Find the area of a kite ABCD if BD= 48cm, AB= 25cm, and BC= 26cm?

1 Answer
Jan 4, 2016

S=408cm^2S=408cm2

Explanation:

The formula of a kite's area is
S=(1/2)d1*d2S=(12)d1d2
where
d1="kite's long diagonal"d1=kite's long diagonal
d2="kite's short diagonal"d2=kite's short diagonal

In the problem, be noticed that once the diagonal BD has an endpoint in B (where 2 segments of different sizes, AB and BD, meet), this diagonal is divided in two equal parts by the other diagonal. Calling E the point where the two diagonals intercept each other, we have:
BD=2*DE=48BD=2DE=48 => DE=24DE=24

In the right triangle ABE we can obtain the segment AE
AB^2=DE^2+AE^2AB2=DE2+AE2 => 25^2=24^2+AE^2252=242+AE2 => AE=sqrt(625-576)=sqrt(49)=7AE=625576=49=7

In the right triangle BCE we can obtain the segment CE
BC^2=DE^2+CE^2BC2=DE2+CE2 => 26^2=24^2+CE^2262=242+CE2 => CE=sqrt(676-576)=sqrt(100)=10CE=676576=100=10

In this way we discovered the previously unknown diagonal:
AC=AE+CE=7+10=17AC=AE+CE=7+10=17

Finally,
S=(1/2)(d1+d2)=(48+17)/2=408cm^2S=(12)(d1+d2)=48+172=408cm2