Question

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

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

$dFA= \sqrt{500} \\ dFA=22.36 units$

step 2
find distance AB
$dAB=20-10 \\ dAB=10 units$

step 3
find the distance BC
$dBC= \sqrt{ (y2-y1)^{2} +(x2-x1)^{2} }$
$dBC= \sqrt{ (0-10)^{2} +(30-20)^{2} }$
$dBC= \sqrt{200}$
$dBC=14.14 units$


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

Answer:

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

Step-by-step 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

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

$FA= \sqrt{500} \\ FA=22.36 units$

Distance AB

$AB=20-10=10 units$

Distance BC

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

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

$BC= \sqrt{200}=14.14 units$

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

$ar(AFE)=\frac{1}{2}\times 20\times 20=200 units^{2}$

area of rectangle ABDE

$ar(ABDE)=10\times 20=200 units^{2}$

area of the triangle BDC

$A3=\frac{1}{2}\times 20\times 10=100 units^{2}$

$Areal=200+200+100=500 units^{2}$