two squares in the manner desired.”  In this case the smaller square is preserved intact.  Still I give it as an illustration of a feature of the puzzle.  It is impossible in a problem of this kind to give a quarter-turn to any of the pieces if the pattern is to properly match, but (as in the case of F, in Diagram 4) we may give a symmetrical piece a half-turn—­that is, turn it upside down.  Whether or not a piece may be given a quarter-turn, a half-turn, or no turn at all in these chequered problems, depends on the character of the design, on the material employed, and also on the form of the piece itself.

[Illustration:  Diagram 2]

[Illustration:  Diagram 3]

[Illustration:  Diagram 4]

[Illustration:  Diagram 5]

175.—­ANOTHER PATCHWORK PUZZLE.

The lady need only unpick the stitches along the dark lines in the larger portion of patchwork, when the four pieces will fit together and form a square, as shown in our illustration.

[Illustration]

176.—­LINOLEUM CUTTING.

There is only one solution that will enable us to retain the larger of the two pieces with as little as possible cut from it.  Fig. 1 in the following diagram shows how the smaller piece is to be cut, and Fig. 2 how we should dissect the larger piece, while in Fig. 3 we have the new square 10 x 10 formed by the four pieces with all the chequers properly matched.  It will be seen that the piece D contains fifty-two chequers, and this is the largest piece that it is possible to preserve under the conditions.

[Illustration]

177.—­ANOTHER LINOLEUM PUZZLE.

Cut along the thick lines, and the four pieces will fit together and form a perfect square in the manner shown in the smaller diagram.

[Illustration:  ANOTHER LINOLEUM PUZZLE.]

178.—­THE CARDBOARD BOX.

The areas of the top and side multiplied together and divided by the area of the end give the square of the length.  Similarly, the product of top and end divided by side gives the square of the breadth; and the product of side and end divided by the top gives the square of the depth.  But we only need one of these operations.  Let us take the first.  Thus, 120 x 96 divided by 80 equals 144, the square of 12.  Therefore the length is 12 inches, from which we can, of course, at once get the breadth and depth—­10 in. and 8 in. respectively.

179.—­STEALING THE BELL-ROPES.

Whenever we have one side (a) of a right-angled triangle, and know the difference between the second side and the hypotenuse (which difference we will call b), then the length of the hypotenuse will be

    a squared b
    —–­ + -.
    2b 2

In the case of our puzzle this will be

48 x 48
------- + 11/2 in. = 32 ft. 11/2 in.,
6

which is the length of the rope.

180—­ THE FOUR SONS.