Problems about trapezium

∠DKC=120 degree

更新2:

eelyw says that "relationship b/d = a/c = 7/5

or c/d = a/b = 7/5". I think it is not correct.b　　　a　　　7　 ------- = -------- = ------d　　　c　　　5 thenc　　　a　　　　　　　　　　7 ------- = -------- but it is not equal to -----d　　　b　　　　　　　　　　5

更新3:

Correction: Q(2) ABCD is a isosceles TRAPEZIUM

更新4:

I think that Area of trapezium should be= (bc sin60)/2 + (bd sin120)/2 + (ad sin60)/2 + (ac sin120)/2　　　　　　　　　　　↑　　　　　　　　　　　↓but NOT　　　Area of trapezium = (ab sin60)/2 + (bd sin120)/2 + (ad sin60)/2 + (ac sin120)/2

更新5:

I mean Byb　　　a　　　7　 ------- = -------- = ------d　　　c　　　5 we CANNOT prove thatc　　　a　　　7　 ------- = -------- = ------d　　　b　　　5

Q.1 let ∠KDC = x ABDC

so ∠ABK = ∠KDC = x ∠ACD = 60deg-x 设高为 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 me (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 补充： 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 补充： 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 补充： 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]