Understanding Quadrilaterals - Class 8 Maths - Extra Questions

$${\textbf{Step -1: Explaining what a convex quadrilateral is}}{\text{.}}$$

                  $${\text{A convex quadrilateral is a four - sided polygon with each of its interior angles being }}$$

                   $${\text{less than $180^\circ$  and whose diagonals lie inside the closed shape}}{\text{. }}$$

                   $${\text{The value of interior angles in a convex quadrilateral must be acute, right, or obtuse}}{\text{.}}$$

$${\textbf{Step -2: Explaining why a rectangle is a convex quadrilateral}}{\text{.}}$$

                   $${\text{A rectangle is a four - sided polygon with the value of each of its interior angles}}$$

                   $${\text{of $90^\circ$  which is less than $180^\circ$ , and both the diagonals lie inside the}}$$

                   $${\text{rectangle}}{\text{. Thus a rectangle is a convex quadrilateral}}{\text{.}}$$

$${\textbf{Thus, a rectangle is a convex quadrilateral}}{\text{.}}$$

Consider that  ABCD be the parallelogram with ∠A = ∠B.

We know: Sum of adjacent angles = 180°

Then,

∠A + ∠B = 180º

2∠A = 180º (∠A = ∠B)

∠A = 90º

∠B = ∠A = 90º

∠C = ∠A = 90º (Opposite angles)

Similarly,

∠D = ∠B = 90º (Opposite angles)

As we know,

Opposite angles of a parallelogram are equal.

$$\implies \angle A =\angle C =65^{\circ }$$

 and let us say that, $$\angle B = \angle D =\theta$$

$$\\65^{\circ}+\theta+65^{\circ}+\theta=360^{\circ}\\        2\theta+130^{\circ}=360^{\circ}\\2\theta=230^{\circ}\\\therefore\theta=(\frac{230^{\circ}}{2})=115^{\circ}$$

$$\text{So angles are } 65^{\circ},65^{\circ},115^{\circ}$$ and $$115^{\circ}$$ .

Construction: 

Draw $$DE \bot AC$$ and $$BF \bot AC$$ .

In $$\triangle DEO$$  and $$\triangle BFO$$

$$\angle DEO= \angle BFO$$          ....By construction 

$$\angle DOE= \angle BOF$$          ....Vertically opposite angles 
$$OD = OB$$                     .....Given

$$\triangle DOE \cong \triangle BOF$$     ...SAA test of congruence    ....(1)

$$\therefore DE=BF$$          .... CSCT        ....(2)

and 

$$OE = OF$$            ... CSCT  ....(3)

In $$\triangle DCE$$  and $$\triangle BFA$$  

$$\angle DEC= \angle BFA$$          ..... By construction

$$DC = AB$$         ....Given
$$DE = BF$$          ....From (2)

$$\triangle DEC \cong \triangle BFA$$    .....(RHS) Right Hypotenuse Side test    ....(4)

$$\therefore EC = AF$$       ......C.S.C.T.      ....(5)

Similarly, we can prove $$\triangle DOA \cong \triangle BOC$$    ....(6)

Adding (1) and (2), we get 

$$OE+EC = OF+AF$$

$$\therefore OC = OA$$    .....(7)

(i) Adding (1) & (4), we get,

$$A(\triangle DOE) + A (\triangle DEC) = A (\triangle BOF) + A (\triangle BFA)$$

$$\therefore A(\triangle DOC) = A (\triangle BOA)$$      ....(8)

(ii) Adding (6) and (8), we get

$$A (\triangle DOC) + A (\triangle DOA) = A (\triangle BOA) + A (\triangle BOC)$$
$$\Rightarrow A (\triangle ACD) = A (\triangle ACB)$$

(iii) So, from (7) and given,

$$\Box ABCD$$ is a parallelogram       .....(Diagonals bisect each other)

and $$DA \parallel CB$$      ....Opp. sides of parallelogram are parallel to each other

The adjacent angles are given as $$\left( {2x + 7} \right)^\circ$$ and $$\left( {3x - 2} \right)^\circ$$ .

Since, the sum of the adjacent angles of a parallelogram is $$180^\circ$$ , then,

$$2x + 7 + 3x - 2 = 180$$

$$5x + 5 = 180$$

$$5x = 175$$

$$x = 35$$

$$\left( {2x + 7} \right)^\circ  = \left( {2\left( {35} \right) + 7} \right)^\circ$$

$$= 77^\circ$$

$$\left( {3x - 2} \right)^\circ  = \left( {3\left( {35} \right) - 2} \right)^\circ$$

$$= 103^\circ$$

And since the opposite angles of a parallelogram is equal, so, the angles will be $$77^\circ ,103^\circ ,77^\circ ,103^\circ$$ .

