Question

Find the area of an octagon ABCDEFGH having each side equal
to 5 cm, HC = 11 cm and AP I HC such that AP = 4 cm.​

Answer 1

Given :-

• AB = BC = CD = DE = EF = FG = GH = AH = 5cm.
• HC = 11cm
• AP ⊥ HC
• AP = 4cm

To Find :-

• Area of Octagon ABCDEFGH

Construction :-

Draw a line XY prependicular from FE to GD

Solution :-

Let's discuss the figure into three part i.e., Figure 1, 2, 3.

Here,

• ABCH is a parallelogram and it will be the first part of the figure.
• GHCD is a Rectangle and it will be the second part of the figure.
• GFED is a parallelogram Nd it will be the third part of the figure.

1.) Area of ABCH :-

AB = 5cm (given)

HC = 11cm (given)

AP = 4cm (given)

Area of ABCH :-

${\mathtt{\boxed{\red{Area\:of\:parallelogram\:=\: \frac{1}{2}(Sum\:of\: parallel\:sides)h}}}}$

Substituting the value in the above formula,

$\mapsto{\bold{Area\:ABCH\:=\: \frac{1}{2}(AB+HC)AP}}$

$\mapsto{\bold{Area\:ABCH\:=\: \frac{1}{2}(5+11)4}}$

$\mapsto{\bold{Area\:ABCH\:=\: \frac{1}{2}×14×4}}$

$\mapsto{\bold{Area\:ABCH\:=\: \frac{1}{\cancel{2}}×14×{\cancel{4}}}}$

$\mapsto{\bold{Area\:ABCH\:=\:14×2}}$

$\huge{\mathtt{\fcolorbox{green}{crimson}{Area\:ABCH\:=\:28cm²}}}$

2.) Area of GHCD :-

GH = 5cm (given)

HC = 11cm (given)

Area of GHCD :-

${\mathtt{\boxed{\green{Area\:GHCD\:=\:(Length×Breadth)}}}}$

Substituting the value in the above formula,

$\mapsto{\bold{Area\:GHCD\:=\:(GH×HC)}}$

$\mapsto{\bold{Area\:GHCD\:=\:(5×11)}}$

$\huge{\mathtt{\fcolorbox{orange}{purple}{Area\:GHCD\:=\:55cm²}}}$

3.) Area of GFED :-

GD = 11cm (∵ GHCD is a Rectangle, HC = GD = 11cm)

FE = 5cm (given)

XY = 4cm (Construction)

Area of GFED :-

${\mathtt{\boxed{\red{Area\:of\:parallelogram\:=\: \frac{1}{2}(Sum\:of\: parallel\:sides)h}}}}$

Substituting the value in the above formula,

$\mapsto{\bold{Area\:GFED\:=\: \frac{1}{2}(GD+FE)XY}}$

$\mapsto{\bold{Area\:GFED\:=\: \frac{1}{2}(5+11)4}}$

$\mapsto{\bold{Area\:GFED\:=\: \frac{1}{2}×14×4}}$

$\mapsto{\bold{Area\:GFED\:=\: \frac{1}{\cancel{2}}×14×{\cancel{4}}}}$

$\mapsto{\bold{Area\:GFED\:=\:14×2}}$

$\huge{\mathtt{\fcolorbox{pink}{darkblue}{Area\:GFED\:=\:28cm²}}}$

Now,

Area of Octagon ABCDEFGH = Area ABCH + Area GHCD + Area GFED

=> 28cm² + 55cm² + 28cm²

$\large{\mathtt{\fcolorbox{lime}{black}{Area\:of\:Octagon\:ABCDEFGH\:=\:111cm²}}}$

Answer 2

This is the Correct Answer...