In $\triangle \mathrm{ABC}, \mathrm{D}$ is a point on side $\mathrm{AB}$ such that $\mathrm{BD}=3 \mathrm{~cm}$ and $\mathrm{DA}=4 \mathrm{~cm}$. $\mathrm{E}$ is a point on $\mathrm{BC}$ such that $\mathrm{DE} \| \mathrm{AC}$,. Then Area of $\triangle \mathrm{BDE}$ : Area of trapezium $\mathrm{ACED}=$
Answer & explanation
Correct answer: option 4
Here, \(\Delta \)ABC is similar to \(\Delta \)BDE [DE is parallel to AC and \(\angle\)B is common for two triangles]
So,
\(\Delta \)ABC/\(\Delta \)BDE = \( { 7}^{2 } \)/\( { 3}^{2 } \)
= \(\Delta \)ABC/\(\Delta \)BDE = \(\frac{49}{9}\)
ACED = \(\Delta \)ABC - \(\Delta \)BDE
= 49 - 9 = 40
Therefore, ratio is 9 : 40.