geometry
posted by karan .
the sides of a quadrilateral abcd are 6cm,8cm,11cm and 12cm respectively,and the angle between the first two sides is a right angle.find its area.

draw the diagonal to show the rightangled triangle,
the hypotenuse of that using Pythagoras is 10
so the area of the right_angled triangle is
(1/2)(6)(8) = 24
for the other triangle with sides, 10, 11, and 12
use Heron's formula
Area = √(s(sa)(sb)(sc))
where s = (1/2)(perimeter) = 16.5
sa = 16.510 = 6.5
sb = 16.5  11 = 5.5
sc = 16.5  12 = 4.5
Area = √(16.5*6.5*5.5*4.5) = 51.92
total area = 24+51.92 = 75.52
check my arithmetic 
ABCD is a square and arc BEC is a semicircle. if AB=10.0 cm,find the area of the area of the figure.