Pyramid Surface Area: Solve Geometry Problems

Hey guys! Today we're diving into a couple of geometry problems that are super important if you're working with pyramids. We've got two brain teasers that will test your understanding of surface area and those tricky dihedral angles. So, grab your notebooks, and let's break down how to find the lateral surface area of a pyramid when its base is an isosceles triangle. We'll also touch on another related problem, ensuring you've got a solid grasp of these concepts.

Problem 35: Finding the Lateral Surface Area of a Pyramid

Alright, let's get straight into the first challenge, Problem 35. This one asks us to find the lateral surface area of a pyramid. The key info we're given is that the base of our pyramid is an isosceles triangle with sides measuring 5 cm, 5 cm, and 8 cm. On top of that, we know that each dihedral angle of the pyramid at the base edge is 30°. This might sound a bit intimidating, but trust me, if we break it down step by step, it's totally manageable. First things first, let's visualize our pyramid. We have a base that's an isosceles triangle. Remember, an isosceles triangle has two sides of equal length, which are 5 cm in this case, and a base of 8 cm. Now, the pyramid part means we have a point (the apex) above this triangle, and lines connecting that apex to each vertex of the base triangle. The lateral surface area is essentially the sum of the areas of all those triangular faces that connect the base to the apex. We're not including the base area itself when we talk about lateral surface area. The dihedral angle information is crucial here; it's the angle between two intersecting planes – in this case, the plane of one of the lateral faces and the plane of the base. Having each of these angles be 30° tells us a lot about the pyramid's symmetry and shape.

To tackle this, we need to find the area of each of the three lateral faces. Since the base is an isosceles triangle, and the dihedral angles at the base edges are equal, the pyramid is likely a right pyramid (meaning the apex is directly above the centroid of the base), and its lateral faces will be congruent isosceles triangles. To find the area of a triangle, we use the formula: (1/2) * base * height. For the lateral faces, the 'base' of each triangular face will be one of the sides of our base triangle (either 5 cm or 8 cm), and the 'height' we need is the slant height of the pyramid relative to that base edge. This slant height is not the same as the height of the pyramid itself. It's the height of each lateral triangular face, measured from the midpoint of a base edge up to the apex. Finding this slant height will be our main mission. We can use trigonometry and the given dihedral angle (30°) to figure this out. Imagine drawing a line from the apex perpendicular to one of the base edges; this line represents the slant height. Now, consider the right triangle formed by this slant height, the altitude from the base edge's midpoint to the apex (which is the height of the pyramid), and a line segment connecting the base edge's midpoint to the foot of the slant height on the base edge. The dihedral angle of 30° comes into play here, relating the pyramid's height to the slant height. We can use the sine or cosine of 30° depending on how we've set up our right triangle. Let's assume we drop a perpendicular from the apex to the midpoint of the 8 cm base edge. This forms a right-angled triangle where the hypotenuse is the slant height we need, one leg is the pyramid's height, and the other leg is the distance from the midpoint of the 8 cm edge to the point where the slant height hits the base edge. The dihedral angle is between the lateral face and the base. By carefully constructing a right triangle involving the dihedral angle, we can find the slant height. For instance, if we consider the apothem of the base triangle (the perpendicular distance from the center to a side) and the pyramid's height, we can form a right triangle where the dihedral angle is one of the acute angles. However, the problem gives us dihedral angles at each base edge. This simplifies things, meaning we can focus on one representative slant height if the pyramid is symmetric. Let's redraw our base triangle. The sides are 5, 5, and 8. We can find the height of this isosceles triangle using the Pythagorean theorem. Drop a perpendicular from the vertex between the two 5 cm sides to the 8 cm base. This bisects the 8 cm base, creating two right triangles with hypotenuses of 5 cm and one leg of 4 cm. So, the height (h_base) of the base triangle is sqrt(5^2 - 4^2) = sqrt(25 - 16) = sqrt(9) = 3 cm. The area of the base triangle is (1/2) * base * height = (1/2) * 8 * 3 = 12 sq cm. Now, back to the lateral surface area. We need the slant heights. Let's denote the slant height to the 8 cm base edge as $l_1$ and the slant heights to the 5 cm base edges as $l_2$. The problem states each dihedral angle at the base edge is 30°. This implies the pyramid is quite regular. Let's focus on the slant height $l_1$ corresponding to the 8 cm base edge. We can form a right triangle where the angle is 30°. One leg of this triangle is the apothem of the base triangle related to the 8 cm side. The apothem is the distance from the triangle's centroid to the midpoint of a side. To find the centroid, we need the medians. The median to the 8 cm side is also its altitude, which we found to be 3 cm. The centroid divides the median in a 2:1 ratio. So, the distance from the vertex to the centroid is (2/3)*3 = 2 cm, and the distance from the centroid to the midpoint of the 8 cm side (this is the apothem, let's call it $a_1$) is (1/3)*3 = 1 cm. Now, consider the right triangle formed by the pyramid's height (H), the apothem $a_1$, and the slant height $l_1$. The angle between the slant height $l_1$ and the apothem $a_1$ is the dihedral angle, 30°. So, $a_1 = l_1 * cos(30°)$. This gives us $1 = l_1 * (\sqrt{3}/2)$, so $l_1 = 2/\sqrt{3}$ cm. The area of the lateral face corresponding to the 8 cm base edge is (1/2) * 8 * $l_1$ = 4 * $(2/\sqrt{3})$ = $8/\sqrt{3}$ sq cm. Now for the slant heights $l_2$ to the 5 cm sides. We need the apothems to the 5 cm sides. Let's find the height of the base triangle from one of the 5 cm sides. This is a bit more complex. However, the problem says each dihedral angle is 30°. This strongly suggests the pyramid is a right pyramid and its lateral faces are congruent. Let's re-read carefully: "кожний двогранний кут піраміди при ребрі основи дорівнює 30°" - "each dihedral angle of the pyramid at the base edge is equal to 30°". This indeed implies symmetry. If all dihedral angles at the base edges are equal, then the lateral faces must have the same slant height. Let's call this slant height $l$. The base triangle has sides 5, 5, 8. The perimeter of the base is 5 + 5 + 8 = 22 cm. The area of the lateral surface is the sum of the areas of the three lateral triangles. If $l$ is the slant height, and the lateral faces are isosceles triangles with base lengths corresponding to the base edges, we need to be careful. Let's assume the pyramid is a right pyramid. The apex is directly above the circumcenter or incenter of the base triangle. Given the isosceles base and equal dihedral angles at all base edges, this means the apex must be directly above the incenter of the base triangle. The distance from the incenter to each side is the inradius ($r$). The dihedral angle is the angle between a lateral face and the base. Let $H$ be the height of the pyramid, and $l$ be the slant height (the height of a lateral face). The inradius $r$ is the apothem of the base triangle when considering the dihedral angle. The area of the base triangle is 12 sq cm. The semi-perimeter $s = (5+5+8)/2 = 11$ cm. The inradius $r = Area / s = 12 / 11$ cm. Now, consider the right triangle formed by the pyramid's height $H$, the inradius $r$, and the slant height $l$. The angle at the base, between $r$ and $l$, is the dihedral angle, 30°. So, we have $r = l * cos(30°)$. Substituting the value of $r$: $12/11 = l * (\sqrt{3}/2)$. Solving for $l$: $l = (12/11) * (2/\sqrt{3}) = 24 / (11\sqrt{3})$. Now we have the slant height $l$. The lateral surface area is the sum of the areas of the three lateral faces. The base edges are 5 cm, 5 cm, and 8 cm. The area of the lateral face with base 8 cm is (1/2) * 8 * $l$. The area of each lateral face with base 5 cm is (1/2) * 5 * $l$. So, the total lateral surface area ($A_{lateral}$) is: $A_{lateral} = (1/2) * 8 * l + 2 * [(1/2) * 5 * l] = 4l + 5l = 9l$. Substituting the value of $l$: $A_{lateral} = 9 * (24 / (11\sqrt{3})) = 216 / (11\sqrt{3})$. To rationalize this, multiply numerator and denominator by $\sqrt{3}$: $A_{lateral} = (216\sqrt{3}) / (11 * 3) = (216\sqrt{3}) / 33$. We can simplify this by dividing 216 and 33 by 3: $A_{lateral} = (72\sqrt{3}) / 11$ sq cm. This is a beautiful result! It highlights how understanding the properties of the base and the meaning of dihedral angles allows us to calculate complex surface areas step-by-step. We used the inradius because the equal dihedral angles at all base edges pointed towards the apex being equidistant from all base edges, which is a property of the incenter for a triangle.

