DB = √( 15² - 9² )
DB = √( 225 - 81 )
DB = √( 144 )
DB = 12 cm
DE = √( 10² - 6² )
DE = √( 100 - 36 )
DE = √( 64 )
DE = 8 cm
BE = DB - DE
BE = 12 - 8
BE = 4 cm
..
terdapat beberapa triple pythagoras sederhana yang perlu kita hafalkan sebagai efisiensi waktu dalam mengerjakan persoalan segitiga siku-siku, diantaranya:
[tex] \begin{array}{ | c|c|c | } \hline \underline{\bold{a}}&\underline{\bold{b}}&\underline{\bold{c}}\\ \hline 3&4&5\\ \hline 5&12&13\\ \hline 8&15&17\\ \hline 7&24&25\\ \hline 9&40&41\\ \hline 11&60&61\\ \hline 20&21&29\\ \hline 12&35&37 \\ \hline \end{array} \: \: \boxed{\begin{array}{l||r} a^2=c^2-b^2& a=\sqrt{c^2-b^2}\\b^2=c^2-a^2& b=\sqrt{c^2-a^2}\\ c^2=a^2+b^2& c=\sqrt{a^2+b^2}\end{array}}[/tex]
Dimana:
Berlaku Kelipatan seperti 6, 8, 10 sebagai kelipatan 2 dari 3, 4, 5
[tex]\begin{aligned} panjang~BE &= \sqrt{AD^2 - AB^2} - \sqrt{CE^2 - DC^2} \\&= \sqrt{(15~cm)^2 - (9~cm)^2} - \sqrt{(10~cm)^2 - (6~cm)^2} \\&= \sqrt{(15~cm - 9~cm) \times (15~cm + 9~cm)} - \sqrt{(10~cm - 6~cm) \times (10~cm + 6~cm)} \\&= \sqrt{6~cm \times 24~cm} - \sqrt{4~cm \times 16~cm} \\&= \sqrt{144~cm^2} - \sqrt{64~cm^2} \\&= 12~cm - 8~cm \\&= \boxed{\bold{\underline{4~cm}}} \end{aligned}[/tex]
Opsi Jawaban : B
[tex]\boxed{\colorbox{ccddff}{Answered by Danial Alf'at | 23 - 05 - 2023}}[/tex]
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Verified answer
DB = √( 15² - 9² )
DB = √( 225 - 81 )
DB = √( 144 )
DB = 12 cm
DE = √( 10² - 6² )
DE = √( 100 - 36 )
DE = √( 64 )
DE = 8 cm
BE = DB - DE
BE = 12 - 8
BE = 4 cm
Teorema Pythagoras
..
terdapat beberapa triple pythagoras sederhana yang perlu kita hafalkan sebagai efisiensi waktu dalam mengerjakan persoalan segitiga siku-siku, diantaranya:
[tex] \begin{array}{ | c|c|c | } \hline \underline{\bold{a}}&\underline{\bold{b}}&\underline{\bold{c}}\\ \hline 3&4&5\\ \hline 5&12&13\\ \hline 8&15&17\\ \hline 7&24&25\\ \hline 9&40&41\\ \hline 11&60&61\\ \hline 20&21&29\\ \hline 12&35&37 \\ \hline \end{array} \: \: \boxed{\begin{array}{l||r} a^2=c^2-b^2& a=\sqrt{c^2-b^2}\\b^2=c^2-a^2& b=\sqrt{c^2-a^2}\\ c^2=a^2+b^2& c=\sqrt{a^2+b^2}\end{array}}[/tex]
Dimana:
Berlaku Kelipatan seperti 6, 8, 10 sebagai kelipatan 2 dari 3, 4, 5
Penyelesaian Soal
[tex]\begin{aligned} panjang~BE &= \sqrt{AD^2 - AB^2} - \sqrt{CE^2 - DC^2} \\&= \sqrt{(15~cm)^2 - (9~cm)^2} - \sqrt{(10~cm)^2 - (6~cm)^2} \\&= \sqrt{(15~cm - 9~cm) \times (15~cm + 9~cm)} - \sqrt{(10~cm - 6~cm) \times (10~cm + 6~cm)} \\&= \sqrt{6~cm \times 24~cm} - \sqrt{4~cm \times 16~cm} \\&= \sqrt{144~cm^2} - \sqrt{64~cm^2} \\&= 12~cm - 8~cm \\&= \boxed{\bold{\underline{4~cm}}} \end{aligned}[/tex]
Opsi Jawaban : B
[tex]\boxed{\colorbox{ccddff}{Answered by Danial Alf'at | 23 - 05 - 2023}}[/tex]