2018 Q16 AF : FE = 8:5 Area of ABF : BFE = 8:5 (same height triangle, area = base ratio) Area of BFE = 120 (5/8) = 75 Area of AFD = 75 * (8/5)^2 = 192 (area of similar triangle ratio = side ratio ^2) Let the height of the parallelogram be y In ABD, 8k (y) (1/2) = 192 + 120 = 312 In ABE 5k (y) (1/2) = 195 CDEF = 8ky - 312 - 195 + 120 = 312*2 - 312 - 195 + 120 = 237
The explanation for the last step of 3rd problem i.e. A1:A2 = 7:5 should not be area ratio of hour-glass similar triangles (A1 & A2 do not form hour-glass.) but area ratio of adjacent triangles on DB where DH:HB = 7:5.
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好用！
我係學生家長，不得不比個👍你。
教得好好！！感謝！
為甚麼沒有了2018 的Q16?
相似三角形面積 = 邊長比 ^2 這個公式很多人都懂
同高三角形 面積比 = 底比例, 同底三角形面積比 = 高比例; 這2個要背的定律是中國中考必考
計2019 Q16 我是用另外解法，先求 AX in terms of EB (2k); AX = 24/k 然後求 三角型面積 DXA = (24/k)(12k)(1/2) = 144 之後 平行四邊型面積 = (144+24)*2 = 336
不規則4邊形面積 = 336 - 168 - (9k/12k)^2 x 144 = 336 -168 - 81 = 87
感謝老師
2018 Q16
AF : FE = 8:5
Area of ABF : BFE = 8:5 (same height triangle, area = base ratio)
Area of BFE = 120 (5/8) = 75
Area of AFD = 75 * (8/5)^2 = 192 (area of similar triangle ratio = side ratio ^2)
Let the height of the parallelogram be y
In ABD, 8k (y) (1/2) = 192 + 120 = 312
In ABE 5k (y) (1/2) = 195
CDEF = 8ky - 312 - 195 + 120 = 312*2 - 312 - 195 + 120 = 237
The explanation for the last step of 3rd problem i.e. A1:A2 = 7:5 should not be area ratio of hour-glass similar triangles (A1 & A2 do not form hour-glass.) but area ratio of adjacent triangles on DB where DH:HB = 7:5.
Very good
正呀喂
o(^▽^)o 感謝
？那個是上一條片，老師你拍的和RUclips播的，是沒共通次序啊
ruclips.net/video/SwCXpMc3U9c/видео.html