7.
`T = \frac{AM * BC}{2} = AM * BM = 105 cm^2`

`AM = \frac{105}{BM} = \frac{105}{7} = 15 cm`

`AB = \sqrt{AM^2 + BM^2} = \sqrt{15^2 + 7^2} = \sqrt{225 + 49} = \sqrt{274} = \color{red}16.552945 cm`

8.
`K = c * 4 = 40 cm^2`
`c = \frac{40}{4} = 10 cm`

`a^2 + b^2 = c^2`
`(2*b)^2 + b^2 = 10^2`
`2^2 * b^2 + b^2 = 10^2`
`4 * b^2 + b^2 = 100`
`5 * b^2 = 100`
`b^2 = \frac{100}{5} = 20`
`b = \sqrt{20} = 4.472136`

`a = 2 * b = 2 * 4.4721361 = 8.944272`

`T_{AOB} = \frac{a * b}{2} = \frac{4.472136 * 8.94427}{2} = \frac{40}{2} = 20 cm^2`

`T_{ABCD} = T_{AOB} * 4 = 20 * 4 = \color{red}80 cm^2`

9.
`FO = \sqrt{AF^2 - AO^2} = \sqrt{105^2 - 100^2} = \sqrt{11025 - 10000} = \sqrt{1025} \approx \color{red}32.015621`

11.
(Kissé furcsán hangzik, hogy dőlésnek nevezték a cm-ben megadandó eredményt, de szerintem erre gondoltak.)

`AF = \sqrt{DF^2 - AD^2} = \sqrt{26^2 - 24^2} = \sqrt{676 - 576} = \sqrt{100} = \color{red}10 cm`

Mivel a (10, 24, 26) pitagoraszi számok amelyekből kettőt már ismertünk, a harmadikat táblázatból is kikereshettük volna. Lásd: https://hu.wikipedia.org/wiki/Pitagoraszi_sz%C3%A1mh%C3%A1rmasok_list%C3%A1ja

18.
`BE = \frac{AB - CD}{2} = \frac{36 - 20}{2} = \frac{16}{2} = 8 cm`

`AE = AB - BE = 28 cm`

`CE = \sqrt{BC^2 - BE^2} = \sqrt{10^2 - 8^2} = \sqrt{100 - 64} = \sqrt{36} = 6 cm`

`AC = \sqrt{AE^2 + CE^2} = \sqrt{28^2 + 6^2} = \sqrt{784 + 36} = \sqrt{820} \approx \color{red}28.635642 cm`

20.
`AO = \frac{AC}{2} = \frac{24}{2} = 12`

`BO = \sqrt{AB^2 - AO^2} = \sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = 5`

`BD = BO * 2 = 5 * 2 = \color{red}10`