We can see that (5 + 4)² = 81
(7 + 9)² = 256
(6 + 15)² = 441
Therefore, (12 + 13)² = 625
Hence, option (D) is the correct answer.

We can see that, 5 × 3 + 4 = 19
Similarly, 6 × 4 + 5 = 29
Therefore, 7 × 5 + 6 = 41
Hence, option (C) is the correct answer.

We can see that, (5 × 6 × 8) + (7 × 4 × 9) = 240 + 252 = 492
Similarly, (7 × 5 × 4) + (6 × 8 × 9) = 140 + 432 = 572
Therefore, (4 × 3 × 5) + (7 × 2 × 5) = 60 + 70 = 130
Hence, option (B) is the correct answer.

We can see that, from second coloumn, 22 + 2 - 23 = 1
Similarly, from third column, 40 + 5 - 43 = 2
Therefore, 21 + 1 - 20 = 2
Option (A) is the correct answer.

We can see that, (3² + 6²) − (2² + 4²) = (9 + 36) − (4 + 16) = 45 − 20 = 25
Similarly, (7² + 11²) − (8² + 6²) = (49 + 121) − (64 + 36) = 170 − 100 = 70
Let the missing number is y, (4² + 1²) − (2² + 5²) = (16 + 1) − (4 + 25) = y
or, y = 17 − 29
y = −12
Hence, option (A) is the correct answer.

We can see that, 7 × 2 + 2 = 16
16 × 2 + 2 = 34
Therefore, the missing term is, 34 × 2 + 2 = 70
Hence, option (D) is the correct answer.

The given figure contains numbers from 1 to 6 in three alternate segments, the smaller number being towards the outside and the number 14 to 19 in the remaining three alternate segments with the smaller number towards the inside. Hence, option (B) is the correct answer.

We can see that in figure (A), 10 − 4 = 6; 18 − 10 = 8; 18 − 4 = 14
Similarly, in figure (B), 14 − 8 = 6; 22 − 14 = 8; 22 − 8 = 14
Therefore, 11 − 5 = 6, 15 − 11 = 4; the missing term is 15 − 5 = 10
Hence, option (A) is the correct answer.

The sum of numbers at the extremities of the three line segments in each figure is equal.
Thus, in figure (A), (39 + 33) = (29 + 43) = (27 + 45)
In figure (B), (42 + 31) = (29 + 44) = (30 + 43)
Let the missing term in figure (C) is y, then, (y + 10) = (59 + 20) = (40 + 39) = 79
y + 10 = 79
y = 79 − 10
y = 69
Hence, option (C) is the correct answer.