[ Special cases of tetrahedrons (other).]

The base of the pyramid is an isosceles triangle with a base of 6 and a height of 9. Each lateral edge is 13. Find the volume of the pyramid.
Let $ABCD$ be a triangular pyramid with base $ABC$, $DA=DB=DC=13$, $BC=6$, $AK=9$ ($K$ is the midpoint of $BC$). From the right triangle $AKC$, we find that

$$
AC=\sqrt{AK^{2}+CK^{2}}=\sqrt{81+9}=3\sqrt{10}, \sin \angle ACK=\frac{AK}{AC}=\frac{3}{\sqrt{10}}
$$

Since the lateral edges of the pyramid are equal, its height passes through the center $O$ of the circle circumscribed around the base. Let $R$ be the radius of this circle. Then

$$
R=\frac{AB}{2 \sin \angle ACCB}=\frac{\frac{3\sqrt{10}}{10}}{\sqrt{10}}=5
$$

From the right triangle $AOD$, we find that

$$
DO=\sqrt{AD^{2}-OA^{2}}=\sqrt{169-25}=12
$$

Therefore,

$$
V_{ABCD}=\frac{1}{3} S_{\triangle ABC} \cdot DO=\frac{1}{3} \cdot \frac{1}{2} BC \cdot AK \cdot DO=\frac{1}{3} \cdot \frac{1}{2} \cdot 6 \cdot 9 \cdot 12=108.
$$

## Answer

108.00

In a triangular pyramid, the lateral edges are pairwise perpendicular and equal to $\sqrt{70}$, $\sqrt{99}$, and $\sqrt{126}$. Find the volume and the area of the base of the pyramid.

## Solution

Let $ABCD$ be the given triangular pyramid $ABCD$, $DA=\sqrt{70}$, $DB=\sqrt{99}$, $DC=\sqrt{126}$, and the edges $DA$, $DB$, and $DC$ are pairwise perpendicular. We will consider $A$ as the vertex of the pyramid $ABCD$. Then its edge $AD$ is perpendicular to two intersecting lines $DB$ and $DC$ in the plane of the face $DBC$. Therefore, $AD$ is the height of the pyramid, and its base is a right triangle $DBC$ with hypotenuse $BC$. Consequently,

$$
V_{ABCD}=\frac{1}{3} S_{\triangle DBC} \cdot AD=\frac{1}{3} \cdot \frac{1}{2} DB \cdot DC \cdot AD=\frac{1}{6} \sqrt{70} \cdot \sqrt{99} \cdot \sqrt{126}=
$$

$$
=\sqrt{35 \cdot 11 \cdot 63}=21\sqrt{55}
$$

From the right triangles $ABD$, $ACD$, and $BCD$ by the Pythagorean theorem, we find that $AB=13$, $AC=14$, and $BC=15$. Then by Heron's formula,

$$
S_{\triangle ABC}=\sqrt{21 \cdot 8 \cdot 7 \cdot 6}=84
$$

## Answer

$21\sqrt{55}; 84$.