Answer:
Area of ABCD = 45.1 cm²
Step-by-step explanation:
From the figure attached,
Area of ABCD = Area of ΔBCD + Area of ΔABD
Area of ΔABD = [tex]\frac{1}{2}(AB)(BD)[/tex]
sin(55°) = [tex]\frac{\text{Opposite side}}{\text{Hypotenuse}}[/tex]
sin(55°) = [tex]\frac{AB}{BD}[/tex]
AB = 10sin(55°)
AB = 8.19 cm
cos(55°) = [tex]\frac{\text{Adjacent side}}{\text{Hypotenuse}}[/tex]
= [tex]\frac{AD}{BD}[/tex]
AD = 10cos(55°)
AD = 5.74cm
Area of ΔABD = [tex]\frac{1}{2}(8.19)(5.74)[/tex]
= 23.51 cm²
Area of ΔBCD = [tex]\frac{1}{2}(\text{Base})(\text{Height})[/tex]
= [tex]\frac{1}{2}(BD)(CE)[/tex]
tan(38°) = [tex]\frac{CE}{BE}[/tex]
BE = [tex]\frac{CE}{\text{tan}(38)}[/tex]
Similarly, DE = [tex]\frac{CE}{\text{tan}(44)}[/tex]
Since, BE + DE = 10 cm
[tex]\frac{CE}{\text{tan}(38)}+\frac{CE}{\text{tan}(44)}=10[/tex]
CE(1.28 + 1.04) = 10
CE(2.32) = 10
CE = 4.31 cm
Area of ΔBCD = [tex]\frac{1}{2}(10)(4.31)[/tex]
= 21.55 cm²
Area of ABCD = Area of ΔBCD + Area of ΔABD
= 21.55 + 23.51
= 45.06
≈ 45.1 cm²
