This chapter is currently out of syllabus
1
JEE Main 2021 (Online) 26th February Evening Shift
+4
-1
Out of Syllabus
The triangle of maximum area that can be inscribed in a given circle of radius 'r' is :
A
An equilateral triangle having each of its side of length $$\sqrt 3$$r.
B
An equilateral triangle of height $${{2r} \over 3}$$.
C
A right angle triangle having two of its sides of length 2r and r.
D
An isosceles triangle with base equal to 2r.
2
JEE Main 2021 (Online) 24th February Evening Shift
+4
-1
Out of Syllabus
Let a, b, c be in arithmetic progression. Let the centroid of the triangle with vertices (a, c), (2, b) and (a, b) be $$\left( {{{10} \over 3},{7 \over 3}} \right)$$. If $$\alpha$$, $$\beta$$ are the roots of the equation $$a{x^2} + bx + 1 = 0$$, then the value of $${\alpha ^2} + {\beta ^2} - \alpha \beta$$ is :
A
$${{69} \over {256}}$$
B
$${{71} \over {256}}$$
C
$$- {{71} \over {256}}$$
D
$$- {{69} \over {256}}$$
3
JEE Main 2020 (Online) 4th September Morning Slot
+4
-1
Out of Syllabus
A triangle ABC lying in the first quadrant has two vertices as A(1, 2) and B(3, 1). If $$\angle BAC = {90^o}$$ and area$$\left( {\Delta ABC} \right) = 5\sqrt 5$$ s units, then the abscissa of the vertex C is :
A
$$1 + 2\sqrt 5$$
B
$$2\sqrt 5 - 1$$
C
$$1 + \sqrt 5$$
D
$$2 + \sqrt 5$$
4
JEE Main 2019 (Online) 12th April Evening Slot
+4
-1
Out of Syllabus
A triangle has a vertex at (1, 2) and the mid points of the two sides through it are (–1, 1) and (2, 3). Then the centroid of this triangle is :
A
$$\left( {{1 \over 3},2} \right)$$
B
$$\left( {{1 \over 3},{5 \over 3}} \right)$$
C
$$\left( {1,{7 \over 3}} \right)$$
D
$$\left( {{1 \over 3},1} \right)$$
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