Question

Катети прямокутного трикутника АВС дорівнюють 9 і 12 см. Через середину гіпотенузи — точку О проведено перпендикуляр до площини трикутника довжиною 6 см. Знайдіть відстані від кінців перпендикуляра до катетів.

Answer 1

Given Information:

- The lengths of the legs (or catheti) of a right triangle ABC are 9 cm and 12 cm. - A perpendicular is drawn from the midpoint of the hypotenuse to the plane of the triangle, with a length of 6 cm.

Solution:

To find the distances from the ends of the perpendicular to the legs of the triangle, we can use the concept of similar triangles.

Let's denote the midpoint of the hypotenuse as point O. We need to find the distances from O to the legs AB and AC.

Using the concept of similar triangles, we can set up the following ratios:

Ratio 1: The ratio of the length of the perpendicular drawn from O to the leg AB to the length of the leg AB is equal to the ratio of the length of the perpendicular drawn from O to the hypotenuse to the length of the hypotenuse.

Let's denote the distance from O to AB as x. The ratio can be written as:

x / 9 = 6 / 15

Simplifying the equation, we get:

x = (6 / 15) * 9

Ratio 2: The ratio of the length of the perpendicular drawn from O to the leg AC to the length of the leg AC is equal to the ratio of the length of the perpendicular drawn from O to the hypotenuse to the length of the hypotenuse.

Let's denote the distance from O to AC as y. The ratio can be written as:

y / 12 = 6 / 15

Simplifying the equation, we get:

y = (6 / 15) * 12

Calculation:

Let's calculate the values of x and y:

x = (6 / 15) * 9 = 3.6 cm y = (6 / 15) * 12 = 4.8 cm

Answer:

The distance from the end of the perpendicular to the leg AB is 3.6 cm, and the distance from the end of the perpendicular to the leg AC is 4.8 cm.

Please let me know if I can help you with anything else.