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Upper primary school mathematics question UPQ92

The diagram below shows a rectangle PRSU, which is made up of square PQTU and rectangle QRST. Figure D, E and F are squares and the area of square F is 1m^{2}. The area of square D is 2/3 of the shaded area in rectangle QRST and QA = AT as shown in the figure below.

Solution:

The area of square E = 3 × 3 = 9m^{2}

The area of square F = 1m^{2 } **==>** Length of square F = 1m

Length of QA = length of AT

Length QT = 1 + 3 + 4 = 8m

The area of the square D = 4 × 4 = 16m^{2}

Given the area of square D is 2/3 of the shaded area in rectangle QRST

The area of triangle QTS = the area of triangle QRS = 3/2 × 16 = 24m^{2}

The shaded area in Square PQTU = 9 + 16 = 25m^{2}

The area of the square PQTU = 8 × 8 = 64m^{2}

The unshaded area = 64 – 25 + 24 = 63m^{2}