Segitiga ABC siku* di A,panjang AB =4x cm,AC=(5x+1)cm dan BC=20 cm.Nyatakan a.keliling dalam x b.luas dalam x

∆ABC siku-siku di A

• keliling ∆
K = AB + AC + BC
K = 4x + (5x + 1) + 20
K = (9x + 21) cm

• luas ∆
L = (1/2)(AB)(AC)
L = (1/2)(4x)(5x + 1)
L = 2x(5x + 1)
L = (10x² + 2x) cm²