by Dudeney, Henry Ernest, 1857-1930
Available in 215 free installments
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176.?LINOLEUM CUTTING.? solution
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.
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177.?ANOTHER LINOLEUM PUZZLE.? solution
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.
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178.?THE CARDBOARD BOX.? solution
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.? solution
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
In the case of our puzzle this will be
which is the length of the rope.
*l2b + b '2
(48 x 48)/ 6 + 1 y 2 i n . = 32 ft. 1 y 2 in.,
180.?THE FOUR SONS.? solution
The diagram shows the most equitable division of the land possible, "so that each son shall receive land of exactly the same area and exactly similar in shape," and so that each shall have access to the well in the centre without trespass on another's land. The conditions do not require that each son's land shall be in one piece, but it is necessary that the two portions assigned to an individual should be kept apart, or two adjoining portions might be held to be one piece, in which case the condition as to shape would have to be broken. At present there is only one shape for each piece of land?half a square divided diagonally. And A, B, C, and D can each reach their land from the outside, and have each equal access to the well in the centre.
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181.?THE THREE RAILWAY STATION S.?solution
The three stations form a triangle, with sides 13,14, and 15 miles. Make the 14 side the base; then the height of the triangle is 12 and the area 84. Multiply the three sides together and divide by four times the area. The result is eight miles and one-eighth, the distance required.
182.?THE GARDEN PUZZLE.
solution
Half the sum of the four sides is 144. From this deduct in turn the four sides, and we get 64, 99, 44, and 81. Multiply these together, and we have as the result the square of 4,752. Therefore the garden contained 4,752
square yards. Of course the tree being equidistant from the four corners shows that the garden is a quadrilateral that may be inscribed in a circle.
183.?DRAWINGASPIRAL? solution
