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.

Answer 1

draw the diagonal to show the right-angled 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(s-a)(s-b)(s-c))
where s = (1/2)(perimeter) = 16.5
s-a = 16.5-10 = 6.5
s-b = 16.5 - 11 = 5.5
s-c = 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

Answer 2

ABCD is a square and arc BEC is a semicircle. if AB=10.0 cm,find the area of the area of the figure.

Answer 3

Your answer is wrong the correct answer is 2100√15m^2

Answer 4

Your answer is wrong the correct answer is 2100 root 15 m square