Just for interest sake, I would like to give another method to solve example 4. My method is based on construction of trapeziaum AECF by adding AE, and trapezium AECG by adding AE and CG independently and use k value (total area divided by total portions) in each construction to calculate actual areas of triangle ECH and triangle ECJ and hence their difference as the answer required. Each construction divides the whole parallelogram into portions, using the area ratio for trapezium (T^2, TB, TB, B^2 where T is top ratio, B is bottom ratio) and area ratio for adjacent triangles (base1:base2) for remaining triangles. The first construction gives 60 portions in total hence the k value for area ratio is 140/60 = 7/3 and area of triangle ECH is 9 portions x 7/3 = 21 sq cm. The second construction gives 84 portions in total hence the k value for area ratio is 140/84 = 5/3 and area of triangle ECJ is 9 portions x 5/3 = 15 sq cm . The answer required is triangle ECH - triangle ECJ = 21 - 15 = 6 sq cm. The advantage of this method is that it is easy to get actual areas of all other divided parts for calculation to get any required answer e.g. area FGJH = 16 portions x 5/3 - 4 portions x 7/3 = 17(1/3) sq cm. The gist of ratio is actual value = k value x ratio value for all parts of the whole.
The simplest solution for example 4: 1. Area of ∆EFG = 140(2/12) = 140/6 (∆://gram area ratio = 2:(3+3+2+2+2)=2:12) 2. Area of ∆FEJ = (140/6)(3/7) = 10 (web at F side ratio 3:4) 3. Area of ∆EHJ = (10)((3/5) = 6 (web at J side ratio 3:2) Triangle in parallelogram: Area of ∆:area of //gram = base side of ∆: total base sides of 2 opposite sides of //gram (by principle of area ratio for equal height triangles)
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06:54 忘了減番BEF的面積(75), 答案應該是A (237 cm^2) 才對🫣🫣 抱歉🫡
演算法帶我黎，動畫整得好有心機！見到認真做教學的同路人，覺得好開心！😀 加油！
多謝支持！大家一齊加油！💪🏾
Just for interest sake, I would like to give another method to solve example 4. My method is based on construction of trapeziaum AECF by adding AE, and trapezium AECG by adding AE and CG independently and use k value (total area divided by total portions) in each construction to calculate actual areas of triangle ECH and triangle ECJ and hence their difference as the answer required. Each construction divides the whole parallelogram into portions, using the area ratio for trapezium (T^2, TB, TB, B^2 where T is top ratio, B is bottom ratio) and area ratio for adjacent triangles (base1:base2) for remaining triangles. The first construction gives 60 portions in total hence the k value for area ratio is 140/60 = 7/3 and area of triangle ECH is 9 portions x 7/3 = 21 sq cm. The second construction gives 84 portions in total hence the k value for area ratio is 140/84 = 5/3 and area of triangle ECJ is 9 portions x 5/3 = 15 sq cm . The answer required is triangle ECH - triangle ECJ = 21 - 15 = 6 sq cm.
The advantage of this method is that it is easy to get actual areas of all other divided parts for calculation to get any required answer e.g. area FGJH = 16 portions x 5/3 - 4 portions x 7/3 = 17(1/3) sq cm. The gist of ratio is actual value = k value x ratio value for all parts of the whole.
The simplest solution for example 4:
1. Area of ∆EFG = 140(2/12) = 140/6 (∆://gram area ratio = 2:(3+3+2+2+2)=2:12)
2. Area of ∆FEJ = (140/6)(3/7) = 10 (web at F side ratio 3:4)
3. Area of ∆EHJ = (10)((3/5) = 6 (web at J side ratio 3:2)
Triangle in parallelogram:
Area of ∆:area of //gram = base side of ∆: total base sides of 2 opposite sides of //gram
(by principle of area ratio for equal height triangles)
好有幫助！不過Example 2嗰個唔係搵CDFE咩👀
So A 😂😂😂
呢類題目其實係大陸小升初中，中考出到爛。
Tactics 都係2類，相似三角形面積比=邊長比二次方。同高三角形面積比等如底長比例
仲有一類係common base 倒轉高比例=面積比例，不過這類是納入競賽題比較少出，而且出的話多數是問/關於不規則三角形
不知將來會否出到相似立方體的問題。
🔥🔥🔥
Example 3 唔係好明點解會加線係AE