Length of kite’s other diagonal is 21 inches
Charlene puts together two isosceles triangles so that they share a base to form a kite
Let Δ ABC and ΔDBC are two isosceles triangles.
Leg of Δ ABC = 10 inches
Leg of ΔDBC = 17 inches
Both have common base BC
Length of BC = 16
From the properties of isosceles triangle we can write that
The altitude of isosceles triangle bisects the base of the isosceles triangle.
From the figure attached we can write
Hence BP = PC = 8 inches
Length of diagonal AD is to be determined by equation (1)
AP +PD = AD .........(1)
From Pythagoras theorem we can write
[tex]\rm H^2 = P^2 +B^2 \\where \\H =Hypotenuse \; of \; the \; right\; triangle \\P = Length\; of\; Perpendicular\; \\B = Length \; of \; Base[/tex]
Applying Pythagoras them we can find Length of AP and PD
[tex]\rm AP =\sqrt{ 10^2 -8^2}\\AP= \sqrt{100-64}\\AP = \sqrt{36} =6[/tex]
Similarly
[tex]\rm PD =\sqrt{17^2- 8^2}\\PD = \sqrt{289-64}\\PD =\sqrt{225} = 15[/tex]
So from equation (1) length of diagonal AD= AP +PD = 15 +6 = 21 inches
hence length of kite’s other diagonal is 21 inches
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