Question

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?

Answer 1

area of ABCD = area of ∆ABD + area of ∆BCD


Perimeter of ∆ABD = (13 + 24 + 13) = 50
semi-perimeter of ∆ABD, s = 25
from Heron's formula,
area of ∆ABD = $\sqrt{s(s-a)(s-b)(s-c)}$
= $\sqrt{25(25-13)(25-24)(25-13)}$
= $\sqrt{25\times12\times1\times12}$
=5 × 12 = 60

similarly,
perimeter of ∆BCD = 20 + 20 + 24 = 64
semiperimeter of ∆BCD = 32
from Heron's formula,
area of ∆BCD = $\sqrt{32(32-20)(32-20)(32-24)}$
= $\sqrt{32\times12\times12\times8}$
= 12 × 2 × 8 = 192

so, area of ABCD = 60 + 192 = 252

now, radius, r = area of ABCD/semiperimeter of ABCD
= 252/{(13 + 13 + 20 + 20)/2}
= 252/33 = 7.64 ≈ 8

hence, radius is 8

Answer 2

ABCD is a kite
so, AC perpendicular to BE
BO=OD=12cm
AO=5cm
OC=16cm
ar (ABCD)=1/2×AC×BD
=252cm
r=252/33=7.63cm
=8