Good afternoon sir/ma'am please solve 2nd part of 15 ques Share with your friends Share 0 Atul Shaw answered this Dear student, Area of the diagram=Area of CDHG+Area of GMNE+Area of KLJI+ Area of EFBA=CD×DH+GM×GE+KL×LJ+AE×EF=12×3+3×9+5×3+3×12[AC=CG+GE+EA=15⇒3+GE+3=15⇒GE=9;And each have equal width across the fig. thus GM=EN=3m]=36+27+15+36=114m2[Area of rectangle=length×breadth]Regards 1 View Full Answer