Answer: Option C. -> Obtuse-angled triangle
According to question,
Given :
∠A = 21°,       ∠C = 38°
As we know that
∠A + ∠B + ∠C
∠B = 180° - 21° - 38°
∠B = 121°
∴ The triangle is obtuse-angled triangle.

Answer: Option D. -> 44°
According to question,
Given :
AB = AC
∠C = ∠A = 2x - 20°
∠B = x°
As we know that
∠A + ∠B + ∠C = 180°
(2x - 20)° + x + (2x - 20)° = 180°
5x = 220°
x = 44°
∴ ∠B = 44°

Answer: Option D. -> 29°
According to question,
Given :
AC = CD
∠BAD = 111°
∠ACB = 80°
∴ ∠ACD = 180° - 80°
∠ACD = 100°
In isosceles triangle ACD
   ∠ACD + ∠CAD + ∠ADC = 180°
   2∠CAD = 180° - 100°
   ∠CAD = 40°
∴ ∠CAB = 111° - 40° = 71°
∴ ∠ABC = 180° - 71° - 80°
   ∠ABC = 29°

Answer: Option B. -> 15 : 8
$$\eqalign{
& \frac{{{\text{area}}\,{\text{of}}\,{\text{triangle}}\,1}}{{{\text{area}}\,{\text{of}}\,{\text{triangle}}\,2}} = \frac{3}{2} \cr
& \Rightarrow \frac{{\frac{1}{2} \times {B_1} \times 4}}{{\frac{1}{2} \times {B_2} \times 5}} = \frac{3}{2} \cr
& \Rightarrow \frac{{{B_1}}}{{{B_2}}} = \frac{3}{2} \times \frac{5}{4} \cr
& \Rightarrow \frac{{{B_1}}}{{{B_2}}} = \frac{{15}}{8} \cr
& \therefore {B_1}:{B_2} = 15:8 \cr} $$

Answer: Option D. -> 13 cm
Given :
AB = 19 cm, AC = 22 cm
∵ BE ⊥ CF (given), [Medians CF & BE are perpendicular to each other]
⇒ In this case
⇒ We know,
   AB2 + AC2 = 5(BC)2
⇒ 192 + 222 = 5(BC)2
⇒ 361 + 484 = 5(BC)2
⇒ 845 = 5(BC)2
⇒ (BC)2 = 169
⇒ BC = 13 cm

Answer: Option C. -> 120°
According to question,
As we know that
   ∠A + ∠B + ∠C = 180°
∴ ∠A = 180° -70° - 50°
   ∠A = 60°
∴ ∠BIC = 90° + $$\frac{1}{2}$$ × 60°
   ∠BIC = 120°

Answer: Option B. -> BD2 + AC2
According to question,
Given :
AD ⊥ BC
∴ In ΔADB
AB2 = BD2 + AD2
AD2 = AB2 - BD2 . . . . . . . . (i)
In ΔADC
AC2 = AD2 + CD2
AD2 = AC2 - CD2 . . . . . . . (ii)
Compare equation (i) and (ii)
AB2 - BD2 = AC2 - CD2
AB2 + CD2 = BD2 + AC2

Answer: Option C. -> 55°
As we know
⇒ The external bisectors of the angle ∠B and ∠C meet at the point O
∠BOC = 90° - $$\frac{{\angle A}}{2}$$
∠BOC = 90° - $$\frac{{70}}{2}$$
∠BOC = 90° - 35°
∠BOC = 55°

Answer: Option C. -> 90°
In ΔABC
∠B = ∠A = 50°
∠ACD = 50° + 50°
∠ACD = 100°
∠ACD is the external angle or ΔABC
∠ACD + ∠CAD + ∠ADC = 180°
∠CAD = ∠ADC
∵ AC = CD
2∠CAD = 180 - 100
∠CAD = 40°
∠BAD = 50° + 40°
∠BAD = 90°

Answer: Option D. -> 3 cm
$$\eqalign{
& \frac{{PQ}}{{LM}} = \frac{{QR}}{{MN}} = \frac{{PR}}{{LN}} \cr
& \left( {\vartriangle PQR\,{\text{and}}\,\vartriangle LMN\,{\text{are}}\,{\text{similar}}} \right) \cr
& \frac{{PQ}}{{LM}} = \frac{{QR}}{{MN}} \cr
& \frac{1}{3} = \frac{{QR}}{9} \cr
& QR = 3\,{\text{cm}} \cr} $$