In trapezoid ABCD, overline{AB} parallel overline{CD}, and AB < CD. The base overline{CD} has a length of 8. The legs overline{AD} and overline{BC} have lengths of 7, and the diagonal overline{BD} has a length of 9. Find the area of the trapezoid. [asy] unitsize(1 cm); pair A, B, C, D; A

The area of the the trapezoid is 40.2cm.

Step-by-step explanation:

check the attached document for the assumed figure of the trapezoid.

First we use cosine rule to determine the angle at C or ∠BCD.

Applying the formula; c²=b²+d²-2bdCos C

9²=8²+7²-2(8)(7) Cos C

81=113-112 Cos C

-32= -112 Cos C

Cos C=32/112= 0.2857

C= Arc Cos 0.2857=73.398°

Next we find the height of the trapezoid /BE/.

And since BEC is a right angled triangle, we use trig ratios.

Sin C= opposite/hypotenuse

sin 73.398°= h/7

so that height= 7 sin 73.398°= 6.7cm

and for /EC/ ⇒Cos 73.398°=Adjacent/Hypotenuse=x/7

x=7 Cos 73.398 =2.0cm

x=/DF/=/EC/=2.0cm.

/AB/= /DC/ - (/DF/+/EC/)

/AB/=8 - (2+2) = 4cm

Hence, Area of trapezoid ABCD=1/2(a+b)h

Area ABCD= 1/2(4+8)cm*6.7cm

Area ABCD= 40.2cm²

answer: i don't understand it

step-by-step explanation: sorry

an

y = 6x -3y -12

y +3y +12 = 6x

=> . x = { 2y / 3 +2 } this is equation of line in terms of x

=> . if line meets the x-axis the co-ordinates of meeting point would be ( x , 0 )

=> . x = 2 ( 0 ) / 3 + 2

=> . x = 2

so, the point is ( 2 , 0 )