This figure is made up of a rectangle BDFE and parallelogram ABCD.
The area of this figure= Area of BDFE+Area of ABCD
Area of parallelogram=base*height and Are of rectangle= length*Width
So, Area of ABCD= CD*BH
And, Area of BDFE=BD*DF
area of this figure= BD*DF+CD*BH
From the figure CD=4 and BH=3
To find BD, Let us apply Pythagorean theorem, [tex] hypotenuse^{2} =adjacent^{2}+ opposite^{2} [/tex], to ΔBHD
[tex] BD^{2} =HD^{2}+ BH^{2} [/tex]
HD=2 and BH=3
[tex] BD^{2} =2^{2}+ 3^{2} [/tex]
[tex] BD^{2} =4+ 9=13 [/tex]
[tex] BD=\sqrt{13} [/tex]
To find DF, Let us apply Pythagorean theorem, [tex] hypotenuse^{2} =adjacent^{2}+ opposite^{2} [/tex], to ΔBHD
[tex] DF^{2} =DG^{2}+ GF^{2} [/tex]
DG=6 and GF=9
[tex] DF^{2} =6^{2}+ 9^{2} [/tex]
[tex] DF^{2} =36+ 81=117 [/tex]
[tex] BD=\sqrt{117}=3\sqrt{13} [/tex]
[tex] BD=\sqrt{13} and DF=3\sqrt{13} [/tex]
area of this figure= BD*DF+CD*BH
area of this figure= [tex] \sqrt{13} [/tex]*3[tex] \sqrt{13} [/tex]+4*3
area of this figure=13*3+12
area of this figure=39+12=51
Area of this figure=51 square units
