Question : In $\Delta \mathrm{ABC}$, a line parallel to side $\mathrm{BC}$ cuts the sides $\mathrm{AB}$ and $\mathrm{AC}$ at points $\mathrm{D}$ and $\mathrm{E}$ respectively and also point $\mathrm{D}$ divides $\mathrm{AB}$ in the ratio of $\mathrm{1 : 4}$. If the area of $\Delta \mathrm{ABC}$ is $200\;\mathrm{cm^2}$, then what is the area (in $\mathrm{cm^2}$) of quadrilateral $\mathrm{DECB}$?
Option 1: 192
Option 2: 50
Option 3: 120
Option 4: 96
Correct Answer: 192
Solution :
Given: $\mathrm{AD:DB=1:4}$ and the area of $\Delta \mathrm{ABC}$ is $200\;\mathrm{cm^2}$.
In $\Delta \mathrm{ABC}$,
$\mathrm{DE}\parallel\mathrm{BC}$
So, $\Delta \mathrm{ADE}\sim\Delta \mathrm{ABC}$,
The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
⇒ $\frac{\text{Area of $\Delta \mathrm{ADE}$}}{{\text{Area of $\Delta \mathrm{ABC}$}}}=(\frac{\mathrm{AD}}{\mathrm{AB}})^2$
⇒ $\frac{\text{Area of $\Delta \mathrm{ADE}$}}{{200}}=(\frac{\mathrm{1}}{\mathrm{5}})^2$
⇒ $\frac{\text{Area of $\Delta \mathrm{ADE}$}}{{200}}=(\frac{\mathrm{1}}{\mathrm{25}})$
⇒ $\text{Area of $\Delta \mathrm{ADE}$}=8\;\mathrm{cm^2}$
The area of quadrilateral $\mathrm{DECB}=\text{Area of $\Delta \mathrm{ABC}$}-\text{Area of $\Delta \mathrm{ADE}$}$
$=200\;\mathrm{cm^2}-8\;\mathrm{cm^2}=192\;\mathrm{cm^2}$
Hence, the correct answer is $192$.
Related Questions
Know More about
Staff Selection Commission Sub Inspector ...
Result | Eligibility | Application | Selection Process | Cutoff | Admit Card | Preparation Tips
Get Updates Brochure
Your Staff Selection Commission Sub Inspector Exam brochure has been successfully mailed to your registered email id “”.
