Problems about trapezium

Q.1
let ∠KDC = x
AB//DC, so ∠ABK = ∠KDC = x
∠ACD = 60deg-x

設高為 h

sinx = h/7
h = 7sinx

sin(60deg-x) = h/5
h = 5sin(60deg-x)

so
7sinx = 5sin(60deg-x)
7/5 = sin(60deg-x)/sinx
7/5 = [sin60deg*cosx - cos60deg*sinx]/sinx
7/5 = 0.8660254038cosx/sinx - 0.5*sinx/sinx
7/5 + 0.5 = 0.8660254038 cotx
2.193931023 = cotx
0.4558028441= tanx
x = 24.50363346 deg

h = 7sinx
h = 2.903256634

AB + DC = 5cos(60deg - x) + 7cosx
AB + DC = 10.44030651

area = 10.44030651*h/2
= 15.15544457

Q.1
Let BK = a, KD= b, AK = c and KC = d.
Therefore, a + b = 7 .............(1) and
c + d = 5..............................(2)
From (1) b(a/b + 1 ) = 7, a/b + 1 = 7/b, therefore, a/b = 7/b -1...........(3)
Similarly from (2), d(c/d + 1) = 5, therefore, c/d = 5/d -1...................(4)
Triangle AKB is similar to triangle KDC (AAA), therefore, c/d = a/b..............(5) that means (3) = (4) That is 7/b -1 = 5/d -1 or 7/b = 5/d or b/d = 7/5.Together with (5) , we get the relationship b/d = a/c = 7/5, or c/d = a/b = 7/5.
Substitute this result into (3) and (4), we get : 7/5 = 7/b -1 and 7/5 = 5/d - 1.
Therefore, b = 35/12 and d= 25/12.
Substitute these result into (1) and (2), we get a = 7-35/12 = 49/12 and c = 35/12.
Area of trapezium = (absin60)/2 + (bdsin120)/2 + (adsin60)/2 + (acsin120)/2
= sqrt3/4(ab + bd + ad + ac) = sqrt3/4(1715/144 +875/144 + 1225/144 + 1715/144) = 2765sqrt3/288.
Let angle KDC = x. From triangle KDC and by sine rule,
b/sin(60-x) = d/sinx
7/sin(60 -x) = 5/sin x
7sin x = 5sinx(60-x)
7sin x = 5sin60cos x - 5cos60sinx
7sinx = 5sqrt3(cosx)/2 - 5sinx/2
14sin x = 5sqrt3(cosx) - 5sin x
19sinx = 5sqrt3(cosx), therefore, tanx = 5sqrt3/19, x =arctan (5sqrt3/19).

2008-06-20 19:11:53 補充：
As calculated, a=49/12, b=35/12, c= 35/12 and d=25/12.
c/d = (35/12)/(25/12) = 35/25 = 7/5.
a/b = (49/12)/(35/12) = 49/35 = 7/5.
b/d = (35/12)/(25/12) = 35/25 = 7/5.
a/c = (49/12)/(35/12) = 49/35 = 7/5.

2008-06-21 13:51:40 補充：
Your area of trapezium is correct. In fact, the length of a, b,c and d are immaterial is calculating the area. Simply using the 4 results: a + b = 7, c + d = 5, 5b =7d and Area = sin60(bc+bd+ad+ac)/2.You can found that area =(7 x 5 xsin60)/2 = 35sqrt3/4 = 15.15

2008-06-21 14:21:20 補充：
For Q.2, I found that area of the isos. trapezium is 98. However, there is no room for me to give you the details here, if 98 is correct and you like to know the details, you may e-mail to [email protected].