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the perimeter of the kite in which A= 49.92yd2

The answer is on this file but I want an explanation on how to get this answer

the perimeter of the kite in which A= 49.92yd2 The answer is on this file but I want an explanation on how to get this answer class=

Verified Answer :

wolf1728

(AD BC) .5  = 49.92
(AD 9.6) .5  = 49.92
Multiplying by 2
AD * 9.6 = 99.84
AD = 10.40
x = 10.40 - 9.0 = 1.4
Hypotenuse BD^2 = Be^2 + x^2
Hypotenuse BD^2 = 4.8^2 + 1.4^2
Hypotenuse BD^2 = 25
Hypotenuse BD = 5

Hypotenuse AB^2 = 9^2 + 4.8^2
Hypotenuse AB^2 = 81 + 23.04
Hypotenuse AB^2 = 104.04
Hypotenuse AB =



10.2176

So perimeter = 5 + 5 + 

10.2176 +

10.2176

Perimeter =



30.4352

(Hmm, that was a "little" bit of work.  LOL)

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