Embibe Experts Solutions for Chapter: Properties of Triangle, Exercise 1: JEE Advanced Paper 2 - 2014
Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Properties of Triangle, Exercise 1: JEE Advanced Paper 2 - 2014
Attempt the free practice questions on Chapter 11: Properties of Triangle, Exercise 1: JEE Advanced Paper 2 - 2014 with hints and solutions to strengthen your understanding. EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS solutions are prepared by Experienced Embibe Experts.
Questions from Embibe Experts Solutions for Chapter: Properties of Triangle, Exercise 1: JEE Advanced Paper 2 - 2014 with Hints & Solutions
Consider a triangle having sides of lengths and opposite to the angles and , respectively. Then which of the following statements is (are) TRUE?

In a triangle the sum of two sides is and the product of the same two sides is . If (where is the third side of the triangle) then the ratio of the inradius to the circumradius of the triangle is

In a , is the largest angle and . Further the incircle of the triangle touches the sides and at and respectively, such that the lengths of and are consecutive even integers. Then possible length(s) of the side(s) of the triangle is (are)

In a triangle , let and . Then the value of is

Let and be positive real numbers. Suppose and are the lengths of the sides of a triangle opposite to its angles and respectively. If , then which of the following statements is/are TRUE?

In a non-right-angled triangle let denote the lengths of the sides opposite to the angles at respectively. The median from meets the side at the perpendicular from meets the side at , and and intersect at If and the radius of the circumcircle of the equals then which of the following options is/are correct?

In a triangle let and the sides and have lengths units, respectively. Then, which of the following statement(s) is (are) TRUE?

In a let be the lengths of sides opposite to the angles respectively and If and area of incircle of the triangle is , then
