Question

3. In a quadrilateral ABCD, AB = AD = 10, BD = 12, CB = CD = 13. Then
A) ABCD is a cyclic quadrilateral
B) ABCD has an in-circle
C) ABCD has both circum-circle and in-circle
D) It has neither a circum-circle nor an in-circle​

Answer 1

ABCD has an in-circle.

Step-by-step explanation:

To find,

AM = $\sqrt{10^{2} - 6^{2} }$ = $\sqrt{64}$ = 8

CM = $\sqrt{13^{2} - 6^{2} }$ = $\sqrt{133}$

In case of incircle,

AB + DC = AD + BC

10 + 13  = 10 + 13

23 = 23

∵ Incircle is possible.

It is not a cyclic quadrilateral as it does not follow the theorem of AC * BD = AB.CD + BC.AD

(8 + $\sqrt{133}$) * 12 ≠ 10 * 13 + 10 * 13

Thus, option B is the correct answer.

Learn more: cyclic quadrilateral

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