An rectangle measures 8 cm by 6 cm. On the shorter sides two equilateral triangles are constructed outside the rectangle. Two right angled isosceles are constructed outside the rectangle with the longer sides as the hypotenuse. Find the total area and perimeter of the figure. 

Answer :

We form our diagram from given information , As :

Here ABCD is a rectangle , AB  =  CD  = 8  cm  and BC  =  AD  6 cm
And
BCE and ADG are equilateral triangle , BC  =  BE  =  CE  = AD  = AG  = DG  =  6 cm
And
ABH and CDF are right angled isosceles triangle , Hypotenuse AB  =  CD  = 8 cm  , Let AH  = BH  = CF = DF  =  x

Now we apply Pythagoras theorem in triangle ABH and get 

AH ² + BH² =  AB²

x ² + x ² = 8²

2 x ² = 64

x ² =  32

x = $\sqrt{32}$

x = 4 $\sqrt{2}$
So,
AH  = BH  = CF = DF  = 4 $\sqrt{2}$

We know Area of rectangle  = Length $\times$ Breadth , So

Area of ABCD  =  8 $\times$  6 =  48 cm²

And

Area of triangle  = $\frac{1}{2}$ $\times$ Height $\times$ Base , So 

Area of triangle ABH  =  Area of triangle CDF = $\frac{1}{2}$ $\times$ 4 $\sqrt{2}$$\times$ 4 $\sqrt{2}$ = 16 cm² 

And

Area of equilateral triangle  = $\frac{\sqrt{3}}{4}\times {\left(Side\right)}^2$ , So 

Area of triangle BCE  =  Area of triangle ADG = $\frac{\sqrt{3}}{4}\times {\left(6\right)}^2 = \frac{\sqrt{3}}{4}\times 36 = 9\sqrt{3} = 9 \times 1.732= 15.588 \approx \mathbf{ }\mathbf{15}\mathbf{.}\mathbf{59}\mathbf{ }\mathit{c}{\mathit{m}}^{\mathbf{2}}$ 

So,

Area of our figure AHBECFDG = Area of ABCD + Area of triangle ABH  +  Area of triangle CDF + Area of triangle BCE  +  Area of triangle ADG

Area of our figure AHBECFDG = 48 + 16 + 16 + 15.59 + 15.59 =  111.18 cm²                       ( Ans )

Perimeter of our figure AHBECFDG = AH  + BH  + BE  + CE  + CF  + DF  + DG  + AG  = 4 $\sqrt{2}$ + 4 $\sqrt{2}$ + 6 + 6 + 4$\sqrt{2}$ + 4 $\sqrt{2}$ + 6 + 6 = 24 + 16$\sqrt{2}$ = 24 + 22.63  = 46.63 cm                               ( Ans )