1. The sides of triangle CAB are in the ratio of 2:3:4. Segment BD is the angle bisector drawn to the shortest side, dividing it into segments AD and DC. What is the length, in inches, of the longer subsegment of side AC if the length of side AC is 10 inches? Express your answer as a common fraction.
2. A right cylindrical oil tank is 15 feet tall and its circular bases have diameters of 4 feet each. When the tank is lying flat on its side (not on one of the circular ends), the oil inside is 3 feet deep. How deep, in feet, would the oil have been if the tank had been standing upright on one of its bases? Express your answer as a decimal to the nearest tenth.
3. Consider the set of points that are inside or within one unit of a rectangular parallelepiped (box) that measures 3 by 4 by 5 units. Given that the volume of this set is {m+n\pi}/p, where m, n, and p are positive integers, and n and p are relatively prime, find m+n+p.
1. use the Angle Bisector Theorem.
2. The volume of oil is the area of a circular cap times the length. So, the the formula for the area of a circular cap. Then if the oil volume is v, the depth would have been (v/60π)*15
3. Consider the problem in 2D, and draw a rectangle. If the rectangle has width x and height y, the area in question is the rectangle plus 4 strips of width 1 on each side, plus 4 quarter-circles of radius 1 at the corners. That is xy + 2x+2y + π
Now extend that to 3D and the volume will be the brick, plus 6 blocks on the faces, of thickness 1, plus a complete sphere of radius 1:
3*4*5 + 2(3*4+4*5+3*5) + 4π/3
= (462+4π)/3
that is incorrect.
1. To find the length of the longer subsegment of side AC, we need to determine the lengths of segments AD and DC.
We are given that the sides of triangle CAB are in the ratio of 2:3:4. Let's assign lengths to the sides of the triangle using a common multiple of 2, 3, and 4. We can choose any specific value, as long as it maintains the same ratio. Let's choose 2 units for the shortest side (AD) and 3 units for the middle side (BD).
Now, we have AD = 2 units and BD = 3 units. Using the angle bisector theorem, we can determine the ratio of the lengths of segments AD and DC. The theorem states that the ratio of the lengths of the segments is equal to the ratio of the lengths of the sides they bisect. In this case, that would be AD:DC = AB:BC.
Since AB is the same as AD (2 units) and BC is the same as BD (3 units), we have AD:DC = 2:3. To find the exact lengths of AD and DC, we can set up a proportion:
2/3 = AD/DC.
Cross-multiplying, we get 3 * AD = 2 * DC.
Since AC = AD + DC (from the triangle), we can substitute for AD: AC = 2 * DC + DC = 3 * DC.
We are told that AC = 10 inches, so 10 = 3 * DC.
Solving for DC, we divide both sides by 3: DC = 10/3.
Therefore, the length of the longer subsegment of side AC (DC) is 10/3 inches or 3.333... (repeating) inches.
2. To find the depth of the oil in the cylindrical tank when it is standing upright on one of its bases, we need to consider the similar triangles formed by the height of the tank and the depth of the oil.
When the tank is lying flat on its side, the depth of the oil is 3 feet. Let's label the length of the oil depth as x when the tank is in its standing position.
We can establish a proportion between the height and depth of the oil for the two different positions of the tank. The proportion would be:
(Height of the tank when upright) / (Depth of the oil when upright) = (Height of the tank when lying down) / (Depth of the oil when lying down).
Using the given measurements, we substitute the values into the proportion:
15 / x = 4 / 3.
To solve for x, we can cross-multiply:
3 * 15 = 4 * x.
Simplifying, we get 45 = 4x.
Dividing both sides by 4, we find x = 11.25 feet.
Therefore, when the tank is standing upright on one of its bases, the oil would be approximately 11.25 feet deep.
3. To find the volume of the set of points that are inside or within one unit of the rectangular parallelepiped, we start by calculating the volume of the parallelepiped. The volume of a rectangular parallelepiped is given by the formula length × width × height.
In this case, the dimensions of the box are 3 by 4 by 5 units. So, the volume of the parallelepiped is 3 × 4 × 5 = 60 cubic units.
Next, we want to find the volume of the set of points that are inside or within one unit of the box. Imagine the box enlarged by one unit on all sides. It forms a new parallelepiped with dimensions 5 by 6 by 7 units.
The volume of this enlarged parallelepiped is 5 × 6 × 7 = 210 cubic units.
The volume of the set of points that are within one unit of the original box is the difference between the volume of the enlarged parallelepiped and the volume of the original parallelepiped: 210 - 60 = 150 cubic units.
Therefore, the volume of the set of points is 150 cubic units, which can be expressed as {150 + 0π}/1. The values m, n, and p in the form {m + nπ}/p are positive integers.
So, m = 150, n = 0, and p = 1.
The sum m + n + p is 150 + 0 + 1 = 151.
Thus, m + n + p = 151.