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

Online Geometry theorems, problems, solutions, and related topics.

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