Question

Find the perimeter and area of a quadrilateral ABCD in which BC = 12 cm, CD = 9 cm, BD = 15cm, DA = 17 cm and ∠ ABD = 90°.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

ABCD is a quadrilateral such that BC = 12 cm, CD = 9 cm, BD = 15cm, DA = 17 cm and ∠ ABD = 90°.

Now, In right triangle ABD

We have,

• AD = 17 cm

• BD = 15 cm

Using Pythagoras Theorem, we have

$\rm :\longmapsto\: {AD}^{2} = {AB}^{2} + {BD}^{2}$

$\rm :\longmapsto\: {17}^{2} = {AB}^{2} + {15}^{2}$

$\rm :\longmapsto\: 289 = {AB}^{2} + 225$

$\rm :\longmapsto\: {AB}^{2} = 289 - 225$

$\rm :\longmapsto\: {AB}^{2} = 64$

$\bf\implies \:AB = 8 \: cm$

Now, We know,

Perimeter of quadrilateral ABCD means sum of all sides of quadrilateral ABCD.

So,

$\red{\sf :\longmapsto\:Perimeter \: of \: quadrilateral \: ABCD}$

$\rm \:  =  \: AB + BC + CD + DA$

$\rm \:  =  \: 8 + 12 + 9 + 17$

$\rm \:  =  \: 46 \: cm$

$\red{\rm\implies \:\boxed{\tt{ \sf \:Perimeter \: of \: quadrilateral \: ABCD} = 46 \: cm}}$

Now, In triangle BDC

We have

$\rm :\longmapsto\:BD = 15 \: cm$

$\rm :\longmapsto\:DC = 9 \: cm$

$\rm :\longmapsto\:BC = 12\: cm$

Now, Consider

$\rm :\longmapsto\: {BC}^{2} + {CD}^{2} = {9}^{2} + {12}^{2} = 81 + 144 = 225 = {BD}^{2}$

$\rm\implies \: \triangle \: BDC \: is \: right \: angle \triangle \: and\: \angle \: C = 90 \degree$

So,

$\purple{\rm :\longmapsto\:Area_{(ABCD)}}$

$\rm \:  =  \: Area_{(ABD)} + Area_{(BCD)}$

$\rm \:  =  \: \dfrac{1}{2} \times \: AB \times BD + \dfrac{1}{2} \times BC \times DC$

$\rm \:  =  \: \dfrac{1}{2} \times \: 8 \times 15 + \dfrac{1}{2} \times 12 \times 9$

$\rm \:  =  \: 60 + 54$

$\rm \:  =  \: 114 \: {cm}^{2}$

$\\ \purple{\rm\implies \:\boxed{\tt{ \: \: Area_{(ABCD)}  =  \: 114 \: {cm}^{2} \: \: }}}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

$\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Breadth\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Breadth\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}$

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

ABCD is a quadrilateral such that BC = 12 cm, CD = 9 cm, BD = 15cm, DA = 17 cm and ∠ ABD = 90°.

Now, In right triangle ABD

We have,

AD = 17 cm

BD = 15 cm

Using Pythagoras Theorem, we have

$\rm :\longmapsto\: {AD}^{2} = {AB}^{2} + {BD}^{2}$

$\rm :\longmapsto\: {17}^{2} = {AB}^{2} + {15}^{2}$

$\rm :\longmapsto\: 289 = {AB}^{2} + 225$

$\rm :\longmapsto\: {AB}^{2} = 289 - 225$

$\rm :\longmapsto\: {AB}^{2} = 64$

$\bf\implies \:AB = 8 \: cm$

Now, We know,

Perimeter of quadrilateral ABCD means sum of all sides of quadrilateral ABCD.

So,

$\red{\sf :\longmapsto\:Perimeter \: of \: quadrilateral \: ABCD}$

$\rm \:  =  \: AB + BC + CD + DA$

$\rm \:  =  \: 8 + 12 + 9 + 17$

$\rm \:  =  \: 46 \: cm$

$\red{\rm\implies \:\boxed{\tt{ \sf \:Perimeter \: of \: quadrilateral \: ABCD} = 46 \: cm}}$

Now, In triangle BDC

We have

$\rm :\longmapsto\:BD = 15 \: cm$

$\rm :\longmapsto\:DC = 9 \: cm$

$\rm :\longmapsto\:BC = 12\: cm$

Now, Consider

$\rm :\longmapsto\: {BC}^{2} + {CD}^{2} = {9}^{2} + {12}^{2} = 81 + 144 = 225 = {BD}^{2}$

$\rm\implies \: \triangle \: BDC \: is \: right \: angle \triangle \: and\: \angle \: C = 90 \degree$

So,

$\purple{\rm :\longmapsto\:Area_{(ABCD)}}$

$\rm \:  =  \: Area_{(ABD)} + Area_{(BCD)}$

$\rm \:  =  \: \dfrac{1}{2} \times \: AB \times BD + \dfrac{1}{2} \times BC \times DC$

$\rm \:  =  \: \dfrac{1}{2} \times \: 8 \times 15 + \dfrac{1}{2} \times 12 \times 9$

$\rm \:  =  \: 60 + 54$

$\rm \:  =  \: 114 \: {cm}^{2}$

$\\ \purple{\rm\implies \:\boxed{\tt{ \: \: Area_{(ABCD)}  =  \: 114 \: {cm}^{2} \: \: }}}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

$\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Breadth\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Breadth\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}$