The convex pentagon ABCDE has a right angle at B, AB = BC and CD = DE = EA = - Maths - Quadrilaterals

Question

The convex pentagon ABCDE has a right angle at B, AB = BC and CD
= DE = EA = 1 cm, AC = 2cm If AC and ED are parallel find the area
of the pentagon

Ankush Jain · Accepted Answer

In âˆ†ABC, by Pythagoras Theorem (AC)2 = (AB)2 + (BC)2 â‡’(2)2 = 2(AB)2 [ As AB = BC ] $\Rightarrow AB = \sqrt{2} cm = BC$ Now, Construct a line DF parallel to CE, Now, DE â•‘ AC and DF â•‘ CE âˆ´ DECF is a â•‘gm in which DE = EC = CF = DF = 1 cm and AD = DF = AF = 1 cm So, âˆ† ADF is a equilateral triangle $∴ Area of Δ ADF = \frac{\sqrt{3}}{4} \cdot (1)^{2} = \frac{\sqrt{3}}{4} Also, area of Δ ADF = \frac{1}{2} \times base \times height \Rightarrow \frac{\sqrt{3}}{4} = \frac{1}{2} \times 1 \times height \Rightarrow height = \frac{\sqrt{3}}{2} cm$ $∴ Area of parallelogram = base \times height = 1 \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2} {cm}^{2}$ $Area of right Δ ABC = \frac{1}{2} \times base × height = \frac{1}{2} \times 1 \times 1 = \frac{1}{2} {cm}^{2}$ âˆ´ Area of convex pentagon = Area of âˆ†ABC + Area of âˆ†ADF + Area of â•‘gm CDEF $= \frac{1}{2} + \frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{2} = \frac{2 + \sqrt{3} + 2 \sqrt{3}}{4} = \frac{3 \sqrt{3} + 2}{4} {cm}^{2}$

The convex pentagon ABCDE has a right angle at B, AB = BC and CD = DE = EA = 1 cm, AC = 2cm. If AC and ED are parallel find the area of the pentagon