State whether the following statement is True or False and justify your answer: $ABCD$ is a cyclic quadrilateral such that $\angle A = 90^{\circ}, \angle B = 70^{\circ}, \angle C = 95^{\circ}$ and $\angle D = 105^{\circ}$.
(B) We know that the sum of opposite angles of a cyclic quadrilateral is always $180^{\circ}$.
In the given quadrilateral $ABCD$,let us check the sum of opposite angles:
$\angle A + \angle C = 90^{\circ} + 95^{\circ} = 185^{\circ}$.
Since the sum of opposite angles is not $180^{\circ}$,the quadrilateral $ABCD$ cannot be a cyclic quadrilateral.
Therefore,the given statement is False.
In the given quadrilateral $ABCD$,let us check the sum of opposite angles:
$\angle A + \angle C = 90^{\circ} + 95^{\circ} = 185^{\circ}$.
Since the sum of opposite angles is not $180^{\circ}$,the quadrilateral $ABCD$ cannot be a cyclic quadrilateral.
Therefore,the given statement is False.
Explore More
Similar Questions
$AB$ and $XY$ are two chords of a circle whose centre is $P$. $A$ line $l$ passing through the centre $P$ bisects chords $AB$ and $XY$ both. Prove that $AB \parallel XY$.
Medium
View Solution$AB$ and $AC$ are two equal chords of a circle. Prove that the bisector of the angle $BAC$ passes through the centre of the circle.
Difficult
View SolutionIf non-parallel sides of a trapezium are equal,prove that it is cyclic.
Difficult
View SolutionBisectors of angles $A, B$ and $C$ of a triangle $ABC$ intersect its circumcircle at $D, E$ and $F$ respectively. Prove that the angles of the triangle $DEF$ are $90^{\circ}-\frac{1}{2} A, 90^{\circ}-\frac{1}{2} B$ and $90^{\circ}-\frac{1}{2} C$.
Medium
View SolutionThe line segment joining any two points on a circle is called a ........ of the circle.
Easy
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo