Question

in the given figure EADF is a rectangle and ABC is a rectangle whose vertices lie on the sides of rectangle as shown AE=22,BE=6,CF=16 and BF=2 then area of triangle ABC is solve in full detail​

Answer 1

EADF is a rectangle and ABC is a triangle AE = 22 , BE = 6 , CF = 16 and BF = 2 then Area of Δ ABC = 70 sq units

Step-by-step explanation:

Area of Δ ABC = Area of Rectangle EADF - Area of Δ AEB - Area of Δ CFB - Area of ΔADC

Area of Rectangle EADF  = EA * EF

EA = 22

EF = EB + BF = 6 + 2 = 8

Area of Rectangle EADF  = 22 * 8 = 176 sq units

Area of Δ AEB = (1/2) AE * EB   = (1/2) * 22 * 6 = 66 sq units

Area of Δ CFB = (1/2) * CF * FB = (1/2) * 16 * 2 = 16  sq units

Area of Δ ACD = (1/2) * AD * CD

AD = EF = 8   ,  CD = FD - FC = 22 - 16 = 6

Area of Δ ACD = (1/2) * 8 * 6 = 24 sq units

Area of Δ ABC = 176 - 66 - 16 - 24

Area of Δ ABC = 70 sq units