Question

Из точки к плоскости проведены две наклонные, длинны проекций которых относятся как 1:2. найдите квадрат длинны проекции меньшей наклонной, если длины наклонных равны 8 и 10.. объясните пожалуйста как это решать

Answer 1

Обозначим квадрат длины меньшей проекции как х, а квадрат расстояния от точки до плоскости как у.
По т. Пифагора в двух тр-ках, образованных наклонными, их проекциями и высотой точки над плоскостью, квадрат высоты равен: у=8²-х в первом тр-ке,
у=10²-2²х во втором тр-ке. (длина второй проекции в два раза больше первой. По условию их отношение 1:2, а квадрат проекции 2²х).
Объединим уравнения:
64-х=100-4х,
3х=36,
х=12 - это ответ.

Answer 2

Problem Analysis

We are given two inclined lines that intersect a plane. The lengths of the projections of these lines onto the plane are in a 1:2 ratio. We need to find the square of the length of the projection of the shorter inclined line, given that the lengths of the inclined lines are 8 and 10.

Solution

Let's assume the lengths of the projections of the inclined lines onto the plane are x and 2x, where x is the length of the projection of the shorter inclined line.

According to the problem, the lengths of the inclined lines are 8 and 10. Therefore, we can set up the following equation:

x + 2x = 8 + 10

Simplifying the equation:

3x = 18

Dividing both sides by 3:

x = 6

So, the length of the projection of the shorter inclined line is 6.

To find the square of this length, we can simply calculate 6 squared:

The square of the length of the projection of the shorter inclined line is 36.

Explanation

To solve this problem, we used the given information that the lengths of the projections of the inclined lines onto the plane are in a 1:2 ratio. By setting up an equation and solving for the unknown length, we found that the length of the projection of the shorter inclined line is 6. Finally, we squared this length to find the square of the length, which is 36.

Please let me know if anything is unclear or if you have any further questions!