This is a part of set Find the Truth
(I) a, b and c are sides of a triangle, then a+b>c.
(2) If ABC is a right triangle with integer sides, then its inradius is also an integer.
(3) l, m and n are 3 lines such that no 2 are parallel to each other, then there are 2 different points on n which are equedistant to l and m.
(4)If in a triangle ABC, , then ABC is right angled.
(5) Area of triangle ABC is greater than that of PQR, then Perimeter of ABC is greater than PQR.
(6) If ABC and PQR have equal area, AC=PR and AB=PQ, then they both are congruent.
Add the serial numbers of all the true statements from these.
For e.g. If (3), (4) and (6) are true, the answer is 3+4+6=13, if none are true answer is 0.
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(I) TRUE - Triangle inequality law
(2) TRUE - I n r a d i u s i n a a − b − c r i g h t t r i a n g l e i s g i v e n b y r = 2 a + b − c I f b o t h a a n d b a r e o d d , t h e n c i s e v e n . S o a + b − c i s e v e n , s o r i s a n i n t e g e r . I f b o t h a a n d b a r e e v e n , t h e n c i s e v e n . S o a + b − c i s e v e n , s o r i s a n i n t e g e r . I f o n e o f a a n d b i s o d d , t h e n c i s o d d . S o a + b − c i s e v e n , s o r i s a n i n t e g e r .
(3) FALSE - This is false when l, m and n are concurrent and n is not the angle bisector of l and m.
(4) TRUE - Can be directly proved by cosine rule.
(5) FALSE - Let ABC be 1, 1, 1 PQR be 2, 2, 3.99
(6) FALSE - Let ABC be 5,5,6 PQR be 5,5,8