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.
$$=\frac{1}{2}n\cdot7$$=$$\frac{7n}{2}$$.
$$\frac{11n}{2}+\frac{7n}{2}=225\ \ \Rightarrow\ 9n=225\ \Rightarrow n=25$$