10,11

Find the lateral surface area of a regular triangular pyramid if its height is 4 and the apothem is 8.
Let $M$ be the center of the base $ABC$ of a regular triangular pyramid $ABCD$, and $K$ be the midpoint of $BC$. Then $DM$ is the height of the pyramid, $DK$ is the apothem of the pyramid, and $AK$ is the height of the equilateral triangle $ABC$, with $AK$ passing through point $M$. Denote $AB = BC = AC = a$. Then

$$
AK = \frac{a \sqrt{3}}{2}, \quad KM = \frac{1}{3} AK = \frac{a \sqrt{3}}{6}.
$$

On the other hand, from the right triangle $DKM$, we find that

$$
KM = \sqrt{DK^2 - DM^2} = \sqrt{64 - 16} = 4 \sqrt{3}.
$$

From the equality $\frac{a \sqrt{3}}{6} = 4 \sqrt{3}$, we find that $a = 24$. If $S$ is the lateral surface area of the pyramid, then

$$
S = 3 S_{\triangle BDC} = 3 \cdot \frac{1}{2} AB \cdot DK = 3 \cdot \frac{1}{2} a \cdot DK = 3 \cdot \frac{1}{2} \cdot 24 \cdot 8 = 288
$$

## Answer

288.00

## Problem