Question

Work out area of ABCD

Answer 1

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 = $\frac{1}{2}(AB)(BD)$

sin(55°) = $\frac{\text{Opposite side}}{\text{Hypotenuse}}$

sin(55°) = $\frac{AB}{BD}$

AB = 10sin(55°)

AB = 8.19 cm

cos(55°) = $\frac{\text{Adjacent side}}{\text{Hypotenuse}}$

              = $\frac{AD}{BD}$

AD = 10cos(55°)

AD = 5.74cm

Area of ΔABD = $\frac{1}{2}(8.19)(5.74)$

                        = 23.51 cm²

Area of ΔBCD = $\frac{1}{2}(\text{Base})(\text{Height})$

                        = $\frac{1}{2}(BD)(CE)$

tan(38°) = $\frac{CE}{BE}$

BE = $\frac{CE}{\text{tan}(38)}$

Similarly, DE = $\frac{CE}{\text{tan}(44)}$

Since, BE + DE = 10 cm

$\frac{CE}{\text{tan}(38)}+\frac{CE}{\text{tan}(44)}=10$

CE(1.28 + 1.04) = 10

CE(2.32) = 10

CE = 4.31 cm

Area of ΔBCD = $\frac{1}{2}(10)(4.31)$

                       = 21.55 cm²

Area of ABCD = Area of ΔBCD + Area of ΔABD

                        = 21.55 + 23.51

                        = 45.06

                        ≈ 45.1 cm²