Di berikan balok ABCD EFGH dengan panjang rusuk AB = 20 cm BC = 30 cm dan AE = 40 cm titik N terletak pada diagonal FH dengan rasio FH : NH = 2 : 1 hitunglah jarak AN
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Kucoba jawab. Mungkin Soalnya panjang AN Perhatikan gambar yang kuupload
AP = 1/3 x 20 = 20/3 PO = 2/3 x 30 = 20 AO² = AP²+ PO² = (20/3)² + 20² = 400/9 + 400 = 400/9 + 3600/9 = 4000/9 AO = √(4000/9) AO = (20/3)√10 AN² = AO² + ON² = 4000/9 + 40² = 4000/9 + 1.600 = 4.000/9 + 14.400/9 = 18.400/9 AN = √(18.400/9) AN = (20/3) √46 Jadi panjang AN = (20/3)√46
Perhatikan gambar yang kuupload
AP = 1/3 x 20 = 20/3
PO = 2/3 x 30 = 20
AO² = AP²+ PO²
= (20/3)² + 20²
= 400/9 + 400
= 400/9 + 3600/9
= 4000/9
AO = √(4000/9)
AO = (20/3)√10
AN² = AO² + ON²
= 4000/9 + 40²
= 4000/9 + 1.600
= 4.000/9 + 14.400/9
= 18.400/9
AN = √(18.400/9)
AN = (20/3) √46
Jadi panjang AN = (20/3)√46