Saturday
In a trapezoid ABCD with segment AB parallel to segment CD, the diagonals intersect at point E. the area of the triangle ABE is 32, and the area of the triangle CDE is 50. Find the area of the trapezoid.
2006-09-30 01:17:22 · 4 answers · asked by Sasha 2 in Science & Mathematics ➔ Mathematics
Let:
A1= Area of triangle ABE
A2= Area of triangle CDE
A= Area of Trapezoid ABCD
h1= heigth of triangle ABE
h2= heigth of triangle CDE
H= height of trapezoid ABCD
Values:
A1=32; A2=50
Formulas:
A1= 1/2 ABh1
A2= 1/2 CDh2
A=[(AB+CD)/2]*H
H= h1+h2
Find h1 in A1:
h1=(2*A1)/AB
h1=64/AB ->1st equation
Find h2 in A2:
h2=(2*A2)/CD
h2=100/CD -> 2nd equation
Find H:
H=h1+h2
Subst 1st and 2nd equation
H=(64/AB)+(100/CD)
H=(64+100)/(AB+CD)
H=164/(AB+CD) ->3rd equation
Find A:
A=[(AB+CD)/2]*H
Subst value of H in 3rd equation
A=[(AB+CD)/2]*[164/(AB+CD)]
cancel out (AB+CD)
A=164/2
A=82
2006-09-30 02:12:59 · answer #1 · answered by Dennis T 2 · 1⤊ 0⤋
32+50+39+39=160
i guessed the part about the 39
2006-09-30 08:25:36 · answer #2 · answered by Anonymous · 0⤊ 0⤋
area(EDC)=50
AREA(EAB)=32
The triangles EDC and EAB are similar. A.A.A.
50/32=(c/a)^2
or
c/a=5/4
5/4=h1/h2 the heights are from E to bases.
area of trapezoid is (a+c)/2* (h1+h2)
2006-09-30 08:38:22 · answer #3 · answered by iyiogrenci 6 · 0⤊ 0⤋
Ithink it would be 39
2006-09-30 08:28:50 · answer #4 · answered by Anonymous · 0⤊ 0⤋