fish1

All measurements are correct to the nearest cm.
Find the least possible area of the figure.
My solution:
Maximum absolute error = 1/2 = 0.5 cm
If i cut it into a 10x4 rectangle and a triangle, and work with the triangle after the rectangle, the least possible area= (9.5)(3.5)+(9.5-3.5)(11.5-3.5)
=81.25 cm^2

But if i work with the triangle first, the answer is different
= (9.5-3.5)(11.5-4.5)+(9.5)(4.5)
=84.75 cm^2

Which one is correct? And where have i mistaken?

[ 本帖最後由 fish1 於 2016-3-6 03:06 AM 編輯 ]

peterkcc2015

依同學的切割方法
﹒ ﹒ ﹒ Total area  = Area of lower rect. + Area of upper Δ
所以最小面積時應
﹒ ﹒ ﹒ Least area = Possible least area of rect. + Possible least area of Δ
再個別考慮矩形和三角形面積何時最小。


先計最小矩形面積，這個容易
﹒ ﹒ ﹒ Possible least area of rect. = (Poss. least L)(Poss. least H)
﹒ ﹒ ﹒ = (9.5)(3.5) cm²

接著到最小三角形面積，這個要考慮清楚
﹒ ﹒ ﹒ Possible least area of Δ = (Poss. least B)(Poss. least H)/2
B受3cm和10cm邊限際，唯計矩形時已定了10cm邊的實際長度，所以
﹒ ﹒ ﹒ Poss. least B = (9.5 - 3.5) cm
H受4cm和12cm邊限際，唯計矩形時已定了4cm邊的實際長度，所以
﹒ ﹒ ﹒ Poss. least H = (11.5 - 3.5) cm
推導出
﹒ ﹒ ﹒ Possible least area of Δ = (9.5 - 3.5)(11.5 - 3.5)/2 cm²


最後總結得
﹒ ﹒ ﹒ Least area = Possible least area of rect. + Possible least area of Δ
﹒ ﹒ ﹒ = (9.5)(3.5) + (9.5 - 3.5)(11.5 - 3.5)/2
﹒ ﹒ ﹒ = 57.25 cm² §


同學
第一次計三角形面積漏了除以2，所以還差點才正確。
第二次計三角形面積漏了除以2，且三角形面積雖然取了小者但相應地矩形面積就被迫取大者，結果依然有誤。
注意係呢度你可以見到加法分割後由大形狀計起和小形狀計起的分別。

[ 本帖最後由 peterkcc2015 於 2016-3-6 06:45 PM 編輯 ]

fish1

Thank you for your detailed explanation!!!