## Monday, August 25, 2008

### Elearn Geometry Problem 163

See complete Problem 163 at:
www.gogeometry.com/problem/p163_trapezoid_triangle_area.htm

Trapezoid, Diagonals, Triangles, Areas. Level: High School, SAT Prep, College geometry

1. Let A be(0,0), B:(b.h),C:(c,h) & D:(d,0)
It can easily be seen that triangles ABC & DBC have the base and height and hence, Tr ABC =Tr. DBC or S3 + S1 = S4 + S1 or S3 = S4.
Now height of E above AD can be determined as hd/(-b+c+d) and thus, S2 = hd^2/2(-b+c+d) while S1=(c-b)/2 *[h - hd/(-b+c+d))=h(c-b)^2/2(-b+c+d) which gives us S1*S2 =(hd(c-b))^2/4(-b+c+d)^2-(1)
While S3 =(1/2)[cdh/(-b+c+d)- bhd/(-b+c+d)]
=(1/2)hd(c-b)/(-b+c+d) -----------(2)
By (1) & (2) we've, S3^2=S1*S2 or S3 = V(S1*S2) where V=square root.
Now S=S1+S2+S3+S4= S1+S2+2S3 = S1+S2+2*V(S1*S2) Hence, VS = VS1 + VS2
Ajit: ajitathle@gmail.com

2. By JCA
2)
S=S1+S2+S3+S4=S1+S2+2sqrt(S1*S2)=[sqrt(S1)+sqrt(S2)]^2
hence
sqrt(S)=sqrt(S1)+sqrt(S2)

3. Could you explain how you got E's height in a little more detail? I like the proof otherwise.

4. E = AC ∩ BD
Solve for y the equations
hx - cy = o (equation of AC),
hx - (b - d)y - hd = 0 (equation of BD)
at E:(b-d -c)y + hd = 0
or y = hd /(c + d - b)

6. Solution to problem 163.
1) Triangles ABC and BCD have the same basis BC, and equal altitudes, so they have the same area. Hence
S3 = S(ABC) – S(BCE) = S(BCD) – S(BCE) = S4.
Triangles ABE and EBC have the same altitude relative to line AC. Then S3/S1 = AE/EC.
Triangles ABE and AED have the same altitude relative to line BD. Then S2/S3 = DE/EB.
But AED and CEB are similar and AE/EC = DE/EB. Then S3/S1 = S2/S3 and S3 = S4 = sqrt(S1.S2).

2) We have
S = S1 + S2 + S3 + S4 =
= S1 + S2 + 2.sqrt(S1.S2) =
= (sqrt(S1) + sqrt(S2))^2
Then sqrt(S) = sqrt(S1) + sqrt(S2).