5. A triangular playground has sides, in metres, measuring 7,24 and 25 . Inside the playground, a lawn is designed so that the distance from each point on the edge of the lawn to the nearest side is 2 metres. What is the area of the lawn?
SolUTION
First solution
Since the corresponding sides of the two triangles are parallel, there is a centre of enlargement $O$ so that $\frac{O P}{O P^{\prime}}=\frac{O Q}{O Q^{\prime}}=\frac{O R}{O R^{\prime}}=k$
The same is true for the perpendiculars from $O$ to the sides, so $\frac{O X}{O X^{\prime}}=\frac{O Y}{O Y^{\prime}}=\frac{O Z}{O Z^{\prime}}=k$
The width of the path is constant, so $X X^{\prime}=Y Y^{\prime}=Z Z^{\prime}$ and hence $O X=O Y=O Z$
It follows that $O$ is the incentre of both $\triangle P Q R$ and $\triangle P^{\prime} Q^{\prime} R^{\prime}$
The area of $\triangle P^{\prime} Q^{\prime} R^{\prime}$ is 84 and its semiperimeter is 28 . Using the formula $\operatorname{Area}(\Delta)=r s$ we have $r=3$
The width of the path is 2 , so $O X=1$ and $k=\frac{1}{3}$. Hence the area of the lawn is $\frac{28}{3}$
Second solution
Let the ratio of the lengths of the sides of the small triangle to the lengths of the sides of the large triangle be $k$. Then we have $P Q=24 k, Q R=25 k, R P=7 k$ and also $P^{\prime} Z=P^{\prime} Z^{\prime}=2$
Hence $X Q^{\prime}=X^{\prime} Q^{\prime}=22-24 k$ and $Y R^{\prime}=Y^{\prime} R^{\prime}=5-7 k$
Now, evaluating $X^{\prime} R^{\prime}$ in two ways, we have $25-(5-7 k)-(22-24 k)=25 k$ and so $k=\frac{1}{3}$
The rest of the proof follows as in the first solution.
Third solution
This approach focuses on the two small triangles with a vertex at $P$, which are similar to $P^{\prime} Q^{\prime} R^{\prime}$ with scale factor $\frac{2}{7}$
Hence $m=\frac{48}{7}$ and $n=\frac{50}{7}$
However, the parallelogram with opposite vertices at $P$ and $P^{\prime}$ is a rhombus, and now we calculate $P Q$ as $24-2-m-n=8$. The rest follows.