This chapter is currently out of syllabus
1
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
$$2 + \sqrt 5$$
2
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
$$\left( {{1 \over 3},1} \right)$$
3
JEE Main 2019 (Online) 10th April Evening Slot
+4
-1
Out of Syllabus
The angles A, B and C of a triangle ABC are in A.P. and a : b = 1 : $$\sqrt 3$$. If c = 4 cm, then the area (in sq. cm) of this triangle is :
A
2$$\sqrt 3$$
B
4$$\sqrt 3$$
C
$${4 \over {\sqrt 3 }}$$
D
$${2 \over {\sqrt 3 }}$$
4
JEE Main 2019 (Online) 8th April Evening Slot
+4
-1
Out of Syllabus
If the lengths of the sides of a triangle are in A.P. and the greatest angle is double the smallest, then a ratio of lengths of the sides of this triangle is :
A
5 : 9 : 13
B
5 : 6 : 7
C
4 : 5 : 6
D
3 : 4 : 5
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