Question

find the perimeter and area of the quadrilateral ABCD in which AB equal to 17 cm is equal to 9 CM CD is equal to 12 centimetre and angle ACB 90 degree and ac equal to 15 centimetre​

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

Answer 2

Perimeter and area of the quadrilateral ABCD is 46 cm and 114 cm square.