Given a compact set $B \subseteq \mathbb{R}^{d-1}$ ($d ≥ 2$), the cone with basis $B$ and hight $h > 0$ may be defined as:
$$K = \{(x, y) \in \mathbb{R}^{d-1} \times \mathbb{R}: 0 ≤ y ≤ h, x \in (1 - \frac{y}{h})B\}$$
I first want to show that $K$ is compact. Next, I want to show that the following formula holds true for the Lebesgue measure $\lambda_d$ of $K$:
$$\lambda_d(K) = \frac{h}{d} \cdot \lambda_{d-1}(B) \space\space\space(1)$$
(where $\lambda_{d-1}(B)$ is the ($d-1$-dimensional) Lebesgue measure of $B$).
I think the first part can be seen by first noting that $\{y \in \mathbb{R}, 0 ≤ y ≤ h\} = [0, h]$ is compact, and, since $B$ is compact (hence bounded), $\cup_{y \in [0, h]} (1 - \frac{y}{h}) B$ is bounded, and it's closed because for any sequence $(a_n) \subseteq \cup_{y \in [0, h]} (1 - \frac{y}{h}) B$ that converges in $\mathbb{R}^{d-1}$, we have that $(a_n) = (b_n) (c_n)$ for some sequences $(b_n) \subseteq \cup_{y \in [0, h]} \{(1 - \frac{y}{h})\} = [0, 1]$, $(c_n) \subseteq B$, and both $(b_n)$ and $(c_n)$ have a limit in their respective set since both $[0, 1]$ and $B$ are compact, hence, $(a_n)$ converges aswell in $\cup_{y \in [0, h]} (1 - \frac{y}{h}) B$, hence, that set's bounded too, and $K$ is therefore compact because it's compact in each component.
For the second part, I was given the hint that I can (without proof) use the fact that, given a diagonal matrix $S = diag(s_1, ..., s_d) \in \mathbb{R}^{d \times d}$ with positive diagonal entries and a Borel set $A \subseteq \mathbb{R}^d$, then the set $S(A) := \{S x: x \in A\}$ is a Borel set with $\lambda_d(S(A)) = det(S) \cdot \lambda_d(A)\space\space\space(2)$.
This looked indeed useful, but so far, I've failed to apply it to what I'm trying to show. I first thought that I would need to choose $A = B$ for the formula $(2)$, and then cleverly choose a diagonal matrix $S$. But that's already becoming problematic because $\lambda_d(B) = 0$ (because $B$ is a subset of $\mathbb{R}^{d-1}$). So maybe I have to choose $A = B \times [0, 1]$ or something like that, in order for the measure of $A$ to actually be positive? But then I would still need to find a diagonal matrix $S$, and for that, I didn't really have any idea so far. Thanks in advance.
Note that $K$ is the image of $B\times [0,h]$ under the map $(x,y) \to ((1-y/h)x,y).$ Since $B\times [0,h]$ is compact and this map is continuous, $K$ is compact.
Let's use Fubini to get $\lambda_d(K).$ We get
$$\lambda_d(K) = \int_0^h f(y)\, dy$$
where $f(y)$ is the $(d-1)$-dimensional volume of the slice of $K$ at height $y.$ That volume is $\lambda_{d-1} ((1-y/h)B) = (1-y/h)^{d-1}\lambda_{d-1} (B).$ Thus
$$\lambda_d(K) = \int_0^h (1-y/h)^{d-1}\lambda_{d-1} (B)\, dy = \frac{h}{d}\lambda_{d-1} (B).$$