Question

A quadrilateral contains two equal sides measuring 12 cm each with an included right angle. If the measure of the third side is 8 cm and the angle opposite the right angle is 120 degrees, find the fourth side and the area of the quadrlateral. how can I answer this?

Answer 1

let AB=12cm
BC=12cm
CD=8cm
DE=xcm
<ABC=90
<CDA=120
by pythagoras theorem AC=sqrt(12^2+12^2)=12sqrt2
cos <CDA=(8^2+x^2-(12sqrt2)^2)/2*8*x
cos120=cos(180-60)=-cos60=(64+x^2-288)/16x
-1/2*16x=x^2-224
x^2+8x-224=0
x=(-8+-sqrt(64-4*1*-224))/2*1
x=(-8+-sqrt960)/2
x=(-8+2sqrt240)/2 (as xis positive so taking only positive sign)
x=-4+4sqrt15=4(sqrt15-1)

Answer 2

you use cos CDA but i know that it only use if that triangle is right triangle
so it can't be :(

Answer 3

produce CD to E and draw AE perpendicularCE
<EDA=180-120=60
sin 60=AE/AD
sqrt3/2=AE/4(sqrt15-1)
AE=sqt3/2 *4(sqrt15-1)=2(sqrt45-sqrt3)
area=1/2*12*12+1/2*8*2(sqrt45-sqrt3)
solve it.
I have used <ABC for pythagoras theorem.

Answer 4

i have used <CDA for cosine formula ,it can be used for any triangle.

Answer 5

my equation and Draw are not working so i can't make a diagram.

Answer 6

me also
hahah!
i think that figure is kite

Answer 7

u might need more hypotheses !

Answer 8

humm i cant see the sketch XD