Triangles, Squares and Areas from Temple Geometry
The problem presented by the applet below has been plucked from a collection of Temple Geometry Problems . Down below, I'll give three solutions, the first being drawn from another collection .
Problem : Five squares are arranged as in the applet. Show that the area of triangle KMN equals the area of the square BEKH.
This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at http://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.
Solution
Copyright © 1996-2008 Alexander Bogomolny
First of all note that angles ABE and CBJ being complementary , right triangle ABE and BCH are congruent. Therefore, if we denote AB as a and BC as b,
CY = CH = AB = a, and
XA = AE = BC = b.
Solution 1
The construction needed for the first solution is hinted to in the applet.
This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at http://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.
Observe that triangles EFK and GHK are congruent to ABE and BCH. Thus, also
EF = GK = a, and
FK = GH = b.
This makes DF = 2a and GI = 2b. Now, triangles DFK and JKM are congruent, as are triangles GIK and KLN, so that
JK = FK = b,
JM = DF = 2a,
KL = GK = a,
LN = GI = 2b.
We see that
Area(KMN) = Area(JMNL) - Area(JKM) - Area(KLN)
= (a + b)(2a + 2b)/2 - 2a·b/2 - a·2b/2
= (a + b)2 - 2ab
= a2 + b2
= Area(BEKH).
This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at http://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.
Solution 2
The only additional construction needed is line DI, which is not actually shown. The diagram presents several Vecten configurations , from which we conclude that triangles ABE, DEK, BCH, KHI have equal areas, as are triangles KMN and DIK.
Area(KMN) = Area(DIK)
= Area(DXYIK) - Area(DXYI)
= (a2 + b2 + 4ab/2 + a2 + b2 ) - (2a + 2b)(a + b)/2
= 2a2 + 2b2 + 2ab - a2 - 2ab - b2
= a2 + b2
= Area(BEKH).
Solution 3
Compared to the first solution, the second one draws very little on algebra. The third solution suggested by Nathan Bowler continues the progression: on the surface it does not use any algebra at all. I placed it on a separate page .
References
H. Fukagawa, D. Pedoe, Japanese Temple Geometry Problems , The Charles Babbage Research Center, Winnipeg, 1989, #4.2.4
J. Konhauser, D. Velleman, S. Wagon, Which Way Did the Bicycle Go? , MAA, 1996, #50
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Copyright © 1996-2008 Alexander Bogomolny
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