Tolong jawab dengan caranya nomor 13 dengan 14
Diketahui panjang sisi kubus = 1 cm.
Misal :
AE = x
? Mencari nilai x
EF = √ EG^2 + GF^2
= √ x^2 + x^2
= √ 2x^2 = x √2
1 – AE – FD = EF
1 – x – x = x √2
x √2 + 2x = 1
x ( √2 + 2 ) = 1
x = 1 / √2 + 2
= (2 – √2) / 2 // Dirasionalkan
? Luas Irisan
L = L.ABCD – 4 L.EFG
= 1 – 4 . 1/2 . x . x
= 1 – 2 . [ ( 2 – √2 ) / 2 ]^2
= 1 – 2 . ( 4 – 4√2 + 2 ) / 4
= 1 – 2 + 2√2 – 1
= 2√2 – 2 ( E )
Nb. Nomer 13 kurang jelas soalnya