Advertisements
Advertisements
Question
PQRS is a cyclic quadrilateral such that PR is a diameter of the circle. If ∠QPR = 67° and ∠SPR = 72°, then ∠QRS =
Options
41°
23°
67°
18°
Advertisements
Solution
Here we have a cyclic quadrilateral PQRS with PR being a diameter of the circle. Let the centre of this circle be ‘O’.
We are given that `angleQPR` and `angleSPR = 72°` . This is shown in fig (2).
So we see that,
\[\angle QPS = \angle QPR + \angle RPS\]
\[ = 67°+ 72° \]
\[ = 139°\]
In a cyclic quadrilateral it is known that the opposite angles are supplementary.
`angleQPS + angleQRS = 180°`
`angleQRS = 180° - angleQPS`
`= 180° - 139°`
= 41°
APPEARS IN
RELATED QUESTIONS
If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
Prove that the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.
In the given figure, ABCD is a cyclic quadrilateral. Find the value of x.
ABCD is a cyclic quadrilateral in ∠BCD = 100° and ∠ABD = 70° find ∠ADB.
Prove that the circles described on the four sides of a rhombus as diameters, pass through the point of intersection of its diagonals.
Find all the angles of the given cyclic quadrilateral ABCD in the figure.
If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral so formed is cyclic.
If non-parallel sides of a trapezium are equal, prove that it is cyclic.
If P, Q and R are the mid-points of the sides BC, CA and AB of a triangle and AD is the perpendicular from A on BC, prove that P, Q, R and D are concyclic.
If bisectors of opposite angles of a cyclic quadrilateral ABCD intersect the circle, circumscribing it at the points P and Q, prove that PQ is a diameter of the circle.
