+0  
 
0
63
2
avatar

Peter has built a gazebo, whose shape is a regular heptagon, with a side length of 1 unit. He has also built a pathway around the gazebo, of constant width 1 unit, as shown below. (Every point on the ground that is within 1 unit of the gazebo and outside the gazebo is covered by the pathway.) Find the area of the pathway.

 

https://latex.artofproblemsolving.com/f/f/7/ff7af7da5a2907b44a55ae65671c4c5c238f574d.png

 Mar 4, 2020
 #1
avatar
0

The area is 2*pi.

 Mar 4, 2020
 #2
avatar+24366 
+1

Peter has built a gazebo, whose shape is a regular heptagon, with a side length of 1 unit.
He has also built a pathway around the gazebo, of constant width 1 unit, as shown below.
(Every point on the ground that is within 1 unit of the gazebo and outside the gazebo is covered by the pathway.)
Find the area of the pathway.

\(\begin{array}{|rcll|} \hline \alpha +\beta &=& 180^\circ \quad & | \quad \mathbf{\beta = \dfrac{7*180^\circ-360^\circ}{7}} =\dfrac{900^\circ}{7} \\ \alpha +\dfrac{900^\circ}{7} &=& 180^\circ \\ \alpha &=& 180^\circ-\dfrac{900^\circ}{7} \\ \mathbf{\alpha} &=& \mathbf{\dfrac{360^\circ}{7}} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \mathbf{\text{area of the pathway}} \\ \hline &=& 7\times(1\times 1 ) + 7 \times \left( \pi \times 1^2 \times\dfrac{\alpha}{360^\circ} \right) \quad & | \quad \mathbf{\alpha=\dfrac{360^\circ}{7}} \\\\ &=& 7 + 7 \times \left( \pi \times\dfrac{\dfrac{360^\circ}{7}}{360^\circ} \right) \\\\ &=& 7 + 7 \times \left( \pi \times\dfrac{1}{7} \right) \\\\ &=&\mathbf{ 7 + \pi} \\ \hline \end{array}\)

 

laugh

 Mar 4, 2020

38 Online Users

avatar
avatar
avatar
avatar
avatar
avatar
avatar
avatar