Sebuah segitiga siku siku ABC (siku - siku di B) dengan panjang AB = (x + 5) cm dan BC = 3x cm a) Nyatakan luas segitiga ABC dalam x b) Hitung panjang AB dan AC jika diketahui luas segitiga ABC adalah 9 cm²

<p>Diketahui :</p><p>Segitiga siku-siku ABC (siku-siku di B) :</p><ul><li>AB = (x + 5) cm</li><li>BC = 3x cm</li></ul><p>&nbsp;</p><p>Pembahasan :</p><p>A. Luas segitiga ABC dalam x</p><p>&nbsp; &nbsp; &nbsp;L = ½ × AB × BC</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;= ½ &nbsp;× (x + 5) &nbsp;× (3x)</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;= ½ &nbsp;× (3x²+ 15x)</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;= 1,5x² + 7,5x</p><p>&nbsp; &nbsp;&nbsp;</p><p>&nbsp; &nbsp; &nbsp;Jadi luas segitiga ABC = (1,5x² + 7,5x) cm²</p><p>&nbsp;</p><p>B. Menentukan panjang AB dan AC</p><p>&nbsp; &nbsp; &nbsp;L = 9</p><p>&nbsp; &nbsp; &nbsp;1,5x² + 7,5x = 9</p><p>&nbsp; &nbsp; &nbsp; 1,5x² + 7,5x - 9 = 0 &nbsp; &nbsp; (dibagi 1,5)</p><p>&nbsp; &nbsp; &nbsp; &nbsp;x² + 5x - 6 = 0</p><p>&nbsp; &nbsp; &nbsp; (x - 1)(x + 6) = 0</p><p>&nbsp; &nbsp; &nbsp; &nbsp;x₁ = 1 atau x₂ = -6</p><p>&nbsp; &nbsp; &nbsp; &nbsp;x₂ = -6 tidak memenuhi syarat karena tidak ada panjang sisi negatif.</p><p>&nbsp; &nbsp; &nbsp; &nbsp; Kita pakai x = 1 maka :</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;AB = x + 5 = 1 + 5 = <strong>6 cm</strong></p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;BC = 3x = 3(1) = 3 cm</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;AC = √(AB² + BC²)</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; = √(6² + 3²)</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; = √(36 + 9)&nbsp;</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; = √45</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; = <strong>3√5 cm</strong></p><p>&nbsp;</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;</p>