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Task
The base of the pyramid is a right triangle, one of the legs of which is 8 cm, and the radius of the circle circumscribed about it is 5 cm.
The base of the height of this pyramid is the midpoint of the hypotenuse. The height of the pyramid is 12 cm.
Calculate the lateral edges of the pyramid.
Solution.
The base of the pyramid is a right triangle. The center of the circle circumscribed about the right triangle lies on its hypotenuse.
Accordingly, AB = 10 cm, AO = 5 cm.
Since the height ON = 12 cm, the size of the edges AN and NB is
AN2 = AO2 + ON2
AN2 = 52 + 122
AN = √169
AN = 13
Since in a right triangle the hypotenuse is simultaneously the diameter of the circumscribed circle,
and OC will also be the radius of the circumscribed circle,
then
Since we know the value AO = OB = 5 cm and the value of one of the legs of the base (8 cm),
the height dropped onto the hypotenuse will equal
AB = 2R = 10
R = 5
CO = 5 cm
Accordingly, the value of the edge CN will be equal to
CN2 = CO2 + NO2
CN2 = 25 + 144
CN = √169
Answer: 13, 13 , 13
Task
The base of the pyramid is a right triangle, the legs of which are 8 and 6 cm. The height of the pyramid is 10 cm.
Calculate the volume of the pyramid.
Solution.
The volume of the pyramid is found by the formula:
V = 1/3 Sh
The area of the base is found by the formula for finding the area of a right triangle:
S = ab/2 = 8 * 6 / 2 = 24
whence
V = 1/3 * 24 *10 = 80 cm3 .
Regular tetrahedron |
Описание курса
| Prism. Parallelepiped. Cube. Solving problems
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