A square is a plane shape with equal length of sides, while a right triangle is a triangle that has one of its angles to be [tex]90^{o}[/tex]. Thus, the areas of the smaller squares could be:
A. 12 and 43
A square has equal length of sides, so that its area is given as:
Area of a square = length x length
= [tex]l^{2}[/tex]
For the largest square its area = 55 [tex]units^{2}[/tex], so that:
Area = [tex]l^{2}[/tex]
⇒ 55 = [tex]l^{2}[/tex]
l = [tex]\sqrt{55}[/tex]
Now applying the Pythagoras theorem to the right triangle, we have:
[tex]/Hyp/^{2}[/tex] = [tex]/Adj 1/^{2}[/tex] + [tex]/Adj 2/^{2}[/tex]
where hypotenuse = [tex]\sqrt{55}[/tex]
([tex]\sqrt{55}[/tex][tex])^{2}[/tex] = [tex]/Adj 1/^{2}[/tex] + [tex]/Adj 2/^{2}[/tex]
[tex]/Adj 1/^{2}[/tex] + [tex]/Adj 2/^{2}[/tex] = 55
Therefore, the addition of the areas of the smaller squares should be equal to that of the largest square.
Thus from the theorem above, the areas of the smaller squares could be 12 and 43.
i.e 12 + 43 = 55
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