Geometry Problem

Click the figure below to see the complete problem 459 about Right triangle, Squares, Distance, Measurement.

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Complete Problem 459

Level: High School, SAT Prep, College geometry

## Friday, May 21, 2010

### Problem 459: Right triangle, Squares, Distance, Measurement

Labels:
measurement,
right triangle,
square

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cos(GAD)= cos(GAC+CAD)=cos(GAC+45)=cos(GAC)cos(45)-sin(GAC)sin(45). But cos(GAC)=c/b & sin(GAC)=a/b hence cos(GAD)=(c-a)/(b√2). From Tr. GAD using the co-sine rule, we’ve: x^2 = GA^2 +AD^2 -2GA*AD*cos(GAD) = (c+a)^2+2b^2-2*(c+a)( b√2)(c-a)/(b√2)= (c+a)^2 + 2(c^2+a^2) -2*(c+a)(c-a)= c^2 + 2ac + a^2 + 4a^2 = 5a^2 + c^2 + 2ac

ReplyDeleteAjit

Let m projection of AB over AC.

ReplyDeleteLet n projection of BC over AC.

Let h altitude over hypotenuse AC.

Then (T.altitude) h = sqrt (mn)

Then (T.legs) m = c^2/sqrt(a^2+c^2)

n = a^2/sqrt(a^2+c^2)

Let H point of intersection of parallel to AC from G and extension of side DC from C. Then GHD is right triangle on H.

We've the next relations:

CD = AC = m + n = sqrt(a^2 +c^2) (Pt. Tri.ABC)

CH = h + n

GH = h - n

HD = CH + CD = h + 2n + m

Then:

x^2=GH^2 + HD^2 = (h-n)^2 + (h + 2n + m)^2 =

= ... = 6mn + 2*sqrt(mn)*(m+n) + 5n^2 + m^2 = ... =

= 5a^2 + 2ac + c^2

MIGUE.

Let H point on line GF, and HF=GF. Then CG=CF and angle GCH=90.

ReplyDeleteOn triangle CDG and triangle CAH, CD=CA and CG=CH.

angle DCG=90+angle DCH, angle ACH=90+angle DCH.

So, triangle CDG==triangle CAH ---> then, AH=DG=x

On triangle AGH, angle G=90, AG=a+c, GH=2a, AH=x.

Therefore, x^2=(a+c)^2+(2a)^2 ---> x^2 = 5a^2+c^2+2ac

@ third comment by Anonymous:

ReplyDeleteCG will never be equal to CF as CG is the diagonal of the square and CF the side

Pls correct this comment