

In a parallelogram ABCD, Sum of any two consecutive angles is supplementary,
$$\quad \therefore \angle C+\angle D=180\\ or,\angle 120+\angle D=180\\ or,\angle D=180-120\\ or,\angle D=60\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad   (3)$$

In a paralleogram ABCD, opposite angles are equal,
$$\therefore \angle C=\angle A=120\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad (from\quad 2)\\ \quad \angle D=\angle B=60\qquad \qquad \qquad \qquad \qquad \quad \quad         (from\quad 3)$$

Given, ABCD is a parallelogram with $$\angle A=4x+20,\quad \angle B=7y,\quad \angle D=6x+3y-8$$

Let the parallelogram be ABCD Perimeter= $$40 $$cm