Math Problems: Surface Area, Volume & Angles

Hey guys! Today we're diving deep into some awesome geometry problems. We're going to tackle a prism, specifically a right prism with a square base. Get ready to flex those math muscles as we calculate areas, perimeters, edge sums, and even some angles. This stuff might seem a bit daunting at first, but trust me, with a little practice and a clear step-by-step approach, you'll be a geometry whiz in no time! So, grab your calculators, sharpen your pencils, and let's get started on problem number one!

Understanding the Prism: ABCDA'B'C'D'

First things first, let's get a solid understanding of the prism we're working with. We've got a right prism denoted as ABCDA'B'C'D'. What does "right prism" mean? It basically means that the lateral edges (like AA', BB', CC', DD') are perpendicular to the bases (ABCD and A'B'C'D'). Think of it like a perfectly straight box. The base of our prism is a square, which is super handy because all sides are equal and all angles are 90 degrees. We're given that the side length of the square base, AB, is 6 cm. And the height of the prism, represented by the length of a lateral edge like AA', is $6\sqrt{3}$ cm. This height is crucial for many of our calculations, so keep it in mind!

Now, let's break down what we need to calculate:

• a) Area and perimeter of the base, and the area and perimeter of a lateral face.
• b) The sum of all the edges of the prism.
• c) The sine of an angle. (We'll need more information to specify which angle, but we'll get there!)

This problem is a fantastic way to solidify your understanding of basic prism properties and calculations. We'll go through each part methodically, making sure we don't miss any steps. Remember, geometry is all about visualizing the shapes and applying the correct formulas. Don't be afraid to sketch out the prism if it helps you! A good visual can make all the difference.

Part a) Base and Lateral Face Calculations

Alright, let's tackle the first part: the area and perimeter of the base, and then the area and perimeter of a lateral face. These are fundamental calculations that will set us up for the rest of the problem. Since our base is a square with side length AB = 6 cm, we can easily find its properties. The perimeter of a square is simply four times the side length. So, for the base ABCD, the perimeter is $4 \times 6 \text{ cm} = 24 \text{ cm}$. Easy peasy!

Now, for the area of the square base, we square the side length. So, the area of base ABCD is $(6 \text{ cm})^2 = 36 \text{ cm}^2$. Remember to always include your units – centimeters for length and square centimeters for area. It's super important for keeping track of our calculations and ensuring they make physical sense.

Next, we need to consider a lateral face. Since the base is a square and it's a right prism, all the lateral faces are rectangles. Let's take the face ABB'A' as an example. The sides of this rectangle are the side of the base (AB) and the height of the prism (AA'). We know AB = 6 cm and the height AA' = $6\sqrt{3}$ cm. So, the perimeter of this lateral face ABB'A' is $2 \times (\text{AB} + \text{AA}') = 2 \times (6 \text{ cm} + 6\sqrt{3} \text{ cm})$. We can factor out the 6 to get $2 \times 6 (1 + \sqrt{3}) \text{ cm} = 12(1 + \sqrt{3}) \text{ cm}$.

And the area of this lateral face ABB'A' is simply the product of its sides: $\text{AB} \times \text{AA'} = 6 \text{ cm} \times 6\sqrt{3} \text{ cm} = 36\sqrt{3} \text{ cm}^2$. Pretty straightforward, right? We've now calculated the key dimensions for both the base and a lateral face. This foundational knowledge is key to unlocking the rest of the problem's challenges. Keep these values handy as we move on to summing up the edges and diving into angles!

Part b) Sum of the Prism's Edges

Moving on to the second part, guys, we need to find the sum of all the edges of the prism ABCDA'B'C'D'. A prism has two types of edges: edges that form the bases and the lateral edges that connect the two bases. Our prism has a square base, so it has 4 edges on the bottom base (AB, BC, CD, DA) and 4 edges on the top base (A'B', B'C', C'D', D'A'). Since it's a right prism with a square base, all these base edges have the same length, which is 6 cm. So, the total length of the base edges is $4 \times 6 \text{ cm} + 4 \times 6 \text{ cm} = 24 \text{ cm} + 24 \text{ cm} = 48 \text{ cm}$.

