F.2 maths畢氏定理問題

BD^2=AD^2 + AB^2 (pyth.thm.)
1.2^2=2AB^2 (∵AB=AD = side of square)
1.44=2 AB^2
AB^2=0.72
AB= 0.849
i.e required perimeter = 0.849x 4= 3.39
= 3 m (correct to the nearest integer)

2011-12-06 01:27:41 補充：
樓上 ka fung 的單位錯 是m 不是cm
錯單位會被扣單位分
:)

Let y cm the length of each side of the square.

ΔABD is a right-angled triangle.
By Pythagorean theoem:
AB² + AD² = BD²
y² + y² = 1.2²
2y² = 1.44
y² = 0.72
y = √0.72

Perimeter of the square
= 4√0.72 cm
= 3 cm (to the nearest integer)

It is a Square AD=AB=BC=CD 2AD2 = 1.22 (Pythagoras Theorem) 2AD2 = 1.44 AD2 = 0.72 AD = √0.72 The perimeter, = √0.72 X 4= 3 cm ( correct to the nearest integer)