PQRS is a quadrilateral with side PS = 7 cm and QR= 11 cm. $$\angle SPQ$$ and $$\angle QRS$$ are both right angles. If E and F are points on PQ and RS, respectively, and QE is twice SF; SF = n cm and n is an integer, then what is the value of n such that the area of the quadrilateral EQFS is 225 $$cm^{2}$$?

Here, we have to find the area of the  quadrilateral EQFS.

Lets draw a line from S to Q.

From the image we can say that the area of  SFQE = area of $$\triangle\ $$SFQ + area of $$\triangle\ $$SEQ

area of $$\triangle\ $$SFQ= $$\frac{1}{2}\cdot b\cdot h$$

We know that SR and RQ are perpendicular(given in question). So, height of $$\triangle\ $$SFQ is 11.

$$=\frac{1}{2}n\cdot11$$=$$\frac{11n}{2}$$.

area of $$\triangle\ $$SEQ = $$\frac{1}{2}\cdot b\cdot h$$

We know that SP and PQ are perpendicular(given in question). So, height of $$\triangle\ $$SEQ is 7.

$$=\frac{1}{2}n\cdot7$$=$$\frac{7n}{2}$$.

$$\frac{11n}{2}+\frac{7n}{2}=225\ \ \Rightarrow\ 9n=225\ \Rightarrow n=25$$