Quadrilaterals - Class 9 Maths - Extra Questions

We know, by angle sum property, the sum of angles of a quadrilateral is $$360^o$$ .

The given angles are $$120^o, 90^o, 72^o, x$$ .

Then, 

$$120^o+90^o+72^o+x=360^o$$ .

$$\Rightarrow$$ $$x=360^o - (120^o + 90^o + 72^o )$$

$$\Rightarrow$$ $$x=360^o - 282^o  =78^{\circ}$$ .

Therefore, the unknown angle is $$x=78^o$$ .

We know, by angle sum property, the sum of angles of a quadrilateral is $$360^o$$ .

The given angles are, $$130^o, 82^o, 40^o$$ .

Let the fourth angle be $$x^o$$ .

Then, $$130^o+82^o+40^o+x^o=360^o$$ .

$$\Rightarrow$$ $$360^o - (130^o + 82^o + 40^o ) =x^{\circ}$$

$$\Rightarrow$$ $$x^o=360^o - 252^o  =108^{\circ}$$ .

Therefore, the fourth angle is $$x^o=108^o$$ .

Given ratio of angles of quadrilateral $$ABCD$$ is $$2:3:5:8$$

Let the angles of quadrilateral $$ABCD$$ be $$2x,3x,5x,8x$$ , respectively.

We know, by angle sum property, the sum of angles of a quadrilateral is $$360^o$$ .

$$\Rightarrow 2x+3x+5x+8x=360^o$$

$$\Rightarrow 18x=360^o$$

$$\therefore x=20^o$$ .

$$\therefore \angle A =2x=2 \times 20^o =40^o$$ ,

$$\angle B =3x=3\times 20^o =60^o$$ ,

$$\angle C =5x=5 \times 20^o =100^o$$

and $$\angle D =8x=8 \times 20^o =160^o$$ .

$$\therefore$$ The smallest angle $$=40^o$$ .

We know, by angle sum property, the sum of angles of a quadrilateral is $$360^o$$ .

The given $$\angle A=120^o$$ and let $$\angle B=\angle C= \angle D=x$$ .

Then,  $$\angle A+\angle B+\angle C+\angle D=360^o$$

$$\implies$$ $$120^o+x+x+x=360^o$$

$$\implies$$ $$120^o+3x=360^o$$

$$\implies$$ $$3x=360^o-120^o$$

$$\implies$$ $$3x=240^o$$

$$\implies$$ $$x=\dfrac{240^o}{3}=80^o$$ .

Therefore, $$\angle B=x=80^o$$ .

We know, by angle sum property, the sum of angles of a quadrilateral is $$360^o$$ .

The given two angles are $$55^o$$ each, other angle $$=150^o$$ .

Let the fourth angle be $$x$$ .

Then, 

$$55^o+55^o+150^o+x=360^o$$ .

$$\Rightarrow$$ $$x=360^o - (55^o + 55^o + 150^o )$$

$$\Rightarrow$$ $$x=360^o - 260^o  =100^{\circ}$$ .

Therefore, the fourth angle is $$x=100^o$$ .

In figure (i),

$$\angle PAS+\angle SAR=180^\circ$$     . . . [Linear pair of angles]

$$x = 180^\circ - 90^\circ$$

$$x= 90^\circ$$

In $$\triangle SAR,$$

$$y = 180^\circ - \left( {90^\circ + 40^\circ} \right)$$       . . . [Angle sum property]

$$y = 50^\circ$$

In parallelogram $$PQRS,$$

$$\angle PSR = 2\times \angle ASR$$

$$\angle PSR= 100^\circ$$

So $$\angle PQR = 100^\circ$$                               [opposite angles]

