Question

when three squares are joined at their vertices to form a right triangle, the combined area of the two smaller squares is the same as the area of the largest square. which three squares do not support this statement? a 3