Question : In the given figure, the area of isosceles triangle $\text{PQT}$ is $128\;\mathrm{cm^2}$ and $\text{QT = PQ}$, $\text{PQ= 4 PS}$ and $\text{PT || SR}$. Then what is the area (in $\mathrm{cm^2}$ ) of the quadrilateral $\text{PTRS}$?

Option 1: $80$
Option 2: $64$
Option 3: $124$
Option 4: $72$

Question

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

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Answer 1

Correct Answer: $64$

Solution : Given that $\mathrm{PQT}$ is an isosceles triangle with $\mathrm{QT = PQ}$, and $\mathrm{PQ= 4 PS}$.
The height of the triangle $\mathrm{PQT}$ is four times the height of the quadrilateral $\mathrm{PTRS}$.
The area of a triangle $=\frac{1}{2} \times \text{base} \times \text{height}$
The area of the triangle $\mathrm{PQT =\frac{1}{2} \times QT \times PQ} = 128\;\mathrm{cm^2}$
Since $\mathrm{QT = PQ}$,
The area of the triangle $\mathrm{PQT=\frac{1}{2} \times QT^2} = 128\;\mathrm{cm^2}$
$\mathrm{QT} = \sqrt{2 \times 128} = 16\;\mathrm{cm}$
Since $\mathrm{PQ = 4 PS}$
The height of the quadrilateral $\mathrm{PTRS=\frac{QT}{4}} = \frac{16}{4} = 4\;\mathrm{cm}$
The area of a parallelogram (since $\mathrm{PT\parallel SR}$) $=\text{base} \times \text{height}$
The area of the quadrilateral $\mathrm{PTRS=QT \times \frac{QT}{4}} = 16 \times 4 = 64\;\mathrm{cm^2}$
Hence, the correct answer is $64$.