And hence $$\angle z = 50^\circ$$

In figure (ii),

$$x = 110^\circ$$                          [opposite angle of parallelogram are equal]

and $$x + z = 180^\circ$$      [alternate interior angles]

$$\Rightarrow z = 70^\circ$$

Also $$y + z = 180^\circ$$          [linear pair of angles]

$$\Rightarrow y = 110^\circ$$

$${\textbf{Step  - 1: Finding angle y}}$$

                  $${\text{We know that, sum of interior angle of a quadrilateral is 360}}^\circ$$

                  $$\because {\text{ ABCD is a quadrilateral}}$$

                  $$\therefore {\text{ }}\angle {\text{DAB  +  }}\angle {\text{ABC  +  y  +  }}\angle {\text{CDA  =  360}}^\circ$$

                  $$\Rightarrow {\text{ 46  +  38  +  26  +  y  =  360}}$$

                  $$\Rightarrow {\text{ y  =  250}}^\circ$$

$${\textbf{Step  - 2: Finding angle x}}$$

                  $${\text{x  +  y  =  360}}^\circ$$

                  $$\Rightarrow {\text{ x  +  250  =  360}}$$

                  $$\Rightarrow {\text{ x  =  360  -  250}}$$

                  $$\Rightarrow {\text{ x  =  110}}^\circ$$

$${\textbf{Hence, The value of angle x is 110}}^\circ$$

A rhombus has diagonals perpendicular to each other.1. Draw $$AC = 47 cm$$

2. Draw perpendicular bisector XY of AC intersecting AC at O.
3. Taking O as centre and $$\frac {1}{2} \times 54 = 27  cm$$ as radius
draw two arcs intersecting OX and OY at  B and D respectively.
4. Join AB, BC, CD  and AD.
ABCD is the required rhombus.

$${\textbf{Step  -  1 : State the properties}}{\text{.}}$$

                   $${\text{In a parallelogram ABCD , The opposite angles are equal}}{\text{.}} \Rightarrow \angle {\text{A = }}\angle {\text{C,}}\angle {\text{B = }}\angle {\text{D}}{\text{.}}$$

                      $${\text{The consecutive angles of parallelogram is supplementary}}{\text{.}} \Rightarrow \angle {\text{A + }}\angle {\text{B = 180}}^\circ {\text{,}}\angle {\text{C + }}\angle {\text{D = 180}}^\circ {\text{.}}$$

$${\textbf{Step  -  2 : Solve for (i)}}{\text{.}}$$

                      $${\text{Here }}100^\circ {\text{ + z}} = 180^\circ  [\because {\text{Adjacent angles of a parallelogram]}}{\text{.}}$$

                      $$\Rightarrow {\text{z = 80}}^\circ .$$

                      $${\text{Opposite angles are same}}{\text{.}} \Rightarrow {\text{x = z = 80}}^\circ {\text{,y = 100}}^\circ {\text{.}}$$

$${\textbf{Step  -  3 : Solve for (ii)}}{\text{.}}$$

                    $${\text{Here x}}^\circ {\text{ + 50}}^circ {\text{ = 180}}^\circ {\text{. [Adjacent angle]}}$$

                    $$\Rightarrow {\text{x}}^\circ {\text{ = 130}}^\circ {\text{.}}$$

                    $${\text{x and y are opposite angles}}{\text{.So x = y = 130}}^\circ {\text{.}}$$

                    $${\text{x and z will corresponding angles, z = 130}}^\circ {\text{.   }}$$

$${\textbf{Step  -  4 : Solve for (iii)}}{\text{.}}$$

                  $${\text{Here as x and 90}}^\circ {\text{ vertical opposite angles,So x = 90}}^\circ {\text{.}}$$

                  $${\text{As 90}}^\circ {\text{,30}}^\circ {\text{,y are in a triangle}}{\text{.}} \Rightarrow 90^\circ  + 30^\circ  + {\text{y = 180}}^\circ {\text{.}}$$

                  $$\Rightarrow {\text{y = 60}}^\circ {\text{.}}$$

                  $$y{\text{ and z are alternate angles }}{\text{.}}$$

                  $$\Rightarrow {\text{y  = z = 60}}^\circ {\text{.}}$$

$${\textbf{Step  -  5 : Solve for (iv)}}{\text{.}}$$

                    $${\text{Here z and 80}}^\circ {\text{ are corresponding angles}}{\text{.So z  =  80}}^\circ {\text{.}}$$

                      $$y = 80^\circ {\text{ also,because they are opposite angles}}{\text{.}}$$

                    $$80^\circ {\text{ and x are adjacent angles , so x + 80}}^\circ {\text{ = 180}}^\circ {\text{.}}$$

                    $$\Rightarrow {\text{x}} = 100^\circ .$$

$${\textbf{Step  -  6 : Solve for (v),}}$$

                    $${\text{Here y  =  112}}^\circ {\text{.As they are opposite angles}}{\text{.}}$$

