\(OK=\dfrac{a}{2}\)
Áp dụng định lí Pytago vào tam giác vuông \(SOK\), ta có:
\(\begin{array}{l}S{K^2} = S{O^2} + O{K^2}\\ \Rightarrow {d^2} = {h^2} + {\left( {\dfrac{a}{2}} \right)^2}\end{array}\)
- Với \(a=6;h=4\) ta có:
\(\begin{array}{l}d = \sqrt {{4^2} + {{\left( {\dfrac{6}{2}} \right)}^2}} = 5\\{S_đ} = 6.6 = 36\\{S_{xq}} = 2.6.5 = 60\\{S_{tp}} = 60 + 36 = 96\\V = \dfrac{1}{3}.36.4 = 48\end{array}\)
- Với \(h=6;S_đ=256\) ta có:
\({S_đ} = {a^2} = 256 \Rightarrow a = \sqrt {256} = 16\)
\(d = \sqrt {{6^2} + {{\left( {\dfrac{{16}}{2}} \right)}^2}} = 10\)
\({S_{xq}} = 2.16.10 = 320\)
\({S_{tp}} = 320 + 256 = 576\)
\(V = \dfrac{1}{3}.256.6 = 512\)
- Với \(d=15;S_{xq}=720\) ta có:
\({S_{xq}} = 2.a.d = 720 \Rightarrow a = \dfrac{{720}}{{2.15}} = 24\)
\(h = \sqrt {{{15}^2} - {{\left( {\dfrac{{24}}{2}} \right)}^2}} = 9\)
\({S_đ} = 24.24 = 576\)
\({S_{tp}} = 720 + 576 = 1296\)
\(V = \dfrac{1}{3}.576.9 = 1728\)
- Với \(a=32;V=4096\) ta có:
\(V = \dfrac{1}{3}.{a^2}.h = 4096\) \( \Rightarrow h = \dfrac{{3.4096}}{{{{32}^2}}} = 12\)
\(d = \sqrt {{{12}^2} + {{\left( {\dfrac{{32}}{2}} \right)}^2}} = 20\)
\({S_đ} = 32.32 = 1024\)
\({S_{xq}} = 2.32.20 = 1280\)
\({S_{tp}} = 1280 + 1024 = 2304\)
- Với \(d=17;S_{xq}=544\) ta có:
\({S_{xq}} = 2.a.d = 544 \Rightarrow a = \dfrac{{544}}{{2.17}} = 16\)
\(h = \sqrt {{{17}^2} - {{\left( {\dfrac{{16}}{2}} \right)}^2}} = 15\)
\({S_đ} = 16.16 = 256\)
\({S_{tp}} = 544 + 256 = 800\)
\(V = \dfrac{1}{3}.256.15 = 1280\)
Ta điền vào bảng như sau:
