Question

Lines \[PS\] , \[RU\] and \[QT\] intersect at a common point \[O\] as shown in figure. \[\;P\] is joined to \[Q\] , \[R\] to \[S\] and \[T\] to \[U\] to form a triangle. The value of is \[\angle P + \angle Q + \angle R + \angle S + \angle T + \angle U\] is

A. \[{270^ \circ }\]
B. \[{360^ \circ }\]
C. \[{450^ \circ }\]
D. \[{540^ \circ }\]

Answer 1

Hint: In this problem we have to find the sum of the angles.
We will use the fact that the external angle of a triangle is equal to the sum of opposite two internal angles. For example: in the above figure \[\angle POT\] is an external angle of triangle \[\vartriangle POQ\] . Hence \[\angle POT\] can be written as the sum of opposite two internal angles.
i.e. \[\angle POT = \angle OPQ + \angle OQP\]

Complete step by step solution:
This problem is based on triangles. Triangle is a closed figure bounded by a three line segment. A triangle has three vertices, three sides and three angles.
The given diagram is mentioned below,

Let us assume, the Sum of all the angles in a triangle is \[180^\circ \] .
Consider the given question,
We know that the external angle is equal to the sum of opposite two internal angles.
Therefore from the figure we have,
 \[\angle POT = \angle P + \angle Q\] ………………….(1)
 \[\angle ROT = \angle T + \angle U\] …………………(2)
 \[\angle POR = \angle R + \angle S\] …………………..(3)
Adding the equation (1), (2) and (3) , then we get
 \[\angle P + \angle Q + \angle R + \angle S + \angle T + \angle U = \angle POT + \angle ROT + \angle POR\]
We know that the sum of all the angles about a point is equal to \[360^\circ \] .
Therefore, by the point \[O\] we can get
 \[\angle POT + \angle ROT + \angle POR = 360^\circ \]
Hence, \[\angle P + \angle Q + \angle R + \angle S + \angle T + \angle U = {360^ \circ }\]
Therefore, Lines \[PS\] , \[RU\] and \[QT\] intersect at a common point \[O\] as shown in figure. \[\;P\] is joined to \[Q\] , \[R\] to \[S\] and \[T\] to \[U\] to form a triangle.
The value of is \[\angle P + \angle Q + \angle R + \angle S + \angle T + \angle U\] is \[{360^ \circ }\]
Hence option \[B\] is the correct answer.
So, the correct answer is “Option B”.

Note: Triangle is a closed figure bounded by a three line segment.
There are three types of triangle : scalene triangle, equilateral triangle and isosceles triangle.
Sum of all the angle of a triangle is \[180^\circ \]
External angle is equal to sum of opposite two internal angles.
Sum of all the angles about a point is equal to \[360^\circ \] .