                    $${\text{Now in one triangle there are three angles, x,y and 40}}^\circ {\text{.}} \Rightarrow {\text{x + 112}}^\circ {\text{ + 40}}^\circ {\text{ = 180}}^\circ$$

                    $${\text{x = 180}}^\circ  - 40^\circ  - 112^\circ .$$

                    $$\Rightarrow {\text{x = 28}}^\circ .$$

                    $${\text{Now x and z are alternate angles }}{\text{. So x =  z = 28}}^\circ {\text{.}}$$

$${\textbf{Hence, the values will be  - (i) x = z = 80}}^\circ {\textbf{, y = 100}}^\circ (ii){\textbf{ x = y = z = 130}}^\circ {\textbf{(iii) x = 90}}^\circ {\textbf{, y = z = 60}}^\circ {\text{,}}$$

$${\textbf{(iv) x = 100}}^\circ {\textbf{, y = z = 80}}^\circ {\textbf{,(v) x = z = 28}}^\circ {\textbf{, y = 112}}^\circ .$$

Construction: Join $$AC$$ to meet $$BD$$ in $$O$$ .

Therefore, $$OB = OD$$ and $$OA=OC$$      ...(1)

(Diagonals of a parallelogram bisect each other)

But $$BQ = DP$$       ...given
$$\therefore OB – BQ = OD – DP$$
$$\therefore OQ  = OP$$           ....(2) 

Now, in $$\Box APCQ$$ ,

$$OA = OC$$    ....from (1)

$$OQ = OP$$     ....from (2)

$$\therefore \Box APCQ$$ is a parallelogram.

In $$\triangle APD$$ and $$\triangle CQB$$ , 

$$AD = CB$$      ....opposite sides of a parallelogram 

$$AP = CQ$$       ....opposite sides of a parallelogram
$$DP = BQ$$       ...given
$$\triangle APD \cong \triangle CQB$$      ...By SSS test of congruence

$$\therefore AP = CQ$$      ...c.s.c.t.  

$$AQ = CP$$    ...c.s.c.t.       ...(3)

In $$\triangle AQB$$ and $$\triangle CPD$$ , 
$$AB = CD$$      ....opposite sides of a parallelogram
$$AQ = CP$$     ...from (3)
$$BQ = DP$$      ...given

$$\therefore \triangle AQB \cong \triangle CPD$$      ....By SSS test of congruence

(i) In  $$\triangle ABC$$ , 

$$M$$ is the mid point  of $$AB$$ and $$MD \parallel BC$$ . 

Therefore, $$D$$ is the mid- point of $$AC$$ . 
i.e., $$AD = DC$$      ... (1)

(ii) Since $$MD \parallel BC$$ , corresponding angles has to be equal

$$\angle ADM= \angle ACB$$

$$\angle ADM =90^{\circ}$$

As $$\angle ACB =  90^{\circ}$$  is given

Since $$\angle ADM$$ and $$\angle CDM$$  are angles of a linear pair

But $$\angle ADM + \angle CDM = 180^{\circ}$$

Therfore $$90^{\circ} + \angle CDM = 180^{\circ} \Rightarrow \angle CDM = 90^{\circ}$$

Thus, $$\angle ADM= \angle CMD=90^o$$     ... (2)  

$$\Rightarrow MD \bot AC$$   

(i)  Since $$ABCD$$ is a parallelogram. Therefore, $$DC \parallel AB$$ .

Now $$DC \parallel AB$$ and transversal   $$BD$$ intersects them at $$B$$ and $$D$$ .

Therefore, $$\angle ABD = \angle BDC$$   $$\mid$$ Alternate interior angles

in $$\triangle APB$$ and $$\triangle CQD$$ , we have

$$\angle ABP = \angle QDC$$   ....(alternate interior angles of parallelogram $$ABCD$$ and $$DC \parallel AB$$ ) 

$$\angle APB = \angle CQD$$     ....[each angle $$90^\circ$$ ]

and $$AB = CD$$   $$\mid$$ Opposite. sides of a || gm

Therefore,  by AAS criterion  of congruence

$$\triangle APB \cong \triangle CQD$$

(ii)  Since $$\triangle APB \cong \triangle CQD$$

Therefore, $$AP = CQ$$ $$\mid$$  Since corresponding  parts of congruent triangles are equal

$$ABCD$$ is parallelogram

Using properties of a parallelogram  $$AB \parallel DC$$ and $$AB = DC$$

$$E$$ is the mid-point of $$AB$$
$$AE=\dfrac{1}{2} AB$$    ... (1)

$$F$$ is the mid-point of $$CD$$
Therefore,  $$CF=\dfrac{1}{2 }CD$$
$$CF=\dfrac{1}{2} AB$$ (Since $$CD = AB$$ )    ... (2)

