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

Post your solutions or ideas in the comments.

## Monday, August 25, 2008

### Elearn Geometry Problem 163

Labels:
area,
diagonal,
Problem 163,
Problem 164,
trapezoid,
triangle

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Let A be(0,0), B:(b.h),C:(c,h) & D:(d,0)

ReplyDeleteIt 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

By JCA

ReplyDelete2)

S=S1+S2+S3+S4=S1+S2+2sqrt(S1*S2)=[sqrt(S1)+sqrt(S2)]^2

hence

sqrt(S)=sqrt(S1)+sqrt(S2)

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

ReplyDeleteE = AC ∩ BD

ReplyDeleteSolve 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)

The solution is uploaded to the following link:

ReplyDeletehttps://docs.google.com/open?id=0B6XXCq92fLJJV1U1OFlBMjhuN2s

Solution to problem 163.

ReplyDelete1) 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).

Triangle BCE ~ Triangle DAE

ReplyDeleteLet BC:AD=1:k, so height of triangle BCE : height of triangle DAE also =1:k

Let BC=a, AD=ka, height of triangle BCE=b and height of triangle DAE =kb

(1)S1+S3=S1+S4(same base same height)

S3=S4=ab(1+k)/2-ab/2=abk/2

(S1)(S2)=(ab/2)*(abk^2)/2=(abk/2)^2

So S3=S4=sqrt((S1)(S2))

(2)S=(a+ka)(b+bk)/2=[ab(1+k)^2]/2

sqrt(S)=(1+k)*sqrt(ab/2)

From(1), sqrt(S1)=sqrt(ab/2), sqrt(S2)=k*sqrt(ab/2)

By simple calculation, sqty(S)=sqrt(S1)+sqrt(S2)