MCQ: Triangles - 2 - SSC CGL MCQ

In ΔABC,
AD is the internal angle bisector of ∠A.
Using property of internal angle bisector.

Hence, option D is correct.

Given that,
D, E and F are midpoints of BC, CA and AB and P, Q and R are midpoints of EF, FD and DE
we know that,
Area of ΔABC = 4 ΔDEF
But area of ABC = 64 sq. cm.

And area ΔDEF = 4 ΔPQR
⇒ 4 ΔPQR = 16 = 16/4 = 4 sq. units
Hence, option A is correct.

∠PRQ = 60° [∵ ΔPQR is an equilateral]
∠PRS = 180° – 60° = 120°
∠PSR + ∠RPS = 180° – 120° = 60° ...(i)
As QR = RS
∴ PR = RS [∵ ΔPQR is an equilateral]
∴ ∠PSR = ∠RPS
From Eq (i),
2∠PSR = 60°
∴ ∠PSR = 30°
Hence, option A is correct.

let the sides of triangle be a/r, a, ar and since r < 1.

Now, triangle is right angled.
Using Pythagoras theorem

Put r² = x
∴ x² + x – 1 = 0
Applying Sridharacharya rule, we get


Hence, option B is correct.

PS = 9 cm


RT = 6 cm

Hence, option C is correct.

Let AB = x cm

By pythagoras theorem in ΔABD,

We know that,

By pythagoras theorem in ΔBOD,

Given,
Circumradius, OB = 10 cm

Hence,

Hence, option A is correct.

∵ ΔABC – ΔDEF

Hence, option A is correct.

Since ∠A is an obtuse angle in Δ ABC, so

Hence, option C is correct.

Hence, option B is correct.

From Eqs. (i) and (ii),

Hence, option C is correct.

Given, AB = 12 cm, BC = 8 cm,
∠C = 59°
Let ∠A = Θ
∴ ∠B = 180° – (59° + Θ) = 121° – Θ
Now, let us see the choices. If AC = 12 cm, triangle would not be scalene. Hence, option A is ruled out. If AC =
10 cm, AB will become the largest side and ∠C the largest angle. But ∠C = 59°. Hence option B is ruled out. So,
AC is either 14 cm or 16 cm. In any case, ∠B will be the largest angle and ∠A (say Θ) the smallest:
Also, ∠B = 180° – (59° + Θ) = 121° – Θ
By sine formula

∴ sin (121° – Θ) ≈ sin (120° – Θ) = sin 120° cos Θ – cos120° sin Θ

Hence, option C is correct.

In an equilateral triangle, orthocentre, circum-cente, incentre and centroid coincide.
Hence, option C is correct.

ΔABC is right angled at B.
Using Pythagoras theorem,
AC² = AB² + BC²
AC = 10 cm

and in case of right angled triangle, radius lies on h hypotenuse and is the circumcircle of ΔABC.
Radius of circumcircle = 10/2 = 5 cm
Hence, option D is correct.

3x – 4y ≡ 0 ...(i)
5x + 12y ≡ 0 ....(ii)
y – 15 ≡ 0 ...(iii)

From (i) and (ii), A = (0, 0)
From (i) and (iii), B = (20, 15)
From (ii) and (iii), C = (–36, 15)

Let (α, β) be the incentre co-ordinates of ΔABC

Hence, option B is correct.

The line segments joining the mid points of the sides of a triangle form four triangles, each of
which is similar to the original triangle.
Hence, option A is correct.