See complete Problem 69

Square Inscribed in a Triangle. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Monday, May 19, 2008

### Elearn Geometry Problem 69

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similarity,
square,
triangle

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Thalès twice

ReplyDeleteLet the vertices of square are PQRS and BM perpendicular is drawn to BC. area triangle APM=(1/2).s.AM

ReplyDeletearea of triangle SMC=(1/2).s.MC

area of triangle PMS=(1/2).s.s

area of triangle BPS=(1/2).s.(h-s)

area of whole triangle=(1/2).b.h

now

1/2).s.AM+(1/2).s.MC+(1/2).s.s+(1/2).s.(h-s)=(1/2).b.h

and AM + MC=b

which on simplifying prove the relation

With P, Q, R and S vertices of the square and k be the height of the triangle ΔSBR, we have ΔABC ~ ΔSBR. Therefore,

ReplyDeleteFrom h/k = b/s, we have:

s = bk/h = b(h-s)/h = (bh-bs)/h

and sh + bs = bh or s = bh/(b+h)

QED

Let PQRS be the square.

ReplyDeleteArea of triangle ABC= Area of triangle BSR + Area of Trapezium ACRS.

(1/2)bh = (1/2)s*(h-s) + (1/2)(b+s)* (h-s)

bh = (h-s)(2s+b)

on simplification we get the result

Area of Tr. ABC is comprised of the areas of 3 Tr.s and the square.

ReplyDeleteSo s^2 + 1/2s(h-s) + 1/2s(b-s) = 1/2 bh

From which the result follows

Sumith Peiris

Moratuwa

Sri Lanka

Or from similar triangles

ReplyDelete(h-s)/h .= s/b and the result follows