Minta tolong bantuannya saya butuh no 22 dan 23 . Bagi yang bisa

Soal No. 22
Segitiga lancip, jika : a² < b² + c²

(1) a = 15, b = 10 dan c = 12

[tex]{{a^2=15^2=225} \atop {b^2+c^2=10^2+12^2=100+144=244}} \} \ a^2\ \textless \ b^2+c^2...Lancip[/tex]

(2) a = 17, b = 7 dan c = 15

[tex]{{a^2=17^2=289} \atop {b^2+c^2=7^2+15^2=49+225=274}} \} \ a^2 \ \textgreater \ b^2+c^2...Tumpul[/tex]

(3) a = 20, b = 10 dan c = 15

[tex]{{a^2=20^2=400} \atop {b^2+c^2=10^2+15^2=100+225=325}} \} \ a^2 \ \textgreater \ b^2+c^2...Tumpul[/tex]

(4) a = 16, b = 8 dan c = 15

[tex]{{a^2=16^2=256} \atop {b^2+c^2=8^2+15^2=64+225=289}} \} \ a^2\ \textless \ b^2+c^2...Lancip[/tex]

Segitiga Lancip = (1) dan (4).........Jawaban (B)

Soal No. 23

[tex]BD= \sqrt{AB^2+AD^2}=\sqrt{9^2+12^2} = \sqrt{81+144}= \sqrt{225}=15\ cm \\ \\ BC= \sqrt{CD^2-BD^2}=\sqrt{25^2-15^2} = \sqrt{625-225}= \sqrt{400}=20\ cm[/tex]

Keliling ABCD adalah
= AB + BC + CD + AD
= (9 + 20 + 25 + 12) cm
= 66 cm