Question

In the figure, the perimeter of hexagon ABCDEF is approximately ___

units, and its area is ___ square units.
Please Help!!!

Answer 1

Let

`F(-10,0) A(10,10) B(20,10) C(30,0) D(20,-10) E(10,-10)`
we know that

`FA=FE \\ AB=ED \\ BC=CD`

step 1
find distance FA

`dFA= \sqrt{ (y2-y1)^(2) +(x2-x1)^(2) } \\ dFA= \sqrt{ (10-0)^(2) +(10+10)^(2)`


`dFA= √(500) \\ dFA=22.36 units`

step 2
find distance AB

`dAB=20-10 \\ dAB=10 units`

step 3
find the distance BC

`dBC= \sqrt{ (y2-y1)^(2) +(x2-x1)^(2) }`

`dBC= \sqrt{ (0-10)^(2) +(30-20)^(2) }`

`dBC= √(200)`

`dBC=14.14 units`


step 4
find the perimeter
the perimeter is equal to

`P=2*[FA+AB+BC] \\ P=2*[22.36+10+14.14] \\ P=93 units`

step 5
find the area
the area is equal to
area triangle AFE+area rectangle ABDE+area triangle BDC

step 6
find the area of triangle AFE

`A1=20*20/2 \\ A1=200 units^(2)`

step 7
find the area of the rectangle ABDE

`A2=10*20 \\ A2=200 units^(2)`

step 8
find the area of the triangle BDC

`A3=20*10/2 \\ A3=150 units^(2)`

`Area Total=200+200+150 \\ Area Total=550 units^(2)`

Answer 2

The perimeter and area is 93 units and 500 sq units.

Explanation:

Given in figure

`F(-10,0) A(10,10) B(20,10) C(30,0) D(20,-10) E(10,-10)`

Figure shows that

`FA=FE \\ AB=ED \\ BC=CD`

In order to find the perimeter we have to find the length of FA, AB, BC, CD, DE, FE

Distance FA

`FA= \sqrt{ (y2-y1)^(2) +(x2-x1)^(2) } \\ FA= \sqrt{(10-0)^(2) +(10+10)^(2)`

`FA= √(500) \\ FA=22.36 units`

Distance AB

`AB=20-10=10 units`

Distance BC

`BC= \sqrt{ (y2-y1)^(2) +(x2-x1)^(2) }`

`BC= \sqrt{ (0-10)^(2) +(30-20)^(2) }`

`BC= √(200)=14.14 units`

Perimeter is equal to sum of all the sides of polygon

`P=2(FA+AB+BC) \\ P=2(22.36+10+14.14)=93 units`

Now, we have to find the area

The area is equal to

=ar(ΔAFE)+ar(ABDE)+ar(ΔBDC)

area of triangle AFE

`ar(AFE)=(1)/(2)* 20* 20=200 units^(2)`

area of rectangle ABDE

`ar(ABDE)=10* 20=200 units^(2)`

area of the triangle BDC

`A3=(1)/(2)* 20* 10=100 units^(2)`

`Areal=200+200+100=500 units^(2)`