Bunuel wrote:

Triangles ABC and ABD share side AB. Triangle ABC has area Q and triangle ABD has area R. If AD is longer than AC, and BD is longer than BC, which of the following could be true?

I. R > Q

II. R = Q

III. R < Q

(A) I only

(B) III only

(C) I and II only

(D) I and III only

(E) I, II and III

if we consider angle ABC=ABD=90 (points C, B, D are on straight line) then by area formula we can say that R>Q as as BD >BC

if the rest two sides are greater than the respective sides of other triangle and third side is same. Always the area of the triangle with bigger sides will be higher than another.

ABC = sides are x,y,z ABD = sides are x,p,q (p>y and q>z)

Sc = (x+y+z)/2 and Sd = (x+p+q)/2

\(Ac = \sqrt{(Sc (Sc-x)(Sc-y)(Sc-z))}\)

\(Ad = \sqrt{(Sd (Sd-x)(Sd-p)(Sd-q))}\)

here

Sd>Sc

Sd-x>Sc-x

Sd-p>Sc-y

Sd-q>Sc-z

Hence irrespective of length of sides Area of ABD(R)>ABC(Q)

Please correct me if i am wrong

A

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