diketahui balok ABCD.EFGH dengan panjang AB = 20 cm AD = 12 cm, dan panjang diagonal ruang 25 cm. Hitunglah luas permukaan dan volume balok ABCD.EFGH

AB = 20 Cm
AD = 12 Cm
AG ( Diagonal Ruang ) = 25 Cm

Tinggi ( AE ) = √ AG² – ( AB² + AD² )
= √ 25² – ( 20² + 12² )
= √ 625 – ( 400 + 144 )
= √ 625 – 544
= √ 81
= 9 Cm

A. Luas Permukaan = ( 2 x Luas Alas ) + ( Keliling Alas x Tinggi )
= ( 2 x AB x AD ) + ( ( 2 x ( AB + AD ) ) x T )
= ( 2 x 20 x 12 ) + ( ( 2 x ( 20 + 12 ) ) x 9 )
= ( 2 x 240 ) + ( ( 2 x 32 ) x 9 )
= 480 + ( 64 x 9 )
= 480 + 576
= 1.056 Cm²

B. Volume = AB x AD x AG
= 20 x 12 x 9
= 20 x 108
= 2.160 Cm³