1. **Problem Statement:** Find the Lateral Surface Area (LSA) and Total Surface Area (TSA) of a triangular pyramid with base side length $8\sqrt{3}$ cm. 2. **Understanding the problem:** A triangular pyramid (tetrahedron) has 4 triangular faces. The LSA is the sum of the areas of the three triangular faces excluding the base. The TSA is the sum of the LSA and the area of the base. 3. **Given:** Base side length $QR = 8\sqrt{3}$ cm. The triangle $OQR$ is isosceles with $OQ = OR$. 4. **Step 1: Calculate the area of the base triangle $\triangle OQR$.** - Since $OQR$ is isosceles with equal sides $OQ = OR$, and base $QR = 8\sqrt{3}$ cm. - To find the height $h$ of $\triangle OQR$, drop a perpendicular from $O$ to $QR$. - Let midpoint of $QR$ be $M$, then $QM = MR = \frac{8\sqrt{3}}{2} = 4\sqrt{3}$ cm. 5. **Step 2: Calculate height $h$ of $\triangle OQR$.** - Using Pythagoras theorem in $\triangle OMQ$: $$h = \sqrt{OQ^2 - QM^2}$$ - Since $OQ = OR$, but length not given, assume $OQ = OR = 8$ cm (common in such problems with $8\sqrt{3}$ base). - Then: $$h = \sqrt{8^2 - (4\sqrt{3})^2} = \sqrt{64 - 16 \times 3} = \sqrt{64 - 48} = \sqrt{16} = 4$$ cm. 6. **Step 3: Area of base $\triangle OQR$** $$\text{Area}_{base} = \frac{1}{2} \times QR \times h = \frac{1}{2} \times 8\sqrt{3} \times 4 = 16\sqrt{3} \text{ cm}^2$$ 7. **Step 4: Calculate LSA (sum of areas of three lateral faces).** - Each lateral face is a triangle formed by $P$ and one side of base. - Without specific height or slant height, assume the pyramid is regular or equilateral tetrahedron for simplicity. - For a regular tetrahedron with side length $a = 8\sqrt{3}$, each face area is: $$A = \frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} (8\sqrt{3})^2 = \frac{\sqrt{3}}{4} \times 64 \times 3 = \frac{\sqrt{3}}{4} \times 192 = 48\sqrt{3}$$ - Since base area is $16\sqrt{3}$, lateral faces are the other 3 faces, each $48\sqrt{3}$. - LSA = $3 \times 48\sqrt{3} = 144\sqrt{3}$ cm$^2$. 8. **Step 5: Calculate TSA (total surface area).** $$\text{TSA} = \text{LSA} + \text{Area}_{base} = 144\sqrt{3} + 16\sqrt{3} = 160\sqrt{3} \text{ cm}^2$$ **Final answers:** - LSA = $144\sqrt{3}$ cm$^2$ - TSA = $160\sqrt{3}$ cm$^2$