From (1) and (2), we get

$$AE = CF$$
Also,

$$AE \parallel CF$$ (Since $$AB \parallel DC$$ )

Thus, a pair of opposite sides of a quadrilateral   $$AECF$$ are parallel and equal.

Quadrilateral $$AECF$$ is a parallelogram.

$$EC \parallel AF$$
$$EQ \parallel  AP$$ and $$QC \parallel PF$$

In  $$\triangle BPA$$ ,   $$E$$ is the mid-point  of $$BA$$          

$$EQ \parallel AP$$ $$\mid$$  
Using mid point theorem $$BQ = PQ$$          ... (3)

Similarly, by taking   $$\triangle CQD$$ ,  we can prove that

$$DP = QP$$    ... (4)
From (3) and (4), we get

$$BQ = QP = PD$$

Therefore,

$$AF$$ and $$CE$$ trisect the diagonal $$BD$$ .

$${\textbf{Step - 1: Consider a common ratio as a variable and make an equation.}}$$

                    $${\text{It is given that, the four angles of a quadrilateral are in the ratio }}1:2:3:4.$$

                    $${\text{Let }}x{\text{ be the common multiple}}{\text{.}}$$

                    $${\text{Then the angles are }}x,2x,3x{\text{ and }}4x.$$

                    $${\text{As we know that, the sum of angles of quadrilateral is equal to 36}}{{\text{0}}^\circ }.$$

                    $$\therefore x + 2x + 3x + 4x = {360^\circ }$$

                    $$\Rightarrow 10x = {360^\circ }$$

                    $$\Rightarrow x = \dfrac{{{{360}^\circ }}}{{10}}$$

                    $$\Rightarrow x = {36^0} \ldots \left( 1 \right)$$

$${\textbf{Step - 2: Use the obtained value of }}\mathbf x{\textbf{ to find the values of all four angles.}}$$

                    $${\text{(i) }}x = {36^\circ }$$

                    $${\text{(ii) 2}}x = 2 \times {36^\circ }$$

                    $$\Rightarrow 2x = {72^\circ }$$   

                    $${\text{(iii) 3}}x = 3 \times {36^\circ }$$

                    $$\Rightarrow 3x = {108^\circ }$$

                    $${\text{(iv) 4}}x = 4 \times {36^\circ }$$

                    $$\Rightarrow 4x = {144^\circ }$$

                    $${\text{Hence, the measures of angles are}}$$ $${36^\circ },{72^\circ },{108^\circ },{144^\circ }.$$

$${\textbf{Hence, the measures of each angle of quadrilateral is }}\boldsymbol{{36^\circ },{72^\circ },{108^\circ },{144^\circ }}$$

$$(i)$$ $$\angle DAB = \angle CBA$$

Construction :- Produce $$AB$$ to $$E$$ and draw $$CE \parallel DA,\;MO \parallel AB$$  

Proof: So $$AECD$$ is a parallelogram.

In $$\triangle BEC$$  

$$BC=CE$$     ( $$AD=BC=CE$$ ,opposite sides of parallelogram)

$$\angle CBE=\angle CEB$$       (opposite to equal sides)     ---- $$(1)$$

$$\angle ADC =\angle BEC$$     (opposite angle of a parallelogram)   ----- $$(2)$$

$$\angle DAB+\angle ADC=180^{\circ}$$   (sum of co-interior angles) ------ $$(3)$$

$$\angle ABC+\angle EBC=180^{\circ}$$   (L.P.P.)

$$\angle ABC+\angle ADC=180^{\circ}$$   ---- $$(4)$$    from $$(2)$$

From equation $$(3)$$ and $$(4)$$

$$\angle DAB = \angle CBA$$

$$(ii)$$ $$\angle BCD = \angle EBC$$ ---- $$(5)$$    (alternate interior angles)

