ABCDA1B1C1D1-cube, point O lies on ray BC so that BC: CO = 2: 1. Calculate the lateral surface area of the pyramid BAB1O if the total surface area of the cube is 24 cm
BC: CO = 2: 1,
Scuba = 24 cm2,
S side (BAB1O)
When denoting a pyramid, one first calls its top, and then the top of the base. In this way:
S side (BAB1O) = S (BAB1) + S (BB1O) + S (BAO)
Scube = 6а2 (a is the length of the side of the cube), therefore a = = = 2 cm
Since ABCDA1B1C1D1 is a cube, DBAB1, DBB1O, and DBAO are rectangular.
By the condition of the problem BC: CO = 2: 1, therefore BО = 2/2 = 1 cm.
The area of a right-angled triangle is half the product of its legs.
S (BAB1) = BA × BB1 = 2 × 2 = 4 cm2
S (BB1O) = BB1 × BО = 2 × 1 = 2 cm2
S (BAO) = BA × BO = 2 × 1 = 2 cm2
S side (BAB1O) = 4 + 2 + 2 = 8 cm2
Answer: The area of the side surface of the BAB1O pyramid is 8 cm2
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