Предмет: Геометрия, автор: aarr04594

Знайти периметр рожевого прямокутника і периметр початкового прямокутника. Рисунок додано.

Приложения:

zmeura1204: Р=13; Р(початков)=21

aarr04594: Ні, не 13, і не 21.

zmeura1204: Да?

zmeura1204: А скільки?

aarr04594: 7 і 15. Мені цікаво було б побачити розв'язок. І максимально простий, на рівні 5 класу. У мене є свій, але ще пропозиції.

zmeura1204: Так розв'язували?

zmeura1204: Виправила.

Ответы

Автор ответа: zmeura1204

Ответ:

Периметр рожевого прямокутника дорівнює 7од; периметр початкового прямокутника дорівнює 15од

Объяснение:

Відмічу сторони через перемінні а;b;c;d.

Pз.=2b+2c - периметр зеленого прямокутника.

Рс=2b+2d - периметр синього прямокутника

Рб=2а+2d - периметр блакитного прямокутника.

Pр=2а+2с - периметр рожевого прямокутника (необхідно знайти (2а+2с))

Складаємо систему рівнянь

{2b+2c=10

{2b+2d=8

{2a+2d=5

Віднімаємо з третього рівняння друге

{2а+2d=5

{2b+2d=8

_________

2a-2b=-3

Тепер складаємо те що получили, з першим рівнянням.

{2a-2b=-3

{2b+2c=10

_______

2a+2c=7 периметр рожевого прямокутника (Рр=7од)

А периметр початкового периметру

буде

Рпоч=2(а+b)+2(c+d)

Pпоч=2а+2b+2c+2d

Pпоч=(2а+2с)+(2b+2d)

Pпоч=Рр+Рс=7+8=15

________________

Помилилась при розрахунках.

Приложения:

aarr04594: Початкового 15 см.

zmeura1204: Виаравила.

aarr04594: Дякую.

zmeura1204: Будь ласка.

aarr04594: Додам розв'язання. Очевидно, що півпериметр рожевого =10/2+5/2-8/2, тобто рожевий Р=10+5-8=7. Тоді периметр початкового Р=(10+8+5+7):2=15.

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