Question

in the given figure, ABCD is a cyclic quadrilateral. Chord AB is congruent to Chord CB, Chord AD is congruent to chord DC. angle ADC = 3x, angle ABC =2x then find the measure of all angle​

Answer 1

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Let AB be the chord of the given circle with centre O and a radius of 10 cm.

Then AB =16 cm and OB = 10 cm

From O, draw OM perpendicular to AB.

We know that the perpendicular from the centre of a circle to a chord bisects the chord.

∴ BM = (162) cm=8 cm162 cm=8 cm

In the right  ΔOMB, we have:

OB2 = OM2 + MB2   (Pythagoras theorem)

⇒ 102 = OM2 + 82

⇒ 100 = OM2 + 64

⇒ OM2 = (100 - 64) = 36

⇒ OM=36−−√ cm=6 cmOM=36 cm=6 cm

Hence, the distance of the chord from the centre is 6 cm.

Answer 2

Answer:

Join A and C

In ∆ABC

AB=CB=X (Because both are equal let's consider it to be X)

Angle A +Angle B +Angle C=180degree

(Sum of all the interior angles of a triangle is 180degree)

X+2X+X=180degree

4X=180degree

X=45degree

Then Angle ABC=2X=90degree.

In ∆ACD

AD=CD=X (Because both are equal let's consider it to be X)

Angle A+Angle C+Angle D=180degree

(Sum of all the interior angles of a triangle is 180degree)

X+3X+X=180degree

5X=180degree

X=36egree

Then Angle ADC=3X=108degree.