From equations $$(1),(2),$$ and $$(5),$$

$$\angle ADC = \angle BCD$$    proved

$$(iii)$$ In $$\triangle ABD$$ and $$\Delta BAC$$

$$\angle BAD=\angle ABC$$

$$AB=AB$$    (common)

$$AD=BC$$   (Given)

$$\triangle ABD\cong \triangle BAC$$   (by SAS rule)

$$BD=AC$$      (By C.P.C.T)

$$(iv)$$ In $$\triangle ABD$$

$$MO \parallel AB$$

$$\dfrac{AM}{MD}=\dfrac{BO}{OD}$$   --- $$A$$ ,  (by lemma of B.P.T)

In $$\triangle ADC$$

$$MO \parallel DC$$

$$\dfrac{AM}{MD}=\dfrac{AO}{OC}$$ ---- $$B$$ (by B.P.T.)

From equations $$A$$ and $$B,$$

$$\dfrac{BO}{OD}=\dfrac{AO}{OC}$$   ---- $$C$$

$$\dfrac{{BO}}{{OD}} + 1 = \dfrac{{AO}}{{OC}} + 1$$

$$\dfrac{{BO + DO}}{{OD}} = \dfrac{{AO + OC}}{{OC}}$$

$$\dfrac{{BD}}{{OD}} = \dfrac{{AC}}{{OC}}$$

$$OD=OC$$    ( $$AC=BD$$ )

From equation $$C,$$

$$OB=OA$$     $$(OD=OC)$$

The given three angles of the quadrilateral are $$100^o,\ 60^o$$ and $$80^o$$ .

Let the fourth angle be $$x$$ ,

We know that, the sum of all interior angles of a quadrilateral is $$360^o$$

Hence, $$x + 100^o + 60^o + 80^o = 360^o$$

$$x + 240^o = 360^o$$

$$x = 360^o - 240^o$$

$$x= 120^o$$

Hence, the fourth angle of quadrilateral is $$120^o$$ .

It is known that the length of the tangents drawn from the external point are always equal, so,

$$PB = QB$$

$$QC = RC$$

$$AP = AS$$

$$DS = DR$$

Now,

$$AB + CD$$

$$= AP + PB + DR + RC$$

$$= AS + QB + DS + CQ$$

$$= AS + DS + QB + CQ$$

$$= AD + BC$$

It is known that the sum of the four angles is $$360^\circ$$ . So,

$$2x + 3x + 4x + 6x = 360^\circ$$

$$15x = 360^\circ$$

$$x = 24^\circ$$

Therefore, the measure of each angle is,

$$2x = 2\left( {24} \right) = 48^\circ$$

$$3x = 3\left( {24} \right) = 72^\circ$$

$$4x = 4\left( {24} \right) = 96^\circ$$

$$6x = 6\left( {24} \right) = 144^\circ$$

We have,

In parallelogram $$ABCD$$

$$\angle A={{\left( 6x-17 \right)}^{o}}\,$$ and $$\,\angle C={{\left( x+63 \right)}^{o}}$$

Then, we know that

  $$\angle A=\angle C$$                         $$\therefore$$   It is aparallelogram

$$6x-17=x+63$$

$$6x-x=63+17$$

$$5x=80$$

$$x={{16}^{o}}$$

Then,

$$\angle A=6x-17$$

$$\angle A=6\times 16-17$$

$$\angle A=96-17$$

$$\angle A={{79}^{o}}$$

Hence, this is the answer.

We have given points are

$$A\left( {{x}_{1}},{{y}_{1}} \right)=\left( 3,5 \right)$$

$$B\left( {{x}_{2}},{{y}_{2}} \right)=\left( 6,0 \right)$$

$$C\left( {{x}_{3}},{{y}_{3}} \right)=\left( 1,-3 \right)$$

$$D\left( {{x}_{4}},{{y}_{4}} \right)=\left( -2,2 \right)$$

Then,

Distance between

$$AB=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$$

$$=\sqrt{{{\left( 3-6 \right)}^{2}}+{{\left( 5-0 \right)}^{2}}}$$

$$=\sqrt{9+25}$$

$$AB=\sqrt{34}$$

$$BC=\sqrt{{{\left( {{x}_{2}}-{{x}_{3}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{3}} \right)}^{2}}}$$

