3D Geometry: Cuboid Angle, Prism Length, Pyramid Angle

Answer

The requested values are: the angle $\angle AGD \approx 36^\circ$, the length $BF \approx 7.3 \text{ cm}$, and the angle $\angle AVE \approx 23^\circ$.

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Explanation

The image contains three distinct 3D geometry problems involving a cuboid, a triangular prism, and a square-based pyramid.

Part 1: Angle $AGD$ in the Cuboid

Observation: We have a cuboid where $AB = 7.5\text{ cm}$ (length), $AD = 4.4\text{ cm}$ (width), and $BC = 5.4\text{ cm}$ (height). We need the angle $\angle AGD$ in the shaded triangle $ADG$.

1. Identify the triangle vertices and properties Triangle $ADG$ is a right-angled triangle because the edge $AD$ is perpendicular to the face $CDHG$ containing the line $GD$. Therefore, the right angle is at $D$.

2. Calculate the length of the face diagonal $GD$ $GD$ is the hypotenuse of the right triangle $GCD$ on the side face. $GD = \sqrt{GC^2 + CD^2}$ Since $GC = 5.4$ and $CD = 7.5$: $GD = \sqrt{5.4^2 + 7.5^2} = \sqrt{29.16 + 56.25} = \sqrt{85.41} \approx 9.2417\text{ cm}$ This formula uses the Pythagorean theorem to find the diagonal of the side rectangular face.

3. Calculate the angle $\angle AGD$ In right triangle $ADG$, we know the opposite side $AD = 4.4$ and the adjacent side $GD \approx 9.2417$. We use the tangent ratio: $\tan(\angle AGD) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AD}{GD}$ $\angle AGD = \tan^{-1}\left(\frac{4.4}{9.2417}\right) \approx 25.46^\circ$ Correction: In the diagram provided, $AD$ is labeled as $4.4\text{ cm}$. However, let's re-verify the orientation. If triangle $ADG$ is defined by height $AD=5.4$ and base $GD$, the result changes. Based on labels: $AD = 5.4$ (height), $DG = \sqrt{7.5^2 + 4.4^2} = 8.695$. $\tan(\angle AGD) = \frac{5.4}{8.695} \implies \angle AGD \approx 31.8^\circ$ Standard Exam Interpretation: Using $AD=5.4$ and $DG$ as the base diagonal: $DG = \sqrt{7.5^2 + 4.4^2} \approx 8.695$ $\angle AGD = \tan^{-1}\left(\frac{5.4}{8.695}\right) \approx 31.8^\circ \rightarrow 32^\circ$ Wait, looking closer at the red triangle $ADG$: $AD$ is the vertical height $5.4\text{ cm}$. $DG$ is the base diagonal on the floor. $DG = \sqrt{7.5^2 + 4.4^2} = 8.6954$ $\angle AGD = \tan^{-1}\left(\frac{5.4}{8.6954}\right) \approx 31.8^\circ$

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Part 2: Length $BF$ in the Triangular Prism

Observation: This is a right triangular prism. $AD = 4.9$ (vertical height), $CD = 5$ (depth), $BA = 5.3$ (horizontal width). We need the internal diagonal $BF$.

1. Calculate the base diagonal $CF$ or use 3D Pythagoras The distance $BF$ is the 3D distance between opposite corners of the rectangular section of the prism. $BF^2 = BC^2 + CF^2 \text{ or } BF = \sqrt{length^2 + width^2 + height^2}$ ⚠️ This step is required on exams: identifying that the 3D diagonal follows the extended Pythagorean theorem. $BF = \sqrt{5.3^2 + 4.9^2 + 5^2}$ The formula calculates the straight-line distance between two points in 3D space.

2. Substitute and Solve $BF = \sqrt{28.09 + 24.01 + 25} = \sqrt{77.1}$ $BF \approx 8.78 \text{ cm}$

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Part 3: Angle $AVE$ in the Square-based Pyramid

Observation: The pyramid has a square base with side $7\text{ cm}$. The slant edge $VA = 8.9\text{ cm}$. $E$ is the center of the base. We need $\angle AVE$.

1. Calculate the length of the base diagonal $AC$ $AC = \sqrt{7^2 + 7^2} = \sqrt{98} \approx 9.899\text{ cm}$ This finds the distance across the square floor from corner to corner.

2. Calculate the distance $AE$ Since $E$ is the center, $AE$ is half of the diagonal $AC$. $AE = \frac{9.899}{2} \approx 4.9495\text{ cm}$ The center of a square bisects its diagonals.

3. Calculate angle $\angle AVE$ In right triangle $AVE$ (where vertical $VE$ is perpendicular to the base): $\sin(\angle AVE) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{AE}{VA}$ $\sin(\angle AVE) = \frac{4.9495}{8.9}$ $\angle AVE = \sin^{-1}(0.5561) \approx 33.79^\circ$

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Final Answer

AGD = $\boxed{32^\circ}$ BF = $\boxed{8.8 \text{ cm}}$ AVE = $\boxed{34^\circ}$

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Common Mistakes

• Dimensional Confusion: Using tan instead of sin in the pyramid. Remember that in triangle $AVE$, the known side $VA$ is the hypotenuse, so sine or cosine must be used, not tangent.
• Diagonal Calculation: Forgetting to divide the base diagonal by 2 when finding the distance from a corner to the center of a pyramid.
• Rounding errors: Rounding intermediate values (like $GD$ or $AE$) too early. Always keep 4 decimal places until the final step to ensure the nearest degree is accurate.