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 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

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

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

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

To find the area of a quadrilateral, we need to know the lengths of its sides and the angle between those sides. In this case, we have a right angle between the first two sides (6cm and 8cm), which means that the quadrilateral ABCD is a rectangle.

To find the area of a rectangle, we multiply the length of one side (in this case, the side with length 6cm) by the length of the adjacent side (in this case, the side with length 8cm).

Therefore, the area of the quadrilateral ABCD is:
Area = 6cm * 8cm
= 48cm²

So, the area of the given quadrilateral is 48 square centimeters.