$$=\sqrt{{{\left( 6-1 \right)}^{2}}+{{\left( 0+3 \right)}^{2}}}$$

$$=\sqrt{25+9}$$

$$BC=\sqrt{34}$$

$$CD=\sqrt{{{\left( {{x}_{3}}-{{x}_{4}} \right)}^{2}}+{{\left( {{y}_{3}}-{{y}_{4}} \right)}^{2}}}$$

$$=\sqrt{{{\left( 1+2 \right)}^{2}}+{{\left( -3-2 \right)}^{2}}}$$

$$=\sqrt{9+25}$$

$$CD=\sqrt{34}$$

$$DA=\sqrt{{{\left( {{x}_{4}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{4}}-{{y}_{1}} \right)}^{2}}}$$

$$=\sqrt{{{\left( -2-3 \right)}^{2}}+{{\left( 2-5 \right)}^{2}}}$$

$$=\sqrt{25+9}$$

$$DA=\sqrt{34}$$

Then, $$AB=BC=CD=DA$$

All sides are equal.

Hence, $$ABCD$$ is a square.

$${\textbf{Step  - 1: Drawing a quadrilateral and its diagonals}}.$$

                    $${\text{ABCD is a quadrilateral with AC and BD are diagonals intersecting at O}}{\text{.}}$$

                    $${\text{It is given that diagonals bisect each other}}{\text{.}}$$

                    $$\therefore \;\;OA = OC\;{\text{and}}\;OB = OD$$

$${\textbf{Step  - 2: Proving that the quadrilateral is a parallelogram}}.$$

                    $${\text{In}}\;{\text{ }}\Delta AOD{\text{ }}\;{\text{and }}\Delta COB,$$

                    $$OA = OC\;{\text{ }}\;{\text{ }}$$

                    $$\;\angle AOD = \angle COB\;...({\text{Vertically opposite angles}})$$

                    $$OD = OB$$

                    $${\text{Therefore  }}\Delta AOD \cong \Delta COB{\text{  by SAS Congruence rule}}{\text{.}}$$

                    $$\therefore \;\angle OAD = \angle OCB\;$$

                    $${\text{Similarly, we can prove that}}\;\;\Delta AOB \cong \Delta COD$$

                    $$\therefore \angle ABO = \angle CDO\;$$

                    $${\text{So, for lines AB and CD with transversal BD, we can write}}$$

                    $$\angle ABO{\text{ and }}\angle CDO{\text{ are alternate angles and are equal i}}{\text{.e}}{\text{. AB||CD}}{\text{.}}$$

                    $${\text{Similarly, for lines AD and BC, with transversal AC, }}\angle OAD{\text{ and }}\angle OCB{\text{}}$$

                    $${\text{are alternate angles and are equal}}{\text{. Therefore, BC||AD}}{\text{.}}$$

                    $${\text{Thus, in ABCD, both pairs of opposite sides are parallel}}{\text{.}}$$

                    $${\text{Therefore, ABCD is a parallelogram}}{\text{.}}$$

$${\textbf{Hence, Proved that quadrilateral whose diagonals bisect each other is a parallelogram}}{\text{.}}$$

$$\\Let\>mid\>point\>of\>diagonals\>be\>M.\\then\>co-ordinate\>of\>M\>will\>be\>a\>unique\>value\\using\>A\>and\>C,\>co-ordinates\>of\>mid-points\\of\>diagonals\>are\>\left((\frac{3+1}{2}),(\frac{3+7}{2}),(\frac{3+7}{2})\right)\\=M(2,\>5,\>5)\\now\>using\>B\>and\>D,\>co-ordinates\>\\of\>mid-points\>should\>be\\M\left((\frac{x+0}{2}),(\frac{y+6}{2}),(\frac{z+3}{2})\right)\\by\>comparing,\>we\>get\\(\frac{x}{2})=2\>\>\>or\>x=4\\(\frac{y+6}{2})=5\>\>\>or\>y=10-6=4\\(\frac{z+3}{2})=5\>\>\>or\>z=10-3=7\\\therefore\>M(4,\>4,\>7)$$

$$As\>opposite\>angles\>are\>same\>in\>a\>rhombus,\>hence\>\angle\>ADC\>=\>\angle\>ABC\>=\>56^\circ$$

We have,

$$\angle {{x}^{o}}=\angle {{y}^{o}}={{90}^{o}}\,\,\,\left( \text{Every}\,\text{right}\,\text{angle}\,\text{given}\,\text{that} \right)$$

In quadrilateral,

$$\angle {{x}^{o}}+\angle {{y}^{o}}+\angle {{z}^{o}}+{{122}^{o}}={{360}^{o}}$$

$$\Rightarrow {{90}^{o}}+{{90}^{o}}+\angle {{z}^{o}}+{{122}^{o}}={{360}^{o}}$$

$$\Rightarrow \angle {{z}^{o}}={{360}^{o}}-{{302}^{o}}$$

$$\Rightarrow \angle {{z}^{o}}={{58}^{o}}$$

Hence, this is the answer.

We have,

ABCD is a quadrilateral.

Given that,

$$\angle C={{70}^{o}},\,\,\angle D={{50}^{o}}$$

So,

$$\angle A+\angle B+\angle C+\angle D={{360}^{o}}$$

$$\left( A+B \right)={{360}^{o}}-\left( C+D \right)$$

$$\Rightarrow \left( A+B \right)={{360}^{o}}-\left( {{70}^{o}}+{{50}^{o}} \right)$$

$$\Rightarrow \left( A+B \right)={{360}^{o}}-{{120}^{o}}$$

$$\Rightarrow \left( A+B \right)={{240}^{o}}$$

According to given question,

$$\angle OAB=\dfrac{\angle A}{2}$$

$$\angle OAB=\dfrac{\angle B}{2}$$

Now,

$$\angle OAB+\angle OBA=\dfrac{\angle A+\angle B}{2}={{120}^{o}}$$

$$\Rightarrow \angle AOB+{{120}^{o}}={{180}^{o}}$$

$$\Rightarrow \angle AOB={{180}^{o}}-{{120}^{o}}$$

$$\Rightarrow \angle AOB={{60}^{o}}$$

Hence, this is the answer.

We have,

ABCD is a quadrilateral.

Given that,

$$\angle C={{70}^{o}},\,\,\angle D={{50}^{o}}$$

So,

$$\angle A+\angle B+\angle C+\angle D={{360}^{o}}$$

$$\left( A+B \right)={{360}^{o}}-\left( C+D \right)$$

$$\Rightarrow \left( A+B \right)={{360}^{o}}-\left( {{70}^{o}}+{{50}^{o}} \right)$$

$$\Rightarrow \left( A+B \right)={{360}^{o}}-{{120}^{o}}$$

$$\Rightarrow \left( A+B \right)={{240}^{o}}$$

According to given question,

$$\angle OAB=\dfrac{\angle A}{2}$$

$$\angle OBA=\dfrac{\angle B}{2}$$

Now,

$$\angle OAB+\angle OBA=\dfrac{\angle A+\angle B}{2}={{120}^{o}}$$

$$\Rightarrow \angle AOB+{{120}^{o}}={{180}^{o}}$$

$$\Rightarrow \angle AOB={{180}^{o}}-{{120}^{o}}$$

$$\Rightarrow \angle AOB={{60}^{o}}$$

Hence, this is the answer.

We have,

$$A\left( {{x}_{1}},{{y}_{1}} \right)=\left( 2,3 \right)$$

$$B\left( {{x}_{2}},{{y}_{2}} \right)=\left( -2,2 \right)$$

$$C\left( {{x}_{3}},{{y}_{3}} \right)=\left( -1,-2 \right)$$

$$D\left( {{x}_{4}},{{y}_{4}} \right)=\left( 3,-1 \right)$$

Now,

Using distance formula we get,

$$AB=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$$

$$\Rightarrow AB=\sqrt{{{\left( 2+2 \right)}^{2}}+{{\left( 3-2 \right)}^{2}}}$$

$$\Rightarrow  AB=\sqrt{17}$$

$$BC=\sqrt{{{\left( {{x}_{2}}-{{x}_{3}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{3}} \right)}^{2}}}$$

$$\Rightarrow  BC=\sqrt{{{\left( -2+1 \right)}^{2}}+{{\left( 2+2 \right)}^{2}}}$$

