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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See complete Problem 69
Square Inscribed in a Triangle. Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
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
the area of the trapezium ACRS is (1/2)(b+s)*s, since it's height is s
DeleteArea 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
Similar triangle
ReplyDelete