MCQ: Triangles - 1 - SSC CGL MCQ

GA = GB = GC is true only and only for equilateral triangle and here it is not given that ABC is an equilateral triangle. So, only for equilateral triangle.
Hence, it is also not correct.
Hence, option D is correct

In equilateral triangle centroid, incentre, orthocentre coincide at the same point.

∴ Height = 3 × in-radius = 3 × 3 = 9 cm.
Hence, option D is correct.

∠BCA = 60° [∵ ΔPQR is an equilateral]
∠BCD = ∠DCA = 30° [∵ CD is bisector of ∠C of an equilateral triangle]
∠DCE = 180°
∠ACE = 180° – 30° = 150°
∠CAE + ∠CEA = 180° – 150° = 30° ...(i)
Given, AC = CE
∴ ∠CEA = ∠CAE
From Eq. (i),
2∠CAE = 30°
∴ ∠CAE = 15°
Hence, option D is correct.

Since, AB is a straight line and C and D are points such that AC ⊥ AB and BD ⊥ AB.

∴ AC || BD
So, ABCD forms trapezium.
Now, by property of trapezium diagonals intersect each other in the ratio of lengths of parallel sides.

But the value of  Can't be determined.
So, we can't find the value of 

Hence, option D is correct.

In equilateral triangle orthocentre and centroid lie at the same point.
Hence, option C is correct.

AB = 10 cm
∴ BD = AB/2 = 5cm

∠ADB = 90°
By pythagoras theorem in ΔABD,

We know that,

Hence, option B is correct.

In a triangle
Sum to two sides is always greater than 3rd side
i.e., x < 40 ..... (i)
Difference of two sides is always less than 3rd side
i.e., 10 < x ..... (7ii)
From Eqs. (i) and (ii),
10 < x < 40.
Hence, option A is correct.

In an equilateral triangle, centroid, incentre etc lie at the same point.
Hence, option D is correct

Smart way :
Note : Radius of incircle of an equilateral triangle of side 

∴ Required radius of the incircle 

Traditional method :
AB = 6 cm
∴ BD = AB = 3 cm
2
∠ADB = 90°
By pythagoras theorem in ΔABD,

We know that,

Hence, option B is correct.

By hit and trial method,
Put n = 2 in option (d)
= [(2 × 2), (2)² – 1, (2)² + 1] = (4, 3, 5)
Which satisfy Pythagoras theorem and one side differes from hypotenuse by 2 units.
Hence, option D is correct.

Triangle will be equilateral.
Hence, option B is correct.

The median of an equilateral triangle are equal.
Hence, option A is correct.

Since, the triangle is right angled. So, all the three consecutive sides must satisfy Pythagoras theorem.

Hence, 3, 4 and 5 are the sides of triangle which satisfy pythagoras theoram.

Hence, option A is correct.

Let AB = BC = CA = 2x units
We know that a perpendicular from any vertex of an equilateral triangle bisects the opposite side.
∴ BD = CD = x units
∴ AB : BD = 2x : x = 2 : 1.
Hence, option C is correct.

AC = BC = 5 cm

Hence, option C is correct.