ABCD is a rhombus. AC and BC intersect at E. GIven that AD = ( x-2) cm , AC = (2x -12) cm and BD = 16cm. Find the perimeter of rhombus ABCD.
Explain in detail please. i know the answer already.
F.3 Mathz
2011-06-04 8:11 pm
回答 (1)
2011-06-04 9:04 pm
✔ 最佳答案
By property of rhombus,AE = EC = AC/2 = (2x - 12)/2 = (x - 6)cm
BE = ED = BD/2 = 16/2 = 8 cm
∠AED = 90 deg
Consider right-angled ΔAED,
By Pyth. theorem,
AD^2 = AE^2 + BE^2
(x - 2)^2 = (x - 6)^2 + 8^2
x^2 - 4x + 4 = x^2 - 12x + 36 + 64
8x = 96
x = 12
Since AB = BC = CD = AD (property of rhombus)
Perimeter of rhombus ABCD
= 4(x - 2)
= 4(12 - 2)
= 40
2011-06-04 13:05:23 補充：
The question should state that AC and BD intersect at E. But not BC.
參考: Knowledge is power.
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