Question

find the perimeter and area of a quadrilateral ABCD in which BC is equal to 12 CM, CD is equal to 9 cm,BD is equal to 15 cm, DA is equal to 17 cm and angle ABD is equal to 90 degree.

Answer 1

by Pythagoras theorem, 
BC = √AB²−√AC²
=√17² - √15²
=√289 - 225
BC = √64 = 8 cm

Perimeter of quadrilateral ABCD = 17 + 9 + 12 + 8 = 46 cm

Area of triangle △ABC=1/2×base×height
△ABC=12×8 x 15
= 60 cm²

NOW, for the area of the triangle, ACD
we will take a = 15 cm, b = 12 cm and c = 9 cm

S = 12(a+b+c)

S = 12(15+12+9)=36/2=18cm

Area of one triangular  =S√(S−a)(S−b)(S−c)  (Heron's formula)
=18√(18−15)(18−12)(18−9)
=√18×3×6×9
=√18×18×3×3
= 18×3=54cm²


area of quadrilateral ABCD= area of △ABC + area of △AC
= 60 + 54 = 114 cm²

hope it helps!!