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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 P+Q+R+S+T+U is
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A. 270
B. 360
C. 450
D. 540

Answer
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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 POT is an external angle of triangle POQ . Hence POT can be written as the sum of opposite two internal angles.
i.e. POT=OPQ+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,
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Let us assume, the Sum of all the angles in a triangle is 180 .
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,
 POT=P+Q ………………….(1)
 ROT=T+U …………………(2)
 POR=R+S …………………..(3)
Adding the equation (1), (2) and (3) , then we get
 P+Q+R+S+T+U=POT+ROT+POR
We know that the sum of all the angles about a point is equal to 360 .
Therefore, by the point O we can get
 POT+ROT+POR=360
Hence, P+Q+R+S+T+U=360
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 P+Q+R+S+T+U is 360
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
External angle is equal to sum of opposite two internal angles.
Sum of all the angles about a point is equal to 360 .