Question

ABCD is a pentagon . AX , BC and ED are perpendicular to CD find the area of pentagon

Answer 1

$\sf {\orange{\underline{\purple{\underline{Given\:that:}}}}}$

• ABCD is a pentagon.
• AX, BC and ED are perpendicular to CD.

$\sf {\orange {\underline {\pink{\underline{To\:find:}}}}}$

• The area of the Pentagon.

$\sf {\pink {\underline {\blue{\underline{Solution:}}}}}$

The Solution can be found out by two different methods and they are as followed:

$\sf\underline\red{METHOD 1}$

First divide the given pentagon into two trapeziums AXCB and AXDE.

→Area of trapezium ABCX$\rm{=}\dfrac{1}{2}\times{4.5}\times{(22+6)cm}^{2}$

$\rm{=}\dfrac{1}{2}\times{4.5}\times{28cm}^{2}$

$\rm{=63cm}^{2}$

→Area of trapezium AXDE$\rm{=}\dfrac{1}{2}\times{5.5}\times{(22+6)cm}^{2}$

$\rm{=}\dfrac{1}{2}\times{5.5}\times{28cm}^{2}$

$\rm{=77cm}^{2}$

Thus ,the area of Pentagon ABCDE $\rm{=(63+77)cm}^{2}{=140cm}^{2}$

$\sf\underline\red{METHOD~2}$

In this method ,first divide the Pentagon into triangle and name it as ABE and a rectangle and name it BCDE

→Area if triangle ABE$\rm{=}\dfrac{1}{2}\times{10×16}{cm}^{2}$

$\rm{=80cm}^{2}$

→Area if rectangle BCDE$\rm{=(10×6)cm}^{2}{=60cm}^{2}$

→Area of Pentagon ABCDE $\rm{=(80+60)cm}^{2}{=140cm}^{2}$

Therefore,by both the methods,the area we found of the Pentagon $\rm{=140cm}^{2}$

Answer 2

$\sf {\orange{\underline{\purple{\underline{Given\:that:}}}}}$

ABCD is a pentagon.

AX, BC and ED are perpendicular to CD.

$\sf {\orange {\underline {\pink{\underline{To\:find:}}}}}$

The area of the Pentagon.

$\sf {\pink {\underline {\blue{\underline{Solution:}}}}}$

The Solution can be found out by two different methods and they are as followed:

$\sf\underline\red{METHOD 1}$

First divide the given pentagon into two trapeziums AXCB and AXDE.

→Area of trapezium ABCX$\rm{=}\dfrac{1}{2}\times{4.5}\times{(22+6)cm}^{2}$

$\rm{=}\dfrac{1}{2}\times{4.5}\times{28cm}^{2}$

$\rm{=63cm}^{2}$

→Area of trapezium AXDE$\rm{=}\dfrac{1}{2}\times{5.5}\times{(22+6)cm}^{2}$

$\rm{=}\dfrac{1}{2}\times{5.5}\times{28cm}^{2}$

$\rm{=77cm}^{2}$

Thus ,the area of Pentagon ABCDE $\rm{=(63+77)cm}^{2}{=140cm}^{2}$

$\sf\underline\red{METHOD~2}$

In this method ,first divide the Pentagon into triangle and name it as ABE and a rectangle and name it BCDE

→Area if triangle ABE$\rm{=}\dfrac{1}{2}\times{10×16}{cm}^{2}$

$\rm{=80cm}^{2}$

→Area if rectangle BCDE$\rm{=(10×6)cm}^{2}{=60cm}^{2}$

→Area of Pentagon ABCDE $\rm{=(80+60)cm}^{2}{=140cm}^{2}$

Therefore,by both the methods,the area we found of the Pentagon $\rm{=140cm}^{2}$