Question

A rectangular path of length $(x-2)$ ft and width $\frac{7}{2}\text{ft}$ passes perpendicularly through another rectangular path of length $(y-\frac{1}{3})\text{ft}$ and width $\frac{13}{4}\text{ft}$. What will be the overall area of the two paths?

• A. $(y-\frac{1}{3})\times \frac{13}{4}+(x-2)\times \frac{7}{2}-\frac{13}{4}\times \frac{7}{2}$
• B. $(y-\frac{1}{3})\frac{13}{4}+(x-2)\times \frac{7}{2}+\frac{3}{4}\times \frac{7}{2}$
• C. $(y-\frac{1}{3})\times \frac{13}{4}+(x-2)\times \frac{1}{2}+\frac{13}{4}$
• D. $(y-\frac{1}{3})\times \frac{13}{4}+(x-2)\times \frac{7}{2}-\frac{13}{4}$

Answer 1

The correct option is A $(y-\frac{1}{3})\times \frac{13}{4}+(x-2)\times \frac{7}{2}-\frac{13}{4}\times \frac{7}{2}$


The overall area of the two paths = (Area of AFGL
-Area of BENK) + (Area of CDZH - Area of BENK) + Area of BENK
The overall area of the two paths = Area of AFGL - Area of BENK + Area of CDZH

Area of BENK $=\text{BK}\times \text{KN}$
Area of BENK $=\text{AL}\times \text{CD}$ [$\because$ BK = AL and KN = CD]
Area of BENK $=\frac{13}{4}\times \frac{7}{2}$

Area of AFGL$=\text{GL}\times \text{AL}$
Area of AFGL $=(y-\frac{1}{3})\times \frac{13}{4}\text{ft}$

Area of CDZH$=\text{DZ}\times \text{CD}$
Area of CDZH $=(x-2)\times \frac{7}{2}$

Overall area $=(y-\frac{1}{3})\times \frac{13}{4}-\frac{13}{4}\times \frac{7}{2}+(x-2)\times \frac{7}{2}$

$=(y-\frac{1}{3})\times \frac{13}{4}+(x-2)\times \frac{7}{2}-\frac{13}{4}\times \frac{7}{2}$
The overall area of the two paths is $(y-\frac{1}{3})\times \frac{13}{4}+(x-2)\times \frac{7}{2}-\frac{13}{4}\times \frac{7}{2}$.
Note: overall area: Area of rectangle 1 + Area of rectangle 2 - Common area (taken twice).