Answer:
Area of the figure = 42 units²
Step-by-step explanation:
Area of the given figure = Area of rectangle JKIG - (Area of ΔEJD + Area of ΔCKD + Area of ΔCIH + Area of ΔEFG)
Area of rectangle JKIG = JK × JG
= 8 × 8
= 64 units²
Area of ΔEJD = [tex]\frac{1}{2}(EJ)(JD)[/tex]
= [tex]\frac{1}{2}(6)(3)[/tex]
= 9 units²
Area of ΔCKD = [tex]\frac{1}{2}(CK)(KD)[/tex]
= [tex]\frac{1}{2}(5)(2)[/tex]
= 5 units²
Area of ΔCIH = [tex]\frac{1}{2}(CI)(IH)[/tex]
= [tex]\frac{1}{2}(3)(2)[/tex]
= 3 units²
Area of ΔEFG = [tex]\frac{1}{2}(FL)(EG)[/tex]
= [tex]\frac{1}{2}(5)(2)[/tex]
= 5 units²
Area of the given figure = 64 - (9 + 5 + 3 + 5)
= 64 - 22
= 42 units²
