Solution-
From the figure,
AE = 2.4
EB = 2.8
BC = 11.7
Area of rectangle 1 = 8.68 sq.in
[tex]\Rightarrow FH \times HI=8.68[/tex]
[tex]\Rightarrow EB \times HI=8.68[/tex] (∵ sides of the rectangle 2)
[tex]\Rightarrow 2.8 \times HI=8.68[/tex]
[tex]\Rightarrow HI=3.1[/tex]
Area of Triangle 1 = 6.48 sq.in
[tex]\Rightarrow \frac{1}{2}\times AE \times EG= 6.48[/tex]
[tex]\Rightarrow \frac{1}{2}\times AE \times (EF+FG)= 6.48[/tex]
[tex]\Rightarrow \frac{1}{2}\times AE \times (EF+HI)= 6.48[/tex] (∵ sides of the rectangle 1)
[tex]\Rightarrow EF+3.1= 5.4[/tex]
[tex]\Rightarrow EF=2.3[/tex]
[tex]\Rightarrow BH=2.3[/tex] (∵ sides of the rectangle 2)
[tex]BC = BH+HI+IC[/tex]
[tex]\Rightarrow 11.7= 2.3+3.1+IC[/tex]
[tex]\Rightarrow IC=6.3[/tex]
The area of Rectangle 2,
[tex]=EB\times BH =2.8\times 2.3=6.44\ sq.in[/tex]
The area of Triangle 2,
[tex]\frac{1}{2}\times GI \times IC=\frac{1}{2}\times EB \times IC=\frac{1}{2}\times 2.8 \times 6.3=8.82\ sq.in[/tex]
The area of the whole figure = Area of Triangle 1 + Area of rectangle 1 + Area of Triangle 2 + Area of rectangle 2
= 6.48+8.68+8.82+6.44=30.42 sq.in
