Try to use elementary geometry (Euclid's Elements).

See complete Problem 80

Triangle Area, Incircle, Inradius, Sides. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Monday, May 19, 2008

### Elearn Geometry Problem 80

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AI bisects angle A since I is the incenter of Tr. ABC. Quad ADIE is cyclic because angle ADI=angle AEI =90. Thus angle IDE =angle IAE =A/2. Now triangle DIE =S1=rsin(A/2)*rcos(A/2)=r^2*sin(A)/2. Further S = b*csin(A)/2 which gives us S1/S =[r^2*sin(A)/2]/[b*csin(A)/2]=r^2/bc. Thus, S1=(r^2/bc)* S

ReplyDeleteAjit: ajitathle@gmail.com

Solution to prob. 80. The area of tr. ABC is S = 2bc.sen (A), and the area of tr. DIE is S1 = 2r*2.sen (ang DIE). Since quadr. ADIE is cyclic (two opposite internal angles equal 90) then ang (DIE) = 180 – A, so sen (ang DIE) = sen (A). Thus S1/S = r*2/(bc) and S1 = (r*2/bc)S.

ReplyDeletePlease change in my solution to problem 80: S = (1/2)bc.sen (A) and S1 = (1/2)r*2.sen (ang DIE).

ReplyDeleteSorry for that.

Simple geometry solution.

ReplyDeleteIn Tr. ABC, let h be the length of the altitude from B

In Tr. DEI, let m be the length of the altitude from E.

From similar triangles

h/c = m/r …(1),

S = bh/2 …(2),

S1 = rm/2 …(3)

(2) And (3) ; S1 / S = rm/bh = rm/b. (r/mc) since 1/h = r/mc from (1)

So S1 = S. r^2/bc

Sumith Peiris

Moratuwa

Sri Lanka