Now, let's consider the lateral edges. These are the vertical edges connecting the vertices of the bottom base to the corresponding vertices of the top base. These are AA', BB', CC', and DD'. In a right prism, all lateral edges have the same length, which is equal to the height of the prism. We are given that the height AA' is $6\sqrt{3}$ cm. So, there are 4 lateral edges, each measuring $6\sqrt{3}$ cm. The total length of the lateral edges is $4 \times 6\sqrt{3} \text{ cm} = 24\sqrt{3} \text{ cm}$.

To find the sum of all the edges, we just add the total length of the base edges and the total length of the lateral edges. So, the sum of all edges is $48 \text{ cm} + 24\sqrt{3} \text{ cm}$. We can factor out 24 from both terms if we want to write it more compactly: $24(2 + \sqrt{3}) \text{ cm}$. This calculation shows us the total "skeleton" length of our prism. It’s a great exercise to visualize all these lines and add up their lengths. Knowing the sum of edges can be useful in various contexts, like calculating the amount of material needed to build a framework representing the prism, or just as a fundamental property of the shape itself. We're doing great, team! On to the final, slightly trickier part!

Part c) Calculating the Sine of an Angle

Alright, the final piece of the puzzle for this section is calculating the sine of an angle. The problem statement is a little vague here as it doesn't specify which angle. In prism problems, usually, when they ask for the sine of an angle without specifying, they are often referring to the angle between a lateral edge and a diagonal of the base, or an angle between a lateral face and the base, or even an angle between two diagonals of the prism. Given the context of a right prism with a square base, a common angle to calculate is the angle between a lateral edge and a diagonal of the base, or the angle formed by a diagonal of a lateral face and the base. Let's assume they mean the angle between a diagonal of the base and a lateral edge that meets at the same vertex. For example, let's consider the angle between the base diagonal AC and the lateral edge AA'. However, since AA' is perpendicular to the base (because it's a right prism), the angle between AC and AA' is 90 degrees. The sine of 90 degrees is 1. This is a bit too trivial.

Let's consider a more likely scenario: the angle between a diagonal of the prism and one of its edges, or perhaps the angle between a diagonal of the base and a lateral face. A very common angle to calculate in such problems is the angle that a diagonal of the prism makes with the base. Let's consider the diagonal of the prism, say A'C. To find this angle, we often look at the right triangle formed by the diagonal of the base (AC), a lateral edge (AA'), and the prism diagonal (A'C). Let's calculate the length of the base diagonal AC first. Since the base is a square with side 6 cm, the diagonal AC can be found using the Pythagorean theorem: $AC^2 = AB^2 + BC^2 = 6^2 + 6^2 = 36 + 36 = 72$. So, $AC = \sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}$ cm.

Now, consider the right triangle $\triangle$AA'C. The sides are AA' = $6\sqrt{3}$ cm (height) and AC = $6\sqrt{2}$ cm (base diagonal). The hypotenuse is the prism diagonal A'C. We can find A'C using the Pythagorean theorem: $(A'C)^2 = (AA')^2 + (AC)^2 = (6\sqrt{3})^2 + (6\sqrt{2})^2 = (36 \times 3) + (36 \times 2) = 108 + 72 = 180$. So, $A'C = \sqrt{180} = \sqrt{36 \times 5} = 6\sqrt{5}$ cm.

If the question implies the angle between the prism diagonal A'C and the base diagonal AC (let's call this angle $\alpha$), then we can use the right triangle $\triangle$AA'C. In this triangle, the side opposite to $\alpha$ is AA', and the hypotenuse is A'C. Therefore, $\sin(\alpha) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{AA'}{A'C} = \frac{6\sqrt{3}}{6\sqrt{5}} = \frac{\sqrt{3}}{\sqrt{5}} = \frac{\sqrt{15}}{5}$.

Alternatively, if the question implies the angle between the prism diagonal A'C and the lateral edge AA' (let's call this angle $\beta$), then in the same right triangle $\triangle$AA'C, the side opposite to $\beta$ is AC, and the hypotenuse is A'C. Therefore, $\sin(\beta) = \frac{AC}{A'C} = \frac{6\sqrt{2}}{6\sqrt{5}} = \frac{\sqrt{2}}{\sqrt{5}} = \frac{\sqrt{10}}{5}$.

Without explicit clarification on which angle is being asked for, these are two of the most common and relevant angles to calculate in this type of prism problem. Both calculations involve using the Pythagorean theorem and the definition of sine in a right triangle. Remember, guys, understanding what angle you need to find is half the battle in trigonometry problems!