4. imagine an upright cone with its base resting on your horizontal desk. match each plane with the image of the cross section formed by intersecting the plane with the cone.
a. horizontal
b.vertical, through cones topmost point
c. diagonal
5. what is the shape of the cross section formed by intersecting a cube with a vertical plane that passes through opposite edges of the cube? explain how you know.
6. sketch the solid of rotation formed by rotating the given two-dimensional figure using the dashed vertical line as an axis of rotation.
7. calculator required. a rope with a length of 4 meters is tied from a stake in the ground to the top of a tent. it forms a 20 degree angle with the ground. how tall is the tent?
8. calculator required. what is the value of y?

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Accepted Answer

### Problem 4 # Explanation: ## Step1: Match horizontal plane A horizontal plane cutting the cone parallel to its base creates a circular cross-section. So Figure 2 ↔ A. ## Step2: Match vertical apex plane A vertical plane passing through the cone's topmost point (apex) and base diameter creates a triangular cross-section. So Figure 1 ↔ B. ## Step3: Match diagonal plane A diagonal plane (neither parallel to base nor passing through apex) creates an elliptical cross-section. So Figure 3 ↔ C. # Answer: Figure 1 ↔ B. vertical, through cone's topmost point Figure 2 ↔ A. horizontal Figure 3 ↔ C. diagonal --- ### Problem 5 # Brief Explanations: When a vertical plane passes through opposite edges of a cube, the plane intersects 4 faces of the cube. The resulting cross-section has two pairs of parallel sides, and the sides are of equal length in each pair, with angles that are not right angles, forming a rectangle (specifically a rhombus, a type of parallelogram with all sides equal). # Answer: The cross-section is a rectangle (rhombus/parallelogram). The plane cuts 4 cube faces, creating 4 line segments that form a shape with two pairs of parallel, equal-length sides. --- ### Problem 6 # Brief Explanations: When the given open two-sided figure (a pair of line segments opening left toward the axis) is rotated around the dashed vertical axis, each line segment sweeps out a curved surface. The resulting solid is a **hollow truncated (cut-off) cone (frustum)** with an open inner side, or more simply, two nested curved surfaces forming a shape like an empty cone with the tip removed, opening toward the axis. # Answer: The solid of rotation is a hollow frustum of a cone (a truncated cone shape with an open interior, formed by rotating the two line segments around the vertical axis to create two curved, parallel surfaces). --- ### Problem 7 # Explanation: ## Step1: Define trigonometric relation We use the sine function, where $\sin( heta) = \frac{ ext{opposite}}{ ext{hypotenuse}}$. Here, $ heta=20^\circ$, hypotenuse = 4 m, opposite side = tent height $h$. $\sin(20^\circ) = \frac{h}{4}$ ## Step2: Solve for tent height Rearrange to solve for $h$: multiply both sides by 4. $h = 4 imes \sin(20^\circ)$ ## Step3: Calculate value Use a calculator for $\sin(20^\circ) \approx 0.3420$ $h \approx 4 imes 0.3420 = 1.368$ # Answer: The tent is approximately 1.37 meters tall (rounded to two decimal places). --- ### Problem 8 # Explanation: ## Step1: Find angle at F The sum of angles in a triangle is $180^\circ$. Calculate $\angle F$. $\angle F = 180^\circ - 42^\circ - 48^\circ = 90^\circ$ ## Step2: Apply Law of Sines $\frac{y}{\sin(48^\circ)} = \frac{ ext{side opposite } 42^\circ}{\sin(42^\circ)}$. First, label side $FE = y$, $DF = ext{side opposite }48^\circ$, $DE = ext{side opposite }90^\circ$. Wait, correct relation: $\frac{y}{\sin(48^\circ)} = \frac{DE}{\sin(90^\circ)}$, but since $\angle F=90^\circ$, we can use $\sin(42^\circ) = \frac{y}{DE}$ and $\sin(48^\circ) = \frac{DF}{DE}$. To solve for $y$, we need the length of $DE$ or $DF$. Assuming $DE$ is the base, but since it's unlabeled, we use exact trigonometric form: $y = DE imes \sin(48^\circ)$. If we assume $DE$ is the hypotenuse (length let $x$), $y = x \sin(48^\circ)$. If we use the Law of Sines with $\angle F=90^\circ$, $y = DF imes \frac{\sin(48^\circ)}{\sin(42^\circ)}$. Alternatively, if we calculate the ratio: $\frac{\sin(48^\circ)}{\sin(42^\circ)} \approx \frac{0.7431}{0.6691} \approx 1.1106$, so $y \approx 1.11 imes DF$. Wait, correction: In $ riangle DEF$, $\angle D=42^\circ$, $\angle E=48^\circ$, $\angle F=90^\circ$. So $y$ is side $FE$, opposite $\angle D=42^\circ$. Wait no: side $y$ is $DF$, opposite $\angle E=48^\circ$. So Law of Sines: $\frac{y}{\sin(48^\circ)} = \frac{DE}{\sin(90^\circ)}$, so $y = DE imes \sin(48^\circ)$. If we assume $DE$ is, for example, if we take $DE$ as a known length, but since it's not given, we use the trigonometric value. If we calculate numerically, $\sin(48^\circ) \approx 0.7431$, so if $DE$ is the hypotenuse, $y = DE imes 0.7431$. If we use tangent: $ an(48^\circ) = \frac{y}{ ext{side } FE}$, but $ an(48^\circ) = \frac{y}{FE}$, $ an(42^\circ) = \frac{FE}{y}$, so $y = FE imes an(48^\circ)$. # Answer: If $\angle F = 90^\circ$, using the Law of Sines, $y = DE imes \sin(48^\circ) \approx DE imes 0.743$. If we assume $DE$ is 1 unit, $y \approx 0.74$. If we solve for $y$ in terms of the other side, $y = FE imes an(48^\circ) \approx FE imes 1.111$.

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QUESTION IMAGE

Question

Explanation:

Problem 4

Step1: Match horizontal plane

Step2: Match vertical apex plane

Step3: Match diagonal plane

Answer:

Problem 5