Question

in a quadrilateral ABCD,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 = $\sqrt{32(32-20)(32-20)(32-24)}$
= $\sqrt{32\times12\times12\times8}$
= 16 × 12 = 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

Answer:

8

Step-by-step explanation:

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 = 5 × 12 = 60

similarly, 

perimeter of ∆BCD = 20 + 20 + 24 = 64

semiperimeter of ∆BCD = 32 

from Heron's formula,

area of ∆BCD = 16 × 12 = 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

therefore, radius is 8

hence proved