AMC 8 1995 Problem 6

AMC8 1995 年第 6 题

四边形与多边形★★☆☆☆

Figures

I

,

II

, and

III

are squares. The perimeter of

I

is 12 and the perimeter of

II

is 24 . The perimeter of

III

is

图形

I

、

II

和

III

是正方形。

I

的周长是 12，

II

的周长是 24。

III

的周长是

AMC 8 1995 Problem 6 figure

A.9
B.18
C.36
D.72
E.81

正确答案 · Correct Answer

C

解析 · Solution

Solution 1

Since the perimeter of

I

12, each side is

\displaystyle \frac{12}{4} = 3

.

Since the perimeter of

II

is 24, each side is

\displaystyle \frac{24}{4} = 6

.

The side of

III

is equal to the sum of the sides of

I

and

II

. Therefore, the side of

III

is 3 + 6 = 9.

Since

III

is also a square, it has an perimeter of

9\cdot 4 = 36

, and the answer is

C

.

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