Answer: Option D. -> 55°
According to question,
In right angle ΔBAC
∠A + ∠B + ∠C = 180°
∠B = 180° - 55° - 90°
∠B = 35°
In right angle ΔADB
∠ADB + ∠ABD + ∠BAD = 180°
∠BAD = 180° - 35° - 90°
∠BAD = 55°
Alternate
ΔBAC ∼ ΔBDA
∴ ∠BCA = ∠BAD = 55°

Answer: Option C. -> 60°
According to question,
Given : ∠BIC = 120°
∠BIC = 90° + $$\frac{1}{2}$$ ∠A
$$\frac{{\angle A}}{2}$$  = (120° - 90°)
$$\frac{{\angle A}}{2}$$  = 30°
∠A = 60°

Answer: Option B. -> $$\frac{{10\sqrt 3 }}{3}\,cm$$
According to question,
Given :
AB = BC = CA = 10 cm
G = Centroid
AG = 2 units
GD = 1 unit
AD = 3 units = Height
As we know that the height of the equilateral triangle is
$$\eqalign{
& = \frac{{\sqrt 3 }}{2} \times 10 = 5\sqrt 3 \cr
& \therefore 3\,{\text{units}} = 5\sqrt 3 \cr
& \,\,\,\,\,\,1\,{\text{unit}} = \frac{{5\sqrt 3 }}{3} \cr
& \,\,\,\,\,\,2\,{\text{units}} = \frac{{5\sqrt 3 }}{3} \times 2 \cr
& \,\,\,\,\,\,2\,{\text{units}} = \frac{{10\sqrt 3 }}{3} \cr
& \therefore {\text{AG}} = \frac{{10\sqrt 3 }}{3}\,cm \cr} $$

Answer: Option B. -> 2 cm
According to question,
Given :
AB = 6 cm,         BC = 8 cm
In right angle ΔABC
By using Pythagoras theorem
AC2 = AB2 + BC2
AC2 = 62 + 82
AC2 = 36 + 64
AC2 = 100
AC  = 10 cm
In radius
$$\eqalign{
& = \frac{{a + b - c}}{2} \cr
& = \frac{{8 + 6 - 10}}{2} \cr
& = \frac{4}{2} \cr
& = 2\,cm \cr} $$

Answer: Option A. -> AB2 + CD2 = AD2 + BC2
According to question,
In ΔABC
AB2 = AC2 + BC2 . . . . . . . (i)
ΔACD
AD2 = AC2 + CD2
AC2 = AD2 - CD2 . . . . . . . (ii)
Put the value of AC2 in equation (i)
AB2 = AD2 - CD2 + BC2
AB2 + CD2 = AD2 + BC2

Answer: Option C. -> $$2\sqrt 3 $$  cm
In Equilateral triangle
$$\eqalign{
& AG:GD = 2:1 \cr
& AD = \frac{{\sqrt 3 }}{2}a \cr
& AD = \frac{{\sqrt 3 }}{2} \times 6 \cr
& AD = 3\sqrt 3 \cr
& 3 \to 3\sqrt 3 \cr
& 1 \to \sqrt 3 \cr
& AG = 2\,{\text{unit}} = 2\sqrt 3 \,{\text{cm}} \cr} $$

Answer: Option A. -> Inside the triangle
In acute angled triangle orthocenter is always inside the triangle

Answer: Option C. -> 6 cm
Median of right angle
$$\eqalign{
& = \frac{{EF}}{2} \cr
& = \frac{{12}}{2} \cr
& = 6\,{\text{cm}} \cr} $$

Answer: Option A. -> 20 cm
Let angle = x, 2x, 3x
x + 2x + 3x = 180
(∵ Sum of internal angle of a Δ)
6x = 180°
x = 30°
So, angle = 30, 60, 90
Smallest side of Δ
= 1 unit
= 10 cm
Largest side of Δ
= 2 units
= 20 cm

Answer: Option B. -> 15°, 125°
∠C = 180 - (∠A + ∠B)
∠C = 180 - 150
2x = 30
x = 15°
∠BDC = 90° + $$\frac{1}{2}$$ ∠A
∠BDC = 90° + $$\frac{1}{2}$$ × 70°
∠BDC = 90° + 35°
∠BDC = 125°
So value of x and y are = 15°, 125°