Question : In $\triangle$ABC, D and E are two points on the sides AB and AC, respectively, so that DE $\parallel$ BC and $\frac{AD}{BD}=\frac{2}{3}$. Then $\frac{\text{Area of trapezium DECB}}{\text{Area of $\triangle$ABC}}$ is equal to:

Option 1: $\frac{5}{9}$

Option 2: $\frac{21}{25}$

Option 3: $1\frac{4}{5}$

Option 4: $5\frac{1}{4}$

Question : In $\triangle ABC$, $D$ and $E$ are the points of sides $AB$ and $BC$ respectively such that $DE \parallel AC$ and $AD : DB = 3 : 2$. The ratio of the area of trapezium $ACED$ to that of $\triangle DBE$ is:

Option 1: $4:15$

Option 2: $15:4$

Option 3: $4:21$

Option 4: $21:4$

Question : DE is tangent to the circumcircle of $\triangle$ABC at the vertex A such that $DE \parallel BC$. If AB = 17 cm, then the length of AC is equal to:

Option 1: 16 cm

Option 2: 16.8 cm

Option 3: 17.3 cm

Option 4: 17 cm

Question : In a $\triangle ABC$, the median AD, BE, and CF meet at G, then which of the following is true?

Option 1: 4(AD + BE + CF) > 3(AB + BC + AC)

Option 2: 2(AD + BE + CF) > (AB + BC + AC)

Option 3: 3(AD + BE + CF) > 4(AB + BC + AC)

Option 4: AB + BC + AC > AD + BE + CF

Question : In $\triangle$ABC, BD and CE are perpendicular to AC and AB respectively. If BD = CE, then $\triangle$ABC is:

Option 1: Equilateral

Option 2: Isosceles

Option 3: Right–angled

Option 4: Scalene