Let $$x^\circ$$ be the smallest angle of the parallelogram.

According to the question, the largest angle is $$\left( {2x^\circ  - 24^\circ } \right)$$ .

It is known that the sum of the adjacent angles of the parallelogram is $$180^\circ$$ . Then,

$$x^\circ  + 2x^\circ  - 24^\circ  = 180^\circ$$

$$3x^\circ  = 180^\circ  + 24^\circ$$

$$3x^\circ  = 204^\circ$$

$$x^\circ  = 68^\circ$$

Therefore, the largest angle is $$\left( {2\left( {68^\circ } \right) - 24^\circ } \right) = 112^\circ$$ .

The opposite angles of a parallelogram are given as $$\left( {3x - 2} \right)^\circ$$ and $$\left( {50 - x} \right)^\circ$$ .

Since, the opposite angles of a parallelogram are equal, then,

$$3x - 2 = 50 - x$$

$$3x + x = 50 + 2$$

$$4x = 52$$

$$x = 13$$

$$\left( {3x - 2} \right)^\circ  = \left( {3\left( {13} \right) - 2} \right)^\circ$$

$$= 37^\circ$$

Therefore, one angle is $$37^\circ$$ , and its opposite angle is also $$37^\circ$$ .

It is known that the sum of adjacent angles of a parallelogram is $$180^\circ$$ , so,

$${\rm{Second}}\;{\rm{angle}} = 180^\circ  - 37^\circ$$

$$= 143^\circ$$

Thus, the angles are, $$37^\circ$$ , $$143^\circ$$ , $$37^\circ$$ and $$143^\circ$$ .

Refer attached image, given ABCD is a parallelogram

$$AP\parallel CR$$

$$CP=PD$$

Since,  $$\triangle APD\sim \triangle RCD\quad \quad (AAA)$$  

$$\Rightarrow \dfrac { DA }{ DR } =\dfrac { DP }{ DC } \quad \quad \quad \quad \quad \quad (CPCT)$$

$$\Rightarrow \dfrac { DA }{ DR } =\dfrac { 1 }{ 2 } \quad \quad \quad \quad \quad \quad (As\quad P\quad is\quad the\quad midpoint\quad of\quad CD)$$

$$\Rightarrow 2DA=DR$$

$$\Rightarrow 2DA=DA+AR$$

$$\Rightarrow DA=AR$$

Hence Proved DA=AR

Again as,  $$\angle APD=\angle QCD\quad \& \quad \angle BQC=\angle APD\quad (AP\parallel QC)$$ ,

$$\Rightarrow \angle APD=\angle QCD=\angle BQC$$

&

$$\angle RDC=\angle CBQ\quad \quad \quad (Opposite\quad of\quad \angle \quad is\quad \parallel gm)$$

&

$$\angle BCQ=\angle CRD\quad \quad \quad (Properties\quad of\quad \parallel gm)$$

$$\Rightarrow \triangle BQC\sim \triangle DCR\quad \quad (AAA)$$

Now, 

As,

$$\Rightarrow \triangle APD\cong \triangle CBQ$$

$$PD=BQ$$

or  $$\dfrac { CD }{ 2 } =BQ\quad \quad \quad (As\quad P\quad is\quad the\quad midpoint\quad of\quad CD)$$

or  $$\dfrac { AB }{ 2 } =BQ\quad \quad \quad (In\quad \parallel gm\quad AB=CD)$$

$$\Rightarrow$$  Q is the mid-point of AB

Again, 

as,

$$\triangle BQC\sim \triangle DCR$$

$$\dfrac { BQ }{ CQ } =\dfrac { CD }{ CR }$$

$$\Rightarrow 2CQ=CR\quad (As\quad CD=2BQ)$$

$$\Rightarrow 2CQ=CQ+QR$$

$$\Rightarrow CQ=QR$$

Hence Proved CQ=QR

Given that ABCD is a trapezium,

$$AB\parallel CD$$ and $$AD=BC$$

 To prove: $$\angle A=\angle B$$

Construction: Draw $$CE\parallel AD$$ and extend AB to intersect CE at E.

Poof:  As AECD is a parallelogram.

By constriction, CE is parallel to AD and AE is parallel to CD,

$$\therefore \,\,AD=EC$$

But, AD=BC (Given)

$$\therefore \,\,BC=EC$$

$$\Rightarrow \,\angle 3=\angle 4$$ (Angle opposite to equal side are equal)

Now, $$\angle 1+\angle 4={{180}^{0}}$$ (consecutive interior angle of parallelogram )

And $$\angle 2+\angle 3={{180}^{0}}$$ (linear pair )

$$\Rightarrow \angle 1+\angle 4=\angle 2+\angle 3$$    $$\Rightarrow \angle 1=\angle 2\,\,\,\,\,\,\,\,\,\,\,\,\because \angle 3=\angle 4$$ (so get cancelled with each other )

$$\Rightarrow \,\angle A=\angle B$$

Hence, proved .

We have,

A parallelogram $$RING$$ , if $$\angle R={{70}^{o}}$$

Then, solve that,

$$\angle I=?$$

$$\angle N=?$$

$$\angle G=?$$

We know that,

Adjacent angles of parallelogram is supplementary.

Then,

$$\angle R+\angle I={{180}^{o}}$$

$${{70}^{o}}+\angle I={{180}^{o}}$$

$$\angle I={{180}^{o}}-{{70}^{o}}$$

$$\angle I={{110}^{o}}$$

But we know that,

Opposite angles of parallelogram are equal.

Then,

$$\angle R=\angle N$$

$${{70}^{o}}=\angle N$$

$$\angle N={{70}^{o}}$$

And

$$\angle I=\angle G$$

$${{110}^{o}}=\angle G$$

$$\angle G={{110}^{o}}$$

We know that the sum of all angles of the pentagon

$$={{540}^{\circ }}$$

Therefore,

$${{\left( x-5 \right)}^{\circ }}+{{\left( x-6 \right)}^{\circ }}+{{\left( 2x-7 \right)}^{\circ }}+{{x}^{\circ }}+{{\left( 2x-2 \right)}^{\circ }}={{540}^{\circ }}$$

$$7{{x}^{\circ }}-{{20}^{\circ }}={{540}^{\circ }}$$

$$7{{x}^{\circ }}={{560}^{\circ }}$$

$${{x}^{\circ }}={{80}^{\circ }}$$

Hence, this is the answer.

$${\textbf{Step -1: Draw a parallelogram ABCD.}}$$           

                  $${\text{The measurement of two adjacent angles of a parallelogram are in the ratio of 3:2}}{\text{.}}$$  

                  $${\text{Let two adjacent angles be A and B. }}$$

                  $${\text{Let}}$$ $$\angle A = 3x$$    $${\text{and}}$$ $$\angle B = 2x$$   $$\quad \quad \text{......eqn(i)}$$

                  $${\text{Since angles}}$$ $$A$$ $${\text{and}}$$ $$B$$ $${\text{are}}$$ $${\text{Supplementary angles}}{\text{.}}$$

                  $$\therefore \angle A + \angle B = {180^\circ }$$

                  $$\Rightarrow 3x + 2x = {180^\circ }$$

                  $$\Rightarrow 5x = {180^\circ }$$

                  $$\Rightarrow x = \dfrac{{{{180}^\circ }}}{5}$$

                  $$\Rightarrow x = {36^\circ }$$

$${\textbf{Step -2: Find the values of angles }}$$ $$\mathbf A$$ $${\textbf{and}}$$ $$\mathbf B$$ $${\textbf{using the value of}}$$ $$\mathbf x.$$

                  $$\angle A = 3\times{{{36}^\circ }}$$             $$\quad \textbf{[from eqn(i)]}$$

                  $$\Rightarrow \angle A = {108^\circ }$$

                  $${\text{and}}$$ $$\angle B = 2\times {{{36}^\circ }}$$     $$\quad \textbf{[from eqn(i)]}$$

                  $$\Rightarrow \angle B = {72^\circ }$$

                  $${\text{As we know that opposite angles in a parallelogram are equal}}{\text{.}}$$

                  $$\Rightarrow \angle C = {108^\circ } = \angle A$$       $$\quad \left[ {\mathbf \because {\textbf{ Opposite angles of parallelogram}}} \right]$$

                  $$\Rightarrow \angle D = {72^\circ } = \angle B$$   $$\quad \quad \left[ {\because {\textbf{Opposite angles of parallelogram}}} \right]$$

$${\textbf{Hence, angles of the parallelogram are 10}}{{\textbf{8}}^\mathbf {\circ} }\mathbf{,{72\boldsymbol{^\circ} },}{\textbf{10}}{{\textbf{8}}^\mathbf\circ }$$ $${\textbf{and 7}}{{\textbf{2}}^\circ }\textbf.$$