Заполним таблицу, используя данные рисунка и свойство средней линии треугольника.
А.
1) Дано: AB = 8, AC = 12, значит, BM = AB/2 = 8/2 = 4. MN = 1/2 * AC = 1/2 * 12 = 6.
По теореме о периметре треугольника PAMNB = BM + BN + MN = 4 + 2 + 6 = 12.
BC = 2MN = 2*6 = 12.
Р ABC = AB + BC + AC = 8 + 16 + 12 = 36
2) Дано: BM = 10, BN = 4, MN = 8.
AB = 2*BM = 2*10 = 20. AC = 2*MN = 2*8 = 16. BC = 2*BN = 2*4 = 8.
PAMNB = BM + BN + MN = 10 + 4 + 8 = 22.
РАВС = AB + BC + AC = 20 + 8 + 16 = 44.
3) Дано: BC = 28, AC = 16, BM = 11.
AB = 2*BM = 2*11 = 22. MN = 1/2 * AC = 1/2 * 16 = 8, MN = 8.
BN = 1/2 * BC = 1/2 * 28 = 14. РAMNB = BM + BN + MN = 11 + 14 + 8 = 33.
РАВС = AB + BC + AC = 22 + 28 + 16 = 66.
4) Дано: AB = 14, AC = 6, BN = 5.
BC = 2 * BN = 2 * 5 = 10. MN = 1/2 * AC = 1/2 * 6 = 3. BM = 1/2 * AB = 1/2 * 14 = 7.
РАМNB = BM + BN + MN = 7 + 5 + 3 = 15.
РАВС = AB + BC + AC = 14 + 10 + 6 = 30.
5) Дано: AB = 12, AC = 9, MN = 7.
BC = 2 * BN = 2 * 7 = 14.
BM = 1/2 * AB = 1/2 * 12 = 6. BN = 1/2 * BC = 1/2 * 14 = 7.
РАМNB = BM + BN + MN = 6 + 7 + 7 = 20.
РАВС = AB + BC + AC = 12 + 14 + 9 = 35.
| AB | BC | AC | BM | BN | MN | PAMNB | PABC | |
|---|---|---|---|---|---|---|---|---|
| 1) | 8 | 16 | 12 | 2 | 10 | 4 | 8 | 36 |
| 2) | 20 | 8 | 16 | 10 | 4 | 8 | 22 | 44 |
| 3) | 22 | 28 | 16 | 11 | 14 | 8 | 33 | 66 |
| 4) | 14 | 10 | 6 | 7 | 5 | 3 | 15 | 30 |
| 5) | 12 | 14 | 9 | 6 | 7 | 7 | 20 | 35 |
Ответ: смотри таблицу выше.