Question

Answer for the 7th question step by step

Answer 1

Answer:

Let,

• The Distance between point A and B = $\sf d_1$

• The Distance between point B and C = $\sf d_2$

• The Distance between point C and D = $\sf d_3$

• The Distance between point D and A = $\sf d_4$

☯ $\underline{\boldsymbol{According\: to \:the\: Question\:now :}}$

$\dashrightarrow\:\:\sf d_1 =\sqrt{(8 - 5)^{2} +(3 + 1) ^{2}}$

$\dashrightarrow\:\:\sf d_1 =\sqrt{(3)^{2} +(4) ^{2}}$

$\dashrightarrow\:\:\sf d_1 = \sqrt{ 9 + 16}$

$\dashrightarrow\:\:\sf d_1 = \sqrt{25}$

$\dashrightarrow \: \: \blue{\sf d_1 = 5 \: units}$

____________________...

$\dashrightarrow\:\:\sf d_2 =\sqrt{(4 - 8)^{2} +(0 - 3) ^{2}}$

$\dashrightarrow\:\:\sf d_2 =\sqrt{( - 4)^{2} +( - 3) ^{2}}$

$\dashrightarrow\:\:\sf d_2 =\sqrt{16 + 9}$

$\dashrightarrow\:\:\sf d_2 =\sqrt{25}$

$\dashrightarrow \: \: \gray{ \sf d_2 = 5 \: units}$

____________________...

$\dashrightarrow\:\:\sf d_3 =\sqrt{(1 - 4)^{2} +( - 4 - 0) ^{2}}$

$\dashrightarrow\:\:\sf d_3 =\sqrt{( - 3)^{2} +( - 4) ^{2}}$

$\dashrightarrow\:\:\sf d_3 =\sqrt{9 +16}$

$\dashrightarrow\:\:\sf d_3 =\sqrt{25}$

$\dashrightarrow \: \: \red{ \sf d_3 = 5 \: units}$

____________________....

$\dashrightarrow\:\:\sf d_4 =\sqrt{(5 - 1)^{2} +( - 1 + 4) ^{2}}$

$\dashrightarrow\:\:\sf d_4 =\sqrt{( 4)^{2} +(3) ^{2}}$

$\dashrightarrow\:\:\sf d_4 =\sqrt{16 +9}$

$\dashrightarrow\:\:\sf d_4 =\sqrt{25}$

$\dashrightarrow \: \: \green{\sf d_4= 5 \: units}$

$\sf d_1 = d_2 + d_3 + d_4\:$ this condition is possible when Quadrilateral is rohmbus or square.

Now, Let the distance B and D be $\sf D_1$

and A and C be $\sf D_2$ to be rohmbus $\sf D_1\:$ ≠ $\sf D_2$

$:\implies\sf D_1 =\sqrt{(1 - 8)^{2} +( - 4 - 3) ^{2}}$

$:\implies\sf D_1 =\sqrt{( - 7)^{2} +( - 7) ^{2}}$

$:\implies\sf D_1 =\sqrt{98}$

$:\implies \pink{\sf D_1 =7 \sqrt{2} \: units}$

__________________...

$:\implies\sf D_2 =\sqrt{(4 - 5)^{2} +( 0 + 1) ^{2}}$

$:\implies\sf D_2 =\sqrt{(- 1)^{2} +( 1) ^{2}}$

$:\implies\sf D_2 =\sqrt{1 + 1}$

$:\implies \orange{\sf D_1 = \sqrt{2} \: units}$

This means that $\sf D_1\:$ ≠ $\sf D_2$ hence Quadrilateral is rohmbus.