Gọi chiều cao và cạnh đáy lần lượt là $h,a(h,a>0)$
Theo bài ra ta có: $\left\{\begin{array}{l} h=\dfrac{1}{2}a\\ \dfrac{1}{2}(a-2)(h-3)=\dfrac{1}{2}ah-21\end{array} \right.\\ \left\{\begin{array}{l} h=\dfrac{1}{2}a\\ \dfrac{1}{2}ah-\dfrac{3}{2}a-h+3=\dfrac{1}{2}ah-21\end{array} \right.\\ \left\{\begin{array}{l} h=\dfrac{1}{2}a\\ \dfrac{3}{2}a+h=24\end{array} \right.\\ \left\{\begin{array}{l} h=\dfrac{1}{2}a\\ \dfrac{3}{2}a+\dfrac{1}{2}a=24\end{array} \right.\\ \left\{\begin{array}{l} h=\dfrac{1}{2}a\\ 2a=24\end{array} \right.\\ \left\{\begin{array}{l} h=6\\ a=12\end{array} \right.$