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\n \n \n \n \n A. $${270^ \circ }$$ B. $${360^ \circ }$$ C. $${450^ \circ }$$ D. $${540^ \circ }$$

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\n    \n    \n    \n    \n  A. $${270^ \circ }$$  B. $${360^ \circ }$$ C. $${450^ \circ }$$ D. $${540^ \circ }$$

Vedantu Content Team · Accepted Answer

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,\n    \n    \n    \n    \n  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  $$360^\circ $$ .

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 A. 270∘ B. 360∘ C. 450∘ D. 540∘

Lines $P S$ , $R U$ and $Q T$ 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

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