Geometry Problem. Post your solution in the comment box below.

Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below.

## Thursday, April 13, 2017

### Geometry Problem 1331: Two Squares Side by Side, Parallel, Perpendicular, 90 Degrees, Measurement

Labels:
90,
measurement,
parallel,
perpendicular,
square

Subscribe to:
Post Comments (Atom)

HC=2, => BH=6 => AH²=100, HK²+AK²=100 => HK=5√2

ReplyDeleteApplying the knowledge from the previous problems, we have

ReplyDeleteAH=HM=GC=10 ----------(1)

and AHMG is cyclic with m(AHM) = 90-----------(2)

=> AM=10Sqrt(2) ---------(3)

=> K is the center of the circle on which A,H,M,G lie

=> HK=AM/2 = 5sQRT(2)

Problem 1331

ReplyDeleteSuppose that the BC intersects GM in point N then NM=FG=6, or MG=8-6=2.

AM^2=AG^2+MG^2=14^2+2^2=200, or AM=10√2 .So HK=AM/2=5√(2 ).

AB = 8, BH = 6 so AH = 10

ReplyDeleteTr. HAK is right isosceles

Hence HK = 10/sqrt2 = 5sqrt2

Sumith Peiris

Moratuwa

Sri Lanka