This document provides examples of using interpolation to solve for unknown values. In the first example, the formula is used to solve for the value of X when given the points (1, 5.297), (1.2, X), and (1.25, 5.003). In the second multi-step example, the formula is applied multiple times to first solve for values of X1 and X2, in order to then solve for the final unknown value of X3.

1. INTERPOLASI

2. RUMUS INTERPOLASI
CONTOH SOAL I :
1,00 = 5,297
1,20 = X
1,25 = 5,003
BERAPAKAH NILAI X ? ?
X =
( Kiri Tengah -
( Kiri Bawah -
Kanan Atas + X ( Kanan Bawah -
Kiri Atas )
Kiri Atas )
Kanan Atas )

3. Penyelesaian :
X =
( 1,20
-
( 1,25 -
5,297 + X ( 5,003 -
1,00 )
1,00 )
5,297 )
X =
0,20
0,25
5,297 + X ( - 0,294 )
X = 0,80
5,297 + X ( - 294 )
X = 5,297 - 0,2352
= 5,0618

4. CONTOH SOAL II :
2.40 M
2.00 M . 3.00 M
2.00 M . . . 152 . 230
3.98 . . . . . . . . . . . . X1 . . . . X3 . . . . X2
7.00 M . . . 152 . 228
Ditanya : X3 = ?

5. Penyelesaian : Mencari dahulu nilai X1
2.00 M
2.00 M . . . . . 152
3.98 . . . . . . . . . . . . . . . X1
7.00 M . . . . . 152
Ditanya : X1 = ?
X1 =
( 3.98 – 2.00 )
152 + X ( 152 - 152 )
( 7.00 – 2.00 )
X1 = 152 +
1.98
5.00
X ( 0 )
X1 = 152 +
X1 = 152 + 0 = 152
0,396 X ( 0 )

6. Penyelesaian : Mencari nilai X2
3.00 M
2.00 M . . . 230
3.98 . . . . . . . . . . . . . . X2
7.00 M . . . 228
Ditanya : X2 = ?
X2 =
( 3.98 – 2.00 )
230 + X ( 228 - 230 )
( 7.00 – 2.00 )
X2 = 230 +
1.98
5.00
X ( - 2
)
X2 = 230 +
X2 = 230 + ( - 0.792 ) = 229.208
0,396 X ( - 2 0 )

7. Penyelesaian :
2.00 M = 152
2.40 M . . . . . . X3
3.00 M = 229.208

8. X =
( 2.40
-
( 3.00 -
152 + X ( 229.208 -
2.00 )
2.00 )
152 )
X =
0.40
1.00
152 + X ( 77.208 )
X = 152
+
30.8832
= 182.8832 dibulatkan 182.88
X = 0.40
152 + X ( 77.2 )
Penyelesaian :