In Δ ABDUsing the Pythagoras theoremBD2=AB2+AD2By substituting the values262=AB2+242On further calculationAB2=676−576By subtractionAB2=100By taking out the square rootAB=100​So we getBase = AB = 10cmWe know that area of ΔABD=21​×b×hBy substituting the valuesArea of ΔABD=21​×10×24On further calculationArea of ΔABD=120cm2We know that the area of ΔBCD=3​/4a2By substituting the valuesArea of ΔBCD=(1.73/4)(26)2So we get Area of ΔBCD=292.37cm2So we get area of quadrilateral ABCD = Area of Δ ABD + Area of Δ BCDBy substituting the valuesArea of quadrilateral ABCD = 120 + 29237By additionArea of quadrilateral ABCD = 412.37 cm2The perimeter of quadrilateral ABCD = AB + BC + CD + DABy substituting the valuesPerimeter = 10 + 26 + 26 + 24So we getPerimeter = 86cmTherefore, the area is 412.37 cm2 and perimeter is