Problem 36: Another Pyramid Challenge

Now, let's briefly look at Problem 36. While the specifics aren't fully detailed here, it likely presents another scenario involving a pyramid where we need to calculate some aspect of its surface area or volume. The phrasing "Основою піраміди є рівнобедрений" ("The base of the pyramid is isosceles") suggests the base triangle is again isosceles, similar to Problem 35. Depending on the information provided, we might be given different angles, edge lengths, or heights. The approach to solving such problems generally follows a similar pattern: first, understand the shape and dimensions of the base; second, determine the height of the pyramid and its slant heights (if needed for surface area); and third, apply the appropriate formulas. If the base is isosceles, we'll often need to find the height of that base triangle and potentially its inradius or circumradius, depending on whether the pyramid is right and where its apex is located relative to the base. The key takeaway from problems like these is the systematic approach. We break down the 3D shape into 2D components (the base and the lateral faces) and use our knowledge of plane geometry and trigonometry to find the necessary dimensions. Remember, practice is your best friend here. The more problems you solve, the more comfortable you'll become with identifying the right triangles, applying trigonometric functions, and linking the different parts of the pyramid together. Don't be afraid to draw diagrams; they are incredibly helpful for visualizing the relationships between heights, slant heights, and angles. Sometimes, the problem might give you the lateral surface area and ask for the height, or vice-versa. The core principles remain the same: use the given information to find unknown dimensions, and then plug those into the formulas you need. Keep practicing, and you'll master these pyramid puzzles in no time!

So, there you have it, guys! Two problems that, while presenting different specific numbers or details, rely on the same fundamental geometric principles. Mastering the calculation of surface areas for pyramids, especially those with non-standard bases like isosceles triangles, is a key skill in geometry. Keep practicing these concepts, and you'll be acing your exams!