Twin Triangles? Could be Cousins.

Geometry Level 4

Consider the following two statements about two triangles:

P : P: Two Triangles have equal area, and two sides of one are equal to two sides of the other.

Q : Q: They are congruent.

Which of the following statements is true?

(A) P is necessary and sufficient for Q.

(B) P is necessary but not sufficient for Q.

(C) P is sufficient but not necessary for Q.

(D) P is neither necessary nor sufficient for Q.

AYWC?
(D) (B) (A) (C)

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2 solutions

If Q Q is true then P P must be true, so P P is necessary for Q . Q.

However, P P can be true without Q Q being true. For example, suppose we have two isosceles triangles, one with the two equal sides separated by an angle of 6 0 60^{\circ} and one with the same two equal sides being separated by an angle of 12 0 . 120^{\circ}. Then the two triangles have the same area, thus satisfying statement P P , but they are not congruent, hence do not satisfy statement Q . Q. Thus in this case P P does not imply Q Q , and thus P P is not sufficient for Q . Q.

So P P is necessary but not sufficient for Q Q , making option B the correct choice.

Vishnu C
Apr 3, 2015

T a k e a g e n e r a l c a s e w h e r e t h e s i d e s o f a t r i a n g l e a r e a , b , c a n d a , b , c . L e t t h e a n g l e f a c i n g a b e A a n d t h e a n g l e f a c i n g a b e A . A r e a = 1 2 b c s i n A = 1 2 b c s i n A s i n A = s i n A . A = A O R , A = π A . S o , P n e e d n o t n e c e s s a r i l y m e a n A = A . P n e e d n o t m e a n Q . B u t t h e o t h e r w a y a r o u n d , w e c a n s e e t h a t Q d o e s i n d e e d m e a n P i s s a t i s f i e d . \\ Take\quad a\quad general\quad case\quad where\quad the\quad sides\quad \\ of\quad a\quad triangle\quad are\quad a,b,c\quad and\quad a',b,c.\quad Let\quad the\\ angle\quad facing\quad a\quad be\quad A\quad and\quad the\quad angle \quad facing\quad a'\quad be\\ A'.\quad Area=\frac { 1 }{ 2 } bc\quad sinA=\frac { 1 }{ 2 } bc\quad sinA'\\ \Rightarrow sinA=sinA'.\\ \Rightarrow A=A'\quad OR,\quad A=\pi -A'.\\ So,\quad P\quad need\quad not\quad necessarily\quad mean\quad A=A'.\\ \therefore \quad P\quad need\quad not\quad mean\quad Q.\quad But\quad the\quad other\quad way\\ around,\quad we\quad can\quad see\quad that\quad Q\quad does\quad indeed\quad mean\quad P\quad is\\satisfied.\\ \\

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