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In the diagram, /PQ/ = /QR/ and /PR/ = /RS/ = /SP/, calculate the side of < QRS....


Question

In the diagram, /PQ/ = /QR/ and /PR/ = /RS/ = /SP/, calculate the side of < QRS.

Options

A) 150o

B) 120o

C) 90o

D) 60o

The correct answer is C.

Explanation:

Since |PR| = |RS| = |SP|
\(\bigtriangleup\) PRS is equilateral and so < RPS = < PRS = < PSR = 70\(^{\circ}\)
But < PQR + < PSR = 180\(^{\circ}\)(Opposite interior angles of a cyclic quadrilateral)
< PQR + 60 = 180\(^{\circ}\)
< PR = 180 - 60 = 120\(^{\circ}\)
But in \(\bigtriangleup\) PQR, PQ = PR, hence < QPR = < PRQ(Base angles of an Isosceles triangle)
< QPR + < PRQ + < PQR = 180\(^{\circ}\)(Angles in a triangle)
2 < QPR + 120 = 18-
2 < QPR = 180 - 120
QPR = \(\frac{60}{2}\) = 30\(^{\circ}\)
From the diagram, < QRS = < PRQ + < PRS
30 + 60 = 90\(^{\circ}\)

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Dicussion (1)

  • Since |PR| = |RS| = |SP|
    \(\bigtriangleup\) PRS is equilateral and so < RPS = < PRS = < PSR = 70\(^{\circ}\)
    But < PQR + < PSR = 180\(^{\circ}\)(Opposite interior angles of a cyclic quadrilateral)
    < PQR + 60 = 180\(^{\circ}\)
    < PR = 180 - 60 = 120\(^{\circ}\)
    But in \(\bigtriangleup\) PQR, PQ = PR, hence < QPR = < PRQ(Base angles of an Isosceles triangle)
    < QPR + < PRQ + < PQR = 180\(^{\circ}\)(Angles in a triangle)
    2 < QPR + 120 = 18-
    2 < QPR = 180 - 120
    QPR = \(\frac{60}{2}\) = 30\(^{\circ}\)
    From the diagram, < QRS = < PRQ + < PRS
    30 + 60 = 90\(^{\circ}\)

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