Question

The given figure shows a pentagon ABCDE with sides AB and ED parallel to each other. $$\mathrm{\angle }B:\mathrm{\angle }C:\mathrm{\angle }D=5:6:7$$. Using the formula, find the sum of interior angles of the pentagon.

Answer 1

Hint: We know that polygons have no number of sides. Like quadrilaterals have 4 sides, pentagon has 5 sides and hexagon has 6 sides. We have a formula to find the sum of interior angles of these quadrilaterals.
Formula used:
Sum of interior angles of a polygon with n sides is given by, $$\left(n-2\right)\times {180}^{\circ }$$ where n is the number of sides of that polygon.

Complete step by step solution:
Given is a pentagon.
Such that the number of sides is equal to five.
So n=5.
Now since AB is parallel to ED we can say that both angles A and E are right angles.
Now the remaining three angles are in the ratio $$\mathrm{\angle }B:\mathrm{\angle }C:\mathrm{\angle }D=5:6:7$$
We will use the formula now to find the sum of interior angles.
$$\left(n-2\right)\times {180}^{\circ }$$
Now we will put n equals to 5.
$$=\left(5-2\right)\times {180}^{\circ }$$
On calculating we get,
$$=3\times {180}^{\circ }$$
On multiplying we get,
$$={540}^{\circ }$$
Thus the sum of interior angles of the pentagon ABCDE is $$={540}^{\circ }$$
So, the correct answer is “ $${540}^{\circ }$$”.

Note: Note that, here the ratio of the angles is not even required. Because we are not asked to find the measure of any of the angles. If so then we must need them.
Also note that the term n-2 in the formula gives the number of triangles so formed in a polygon. And we know that the sum of all angles of a triangle is $$={180}^{\circ }$$.
Like for a rectangle this formula will be, $$=\left(4-2\right)\times {180}^{\circ }$$ since the number of sides are 4 and sum of interior angles will be, $$={360}^{\circ }$$
Hope this is clear!