Pre RMO 2018(IOQM), Question 2 incircle quadrilateral

Question 2, Pre RMO 2018-19(IOQM):


In a quadrilateral ABCD, it is given that AB = AD = 13, BC= CD = 20, BD = 24. If r is the radius of the circle inscribable in the quadrilateral, then what is the integer closest to r?.

Prerequisites:
1. Incenter, incircle and angle bisectors in triangle/polygons
2. Kite diagonals

Solution:


ABCD is a kite.

So, AC⊥BDAC \perp BD.

So, BO=OD=12 cmBO = OD = 12 \text{ cm}.

AO=132−122=5 cmAO = \sqrt{13^2 - 12^2} = 5 \text{ cm} OC=202−122=16 cmOC = \sqrt{20^2 - 12^2} = 16 \text{ cm} ar(□ABCD)=12×AC×BD\text{ar}(\square ABCD) = \tfrac{1}{2} \times AC \times BD =12×21×24= \tfrac{1}{2} \times 21 \times 24 =252 cm2= 252 \text{ cm}^2 r=ΔS=25233=7.63 cmr = \frac{\Delta}{S} = \frac{252}{33} = 7.63 \text{ cm}

Closest integer =8= 8.

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