Question

The angles of a pentagon in degrees are $$y^{\circ },(y+{20}^{\circ }),(y+{40}^{\circ }),(y+{60}^{\circ })$$ and $$(y+{80}^{\circ }).$$ The smallest angle of the pentagon is:
$$\left(\text{a}\right){88}^{\circ }$$
$$\left(\text{b}\right)\text{ 7}{8}^{\circ }$$
$$\left(\text{c}\right)\text{ 6}{8}^{\circ }$$
$$\left(\text{d}\right)\text{ 5}{8}^{\circ }$$

Answer 1

Hint: To solve the given question, we will first find out what a pentagon is. Then, we will find out the sum of the interior angles in the pentagon with the help of the formula: $$\text{Sum of interior angles}=(n-2)\times {180}^{\circ },$$ where n is the number of sides in the polygon. Then, we will add all the interior angles in the given polygon and equate it to the sum of the interior angles. From here, we will get the value of y which is also the smallest angle in the given pentagon.

Complete step-by-step answer:
Before we solve the question, we will first find out what kind of polygon is a pentagon. A pentagon is a polygon having five sides or edges and five angles. The angles may or may not be equal. In our case, there are five angles of the pentagon of which all are different so the given pentagon is an irregular pentagon.
Let us draw a representative figure of pentagon and all the angles as shown below,

Now, we will find the sum of interior angles in a pentagon. The sum of interior angles in a polygon with n sides is given by the formula,
$$\text{Sum of interior angles}=(n-2)\times {180}^{\circ }$$
In our case, n = 5, so we will get,
$$\Rightarrow \text{Sum of interior angles}=(5-2)\times {180}^{\circ }=3\times {180}^{\circ }$$
$$\Rightarrow \text{Sum of interior angles}={540}^{\circ }$$
Now, the angles which are given in the question should have the sum equal to the sum of interior angles. Thus, we can say that,
$$y^{\circ }+(y+{20}^{\circ })+(y+{40}^{\circ })+(y+{60}^{\circ })+(y+{80}^{\circ })={540}^{\circ }$$
$$\Rightarrow 5y+{200}^{\circ }={540}^{\circ }$$
$$\Rightarrow 5y={540}^{\circ }-{200}^{\circ }$$
$$\Rightarrow y={\displaystyle \frac{{340}^{\circ }}{5}}$$
$$\Rightarrow y={68}^{\circ }$$
Now, we have to find the smallest angle of the pentagon. Among the five angles given in the pentagon, the smallest angle us y. So, the smallest angle of the pentagon will be $${68}^{\circ }.$$

Note: We can also proceed in an alternate way after finding the sum of the interior angles as $${540}^{\circ }.$$ This is as shown. We know that the angles given in the question are in AP with a common difference $${20}^{\circ }.$$ If the AP is of the form: $$a_1,a_2,a_3,.......a_n$$ then the sum is given by
$$S={\displaystyle \frac{n}{2}}[a+a_n]$$
Thus, we have,
$$540={\displaystyle \frac{5}{2}}[y+y+80]$$
$$\Rightarrow 1080=5[2y+80]$$