Question

An equilateral triangle ABC and a point P on the minor arc joining A and B, is chosen. Let x = PA, y = PB and z = PC. (z is larger than both x and y.) Statement-1: Each of the possibilities (x + y) greater than z, equal to z or less than z, is possible for some P. because Statement-2: In a triangle ABC, sum of the two sides of a triangle is greater than the third and the third side is greater than the difference of the two.

• A. Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.
• B. Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.
• C. Statement-1 is true, statement-2 is false.
• D. Statement-1 is false, statement-2 is true.

Answer 1

Answer: (D)

(D) Statement-1 is false, statement-2 is true.

Answer 2

Statement-2 is the Triangle Inequality Theorem, which states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side, and the difference between the lengths of any two sides must be less than the length of the third side. This is a fundamental theorem in geometry and is always true. Therefore, Statement-2 is true.

Statement-1 concerns the relationship between $x = PA$, $y = PB$, and $z = PC$, where P is a point on the minor arc AB of an equilateral triangle ABC. Assuming P lies on the circumcircle of triangle ABC, the quadrilateral PABC is cyclic. By Ptolemy's Theorem for a cyclic quadrilateral PABC: $PA \cdot BC + PB \cdot AC = PC \cdot AB$

Let $s$ be the side length of the equilateral triangle ABC, so $AB = BC = AC = s$. Substituting these into Ptolemy's Theorem: $x \cdot s + y \cdot s = z \cdot s$

Since $s \neq 0$, we can divide by $s$: $x + y = z$

This shows that for any point P on the minor arc AB of the circumcircle, the sum $x+y$ is always equal to $z$. Therefore, the possibilities $(x+y) > z$ and $(x+y) < z$ are impossible. Statement-1, which claims all three possibilities are possible, is false.

Since Statement-1 is false and Statement-2 is true, the correct option is (D).