Geometry Problem

Click the figure below to see the complete problem 519 about Triangle, Squares, Altitude, Perpendicular, Rectangle, Area.

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Complete Problem 519

Level: High School, SAT Prep, College geometry

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## Friday, September 10, 2010

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Problem 519: Triangle, Squares, Altitude, Perpendicular, Rectangle, Area

See also:

Complete Problem 519

Level: High School, SAT Prep, College geometry

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Geometry Problem

Click the figure below to see the complete problem 519 about Triangle, Squares, Altitude, Perpendicular, Rectangle, Area.

See also:

Complete Problem 519

Level: High School, SAT Prep, College geometry

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▲BPH ~ ▲ABH => PB/BH = BH/AB

ReplyDelete▲BMH ~ ▲BHC => BM/BH = BH/BC

=> PB∙AB = BM∙BC

In right triangle ABH with altitude HP we have BH^2=BP.BA=S1 ( relations in right triangles)

ReplyDeleteSimilarly in right triangle BHC we have BH^2=BM.BC= S2

So S1=S2

Peter Tran

Triangle ABH and BHP are similar --->

ReplyDeleteAB:BH=BH:BP ---> S1=ABxBP=BH^2

Triangle BCH and BHM are similar --->

BC:BH=BH:BM ---> S2=BCxBM=BH^2 ---> S1=S2

a trigonometric solution

ReplyDeleteBH=csinA=asinC

BM=BHsinC

BP=BHsinA

S1=c.BP=acsinAsinC

S2=a.BM=acsinAsinC

.-.

Angles BPM = BHM = C;

ReplyDeleteTriangles BPM, BCA are ///.

BP:BC = BM:BA,

BP.BA = BM.BC,

BP.BD = BM.BG

S1 = S2

Problem 519

ReplyDeleteIs BH^2=BP.BA=BP.BD=S_1, BH^2=BM.BC=BM.BG=S_2,Therefore S_1=S_2.

APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE