Hello, lalji!

Here is the first part . . .

Kevin had five rectangular photographs. The ten side lengths of the photographs were:

1 cm, 2 cm, 3 cm, …, 10 cm, (with each occurring once).

Kevin assembled the five photographs with no gaps or overlaps into a square frame of side length 11 cm.

Three of the photographs had size 1 cm x 6 cm, 2 cm x 10 cm and 3 cm x 9cm.

a) Find the side lengths of the other two photographs

and show how the photographs can be put in the frame.

The square frame has an area of: $\displaystyle 11^2 = 121$ cm².

The three given photos have a total area of: $\displaystyle (1\times6) + (2\times10) + (3\times9) \:=\:53$ cm².

This leaves: $\displaystyle 121 - 53 \:=\:68$ cm² to be covered by the other two photos.

The two photos have dimensions $\displaystyle \{4,5,7,8\}$

(a) The only combination which totals 68 cm² is: .$\displaystyle (4\times7) + (5\times8)$

Code:

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