Question

What is the area  of a pentagon with sides of length 15, 20, 27, 24 and 20 units?

Please Answer. :)

Answer 1

From the given data the pentogon is a irregular pentagon.

Let us assume that, the pentagon diagram with different sides.

The sides of the pentagon are EA = 15, AB = 20, BC = 27, CD = 24, and DE = 20.

To find the area of the irregular pentagon, first devide the pentagon into regular triangle and rectancle.

Then use formulas to find the area of the rectangle and triangle, add up  these two areas to find the area of the irregular pentagon.

Observe the diagram,

Pentagon ABCDE is devided into rectangle ABDE and triangle BCD.

Area of the rectangle ABDE ( A₁) = length * breadth.

                                                     = AB * BD

                                                    = 20 * 15

                                                    = 300 square units.
If the sides of the triangle are a, b, c, and the semiperimeter is s(s = (a+b+c)/2), then the area is √ [s * (s-a) * (s-b) * (s-c)].

Let, BC = a, CD = b, and BD = c.

Perimeter of the triangle BCD ( s ) = ( BC + CD + BD)/2 = ( 27 + 24 + 15 )/2 = 66/2 = 33 units.

Area of the triangle ( A₂ ) = √ [s * (s-a) * (s-b) * (s-c)]

                                        = √ [33 * (33-27) * (33-24) * (33-15)]

                                        = √ [33 * 6 * 9 * 18]

                                        = √32076

                                        = 179.09 square units.

Area of the pentagon ( A ) = A₁ + A₂.

                                          = 300 + 179.09

                                          = 479.09 square units.

Therefore, area of the pentagon is 479.09 square units.