
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
409.2k+ views
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 \[360^\circ \] .
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 \[360^\circ \] .
Recently Updated Pages
Master Class 11 Accountancy: Engaging Questions & Answers for Success

Express the following as a fraction and simplify a class 7 maths CBSE

The length and width of a rectangle are in ratio of class 7 maths CBSE

The ratio of the income to the expenditure of a family class 7 maths CBSE

How do you write 025 million in scientific notatio class 7 maths CBSE

How do you convert 295 meters per second to kilometers class 7 maths CBSE

Trending doubts
When people say No pun intended what does that mea class 8 english CBSE

How many ounces are in 500 mL class 8 maths CBSE

In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE

Which king started the organization of the Kumbh fair class 8 social science CBSE

What is BLO What is the full form of BLO class 8 social science CBSE

Advantages and disadvantages of science
