Posted by **karan** on Wednesday, June 22, 2011 at 9:58am.

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.

- geometry -
**Reiny**, Wednesday, June 22, 2011 at 12:47pm
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

- geometry -
**feba**, Tuesday, February 5, 2013 at 8:28pm
ABCD is a square and arc BEC is a semicircle. if AB=10.0 cm,find the area of the area of the figure.

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