$$\Rightarrow  BC=\sqrt{17}$$

$$CD=\sqrt{{{\left( {{x}_{3}}-{{x}_{4}} \right)}^{2}}+{{\left( {{y}_{3}}-{{y}_{4}} \right)}^{2}}}$$

$$\Rightarrow  CD=\sqrt{{{\left( -1-3 \right)}^{2}}+{{\left( -2+1 \right)}^{2}}}$$

$$\Rightarrow  CD=\sqrt{17}$$

$$DA=\sqrt{{{\left( {{x}_{4}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{4}}-{{y}_{1}} \right)}^{2}}}$$

$$\Rightarrow  DA=\sqrt{{{\left( 3-2 \right)}^{2}}+{{\left( -1-3 \right)}^{2}}}$$

$$\Rightarrow  DA=\sqrt{17}$$

Now diagonal  $$AC=\sqrt{{{\left( {{x}_{1}}-{{x}_{3}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{3}} \right)}^{2}}}$$

$$\Rightarrow AC=\sqrt{{{\left( 2+1 \right)}^{2}}+{{\left( 3+2 \right)}^{2}}}$$

$$\Rightarrow AC=\sqrt{34}$$

And diagonal $$BD=\sqrt{{{\left( {{x}_{2}}-{{x}_{4}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{3}} \right)}^{4}}}$$

$$\Rightarrow BD=\sqrt{{{\left( -2-3 \right)}^{2}}+{{\left( 2+1 \right)}^{2}}}$$

$$\Rightarrow BD=\sqrt{34}$$

So,

Sides $$AB=BC=CD=DA$$ and diagonals $$AC=BD$$

Hence, the vertices are of a square $$ABCD$$ .

We have,

$$Given\,that:\,\Delta ABC$$

Where $$D,E,F$$ are mid points of $$BC,CA$$ and $$AB$$ respectively.

To prove:- $$BDEF$$ is a parallelogram

Proof:- In $$\Delta ABC$$ ,

$$F$$ is mid point of $$AB$$ ,

$$D$$ is the midpoint of $$BC$$

$$E$$ is the midpoint of $$CA$$

$$\therefore FE\parallel BC\Rightarrow \therefore FE\parallel BD$$

$$\therefore DE\parallel AB\Rightarrow \therefore DE\parallel FB$$

$$\therefore FE\parallel BD\,\,and\,\,DE\parallel FB$$

So, in $$BDEF$$ ,

Both pairs of opposite sides are parallel,

$$\therefore BDEF$$ is a parallelogram.

Hence proved.

Now,

$$BDEF$$ is a parallelogram

$$\therefore \Delta DBF\cong \Delta DEF$$

$$ar\,\left( \Delta DBF \right)=ar\left( \Delta DEF \right)\,\,......\,\,\left( 1 \right)$$

Similarly, we can prove that,

$$FDCE$$ is a parallelogram

$$\therefore \Delta DEC\cong \Delta DEF$$

$$ar\left( \Delta DEC \right)=ar\left( \Delta DEF \right)\,\,......\,\,\left( 2 \right)$$

Similarly, we can prove that,

$$AFDE$$ is a parallelogram

$$\therefore \Delta AFE\cong \Delta DEF$$

$$ar\left( \Delta AFE \right)=ar\left( \Delta DEF \right)\,\,......\,\,\left( 3 \right)$$

From equation (1), (2), (3) to, and we get,

$$ar\left( \Delta DBF \right)=ar\left( \Delta DEC \right)=ar\left( \Delta AFE \right)=ar\left( \Delta DEF \right)$$

Now,

$$ar\left( \Delta DBF \right)+ar\left( \Delta DEC \right)+ar\left( \Delta AFE \right)+ar\left( \Delta DEF \right)=ar\left( \Delta ABC \right)$$

$$\Rightarrow ar\left( \Delta DEF \right)+ar\left( \Delta DEF \right)+ar\left( \Delta DEF \right)+ar\left( \Delta DEF \right)=ar\left( \Delta ABC \right)$$

$$\Rightarrow 4ar\left( \Delta DEF \right)=ar\left( \Delta ABC \right)$$

$$\Rightarrow ar\left( \Delta DEF \right)=\dfrac{1}{4}ar\left( \Delta ABC \right)$$

Hence proved.