Question

A field is in the shape of a pentagon as shown in the figure. Three of the interior angles of the pentagon are right angles. The remaining two interior angles are congruent. What is the measure of each angle?

Answer 1

Hint: For solving this question firstly we will draw the given figure and consider there is a pentagon ABCDE as per the given configurations. After that, we will take a point O inside the pentagon and join every vertex of it to point O. Then, we will get five triangles and we will use one of the important result that the sum of the interior angles of a triangle is always ${180}^0$ to find the unknown angles easily.

Given:
It is given that there is a field that is in the shape of a pentagon as shown in the figure. Three of the interior angles of the pentagon are right angles. The remaining two interior angles are congruent, and we have to find the measure of each angle.
Now, let there be a pentagon ABCDE in which $\mathrm{\angle }DAB=\mathrm{\angle }CBA=\mathrm{\angle }DEC={90}^0$ and $\mathrm{\angle }EDA=\mathrm{\angle }ECB$ . For more clarity look at the figure given below:

Now, just take a point O inside of the pentagon and join the point O with each vertex of the pentagon as shown in the figure below:

Now, as we know that sum of the interior angles of a triangle is ${180}^0$ so, sum of the interior angles of the $\mathrm{\Delta }AOB,\mathrm{\Delta }BOC,\mathrm{\Delta }COE,\mathrm{\Delta }DOE$ and $\mathrm{\Delta }DOA$ will be ${180}^0\times 5={900}^0$ . Then,
$\begin{array}{rl}& (\mathrm{\angle }AOB+\mathrm{\angle }OAB+\mathrm{\angle }OBA)+(\mathrm{\angle }BOC+\mathrm{\angle }CBO+\mathrm{\angle }BCO)+(\mathrm{\angle }COE+\mathrm{\angle }OEC+\mathrm{\angle }OCE)\\ & +(\mathrm{\angle }DOE+\mathrm{\angle }ODE+\mathrm{\angle }OED)+(\mathrm{\angle }AOD+\mathrm{\angle }OAD+\mathrm{\angle }ODA)={900}^0\\ & \Rightarrow (\mathrm{\angle }AOB+\mathrm{\angle }BOC+\mathrm{\angle }COE+\mathrm{\angle }DOE+\mathrm{\angle }AOD)+(\mathrm{\angle }OAB+\mathrm{\angle }OAD)+(\mathrm{\angle }OBA+\mathrm{\angle }CBO)\\ & +(\mathrm{\angle }BCO+\mathrm{\angle }OCE)+(\mathrm{\angle }OED+\mathrm{\angle }OEC)+(\mathrm{\angle }ODE+\mathrm{\angle }ODA)={900}^0\\ & \Rightarrow (\mathrm{\angle }AOB+\mathrm{\angle }BOC+\mathrm{\angle }COE+\mathrm{\angle }DOE+\mathrm{\angle }AOD)+\mathrm{\angle }DAB+\mathrm{\angle }CBA+\mathrm{\angle }ECB+\mathrm{\angle }EDA+\mathrm{\angle }DEC={900}^0\end{array}$
Now, as the lines are concurrent and intersect at point O. So, the value of the $\mathrm{\angle }AOB+\mathrm{\angle }BOC+\mathrm{\angle }COE+\mathrm{\angle }DOE+\mathrm{\angle }AOD$ will be equal to ${360}^0$ and it is given that $\mathrm{\angle }DAB=\mathrm{\angle }CBA=\mathrm{\angle }DEC={90}^0$ and $\mathrm{\angle }EDA=\mathrm{\angle }ECB$ . Then,
$\begin{array}{rl}& (\mathrm{\angle }AOB+\mathrm{\angle }BOC+\mathrm{\angle }COE+\mathrm{\angle }DOE+\mathrm{\angle }AOD)+\mathrm{\angle }DAB+\mathrm{\angle }CBA+\mathrm{\angle }ECB+\mathrm{\angle }EDA+\mathrm{\angle }DEC={900}^0\\ & \Rightarrow {360}^0+{90}^0+{90}^0+{90}^0+\mathrm{\angle }ECB+\mathrm{\angle }ECB={900}^0\\ & \Rightarrow {630}^0+2\mathrm{\angle }ECB={900}^0\\ & \Rightarrow 2\mathrm{\angle }ECB={270}^0\\ & \Rightarrow \mathrm{\angle }ECB={135}^0\end{array}$
Now, from the above result, we conclude that the value of $\mathrm{\angle }EDA=\mathrm{\angle }ECB={135}^0$ .
Thus, in the given pentagon there will be three angles of ${90}^0$ and two angles will be of ${135}^0$ .


Note: Here, we should first understand what is asked in the question and then proceed in the right direction to the correct answer quickly. Moreover, for objective problems, we should use the formula $(n-2)\times {180}^0$ to find the sum of all interior angles of $n$ sided polygon.