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

# 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.

A) 150o

B) 120o

C) 90o

D) 60o

### 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$$^{\circ}$$

## 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$$^{\circ}$$