Answer:
The distance from the peak of the kite to the intersection of the diagonals is [tex]\sqrt{91}[/tex] inches.
The distance from the intersection of the diagonals to the top of the tail is 4 inches.
Length of the longer diagonal = [tex]\sqrt{91} +4[/tex] inches.
Step-by-step explanation:
Please refer to the attached figure.
It is given that the length of the shorter diagonal = 6 inches.
BD = 6
BE = [tex]\frac{1}{2} BD[/tex]
= 3 inches
In right triangle AEB,
[tex]AE^{2} =AB^{2} -BE^{2}[/tex]
[tex]=10^{2} -3^{2}[/tex]
= 100 - 9
= 91
[tex]AE=\sqrt{91}[/tex]
Hence, the distance from the peak of the kite to the intersection of the diagonals is [tex]\sqrt{91}[/tex] inches.
From the right triangle BEC,
[tex]EC^{2} =BC^{2} -BE^{2}[/tex]
= [tex]5^{2} -3^{2}[/tex]
= 25 - 9
= 16
EC = 4 inches
Hence, the distance from the intersection of the diagonals to the top of the tail is 4 inches.
Length of the longer diagonal = AC = AE + EC = [tex]\sqrt{91} +4[/tex] inches.