Proposed Problem

Click the figure below to see the complete problem 327 about Right triangle Area, Incircle, Circumcircle, Square.

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Complete Problem 327

Level: High School, SAT Prep, College geometry

## Sunday, July 26, 2009

### Problem 327. Right triangle Area, Incircle, Circumcircle, Square

Labels:
area,
circumcircle,
incircle,
right triangle,
square

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Let ID = r and AC = b = 2R, AB = c & BC = a following the usual triangle notation.

ReplyDeleteNow, DE^2 = AD * DC since ADC is also a rt. triangle where AD = R - OD & DC = R + OD.

Hence, DE^2 = (R - OD)(R + OD) = R^2 - OD^2

But OD^2 = OI^2 - ID^2 = (R^2 - 2rR) - r^2 using Euler distance formula for OI. Hence we've: DE^2 = R^2 - (R^2- 2rR)+r^2= r^2+2rR=r^2+br=r(r+b).

In rt. triangle ABC, the in-radius r=(a+c-b)/2 & hence, DE^2 = [(a + c - b)/2]*[(a + c - b)/2 + b]

= [(a + c - b)/2]*[(a + c + b)/2]

= (1/4)*[(a + c)^2 - b^2]

= (1/4)*[a^2 + c^2 + 2ac - b^2]

= (1/4)*(2ac) since a^2 + c^2 = b^2

or DE^2 = ac/2 = Area of Tr. ABC. QED.

Ajit: ajitathle@gmail.com

DE^2=ADxDC;

ReplyDeleteAD=AB-r;

DC=BC-r;

(AB-r)x(BC-r) = ABxBC - rx(AB+BC) -r^2 = ABxBC - [(1/2 rxAB +1/2 rx AD + 1/2 r^2) + (1/2 rxBC + 1/2 x rxDC + 1/2 r^2) - r^2 =Area ABC;

AC is a diameter so AD. DC = DE^2 = (c-r).(a-r) = ( b+c-a)(a+b-c)/4 since r = (a+c-b)/2

ReplyDeleteThis is = to { b^2 - (a-c)^2}/4 = ac/2 = Area of ABC (I used b^2 = a^2 + c^2 to simplify the algebra)

Sumith Peiris

Moratuwa

Sri Lanka