Advertisements
Advertisements
Question
ABCD is a cyclic quadrilateral such that ∠A = 90°, ∠B = 70°, ∠C = 95° and ∠D = 105°.
Options
True
False
Advertisements
Solution
This statement is False.
Explanation:
In a cyclic quadrilateral, the sum of opposite angles is 180°.
Now, ∠A + ∠C = 90° + 95° = 185° ≠ 180°
And ∠B + ∠D = 70° + 105° = 175° ≠ 180°
Here, we see that, the sum of opposite angles is not equal to 180°.
So, it is not a cyclic quadrilateral.
APPEARS IN
RELATED QUESTIONS
Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see the given figure). Prove that ∠ACP = ∠QCD.

Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.
Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.
Prove that the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.
The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm from the centre, what is the distance of the other chord from the centre?
ABCD is a cyclic quadrilateral in BC || AD, ∠ADC = 110° and ∠BAC = 50°. Find ∠DAC.
ABCD is a cyclic quadrilateral in which BA and CD when produced meet in E and EA = ED. Prove that EB = EC.
In the given figure, ABCD is a quadrilateral inscribed in a circle with centre O. CD is produced to E such that ∠AED = 95° and ∠OBA = 30°. Find ∠OAC.

If non-parallel sides of a trapezium are equal, prove that it is cyclic.
The three angles of a quadrilateral are 100°, 60°, 70°. Find the fourth angle.
