Task B-4.7. The base of a triangular pyramid is a triangle with sides $a, b$, and $c$. The length of side $c$ is $7 \, \text{cm}$, $a - b = 5 \, \text{cm}$, and the angle opposite side $c, \gamma = 60^\circ$. The lateral face containing the longest base edge is perpendicular to the base plane and congruent to the base. Calculate the volume of the pyramid and the area of the largest lateral face of the pyramid.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_0148fcdd1c92efebb536g-25.jpg?height=788&width=873&top_left_y=560&top_left_x=583)

Using the Law of Cosines, we will calculate the lengths of sides $a$ and $b$.

$$
\begin{aligned}
& c^{2}=a^{2}+b^{2}-2 a b \cdot \cos \gamma \\
& 49=(b+5)^{2}+b^{2}-2(b+5) b \cdot \frac{1}{2} \\
& b^{2}+5 b-24=0 \\
& b=3 \text{ cm}, \quad a=8 \text{ cm}
\end{aligned}
$$

Since the lateral face $C B D$ is congruent to the base and perpendicular to it, the height of the pyramid is equal to the height of the base, and we have

$$
\sin \gamma=\frac{v}{b} \quad \Rightarrow \quad v=\frac{3 \sqrt{3}}{2} \text{ cm}
$$

The area of the base is $B=\frac{1}{2} a b \sin \gamma=12 \cdot \frac{\sqrt{3}}{2}=6 \sqrt{3}$.

Then the volume of the pyramid is

$$
V=\frac{6 \sqrt{3} \cdot \frac{3 \sqrt{3}}{2}}{3}=9 \text{ cm}^{3}
$$

The area of the lateral face $C B D$ is equal to the area of the base. Since the length of side $b$ is less than the length of side $c$, the lateral face $A B D$ is larger than the lateral face $C A D$, so we will calculate its area.

(1 point)

Let's calculate the length of the unknown edge $x$:

$$
x=v \sqrt{2}=\frac{3 \sqrt{6}}{2} \text{ cm}
$$

The length of the height of the lateral face $A B D$ is

$$
y=\sqrt{c^{2}-\left(\frac{x}{2}\right)^{2}}=\frac{\sqrt{730}}{4}
$$

and its area is

$$
P=\frac{y \cdot x}{2}=\frac{1}{2} \cdot \frac{\sqrt{730}}{4} \cdot \frac{3 \sqrt{6}}{2}=\frac{3}{16} \cdot \sqrt{4380}=\frac{3}{8} \cdot \sqrt{1095} \text{ cm}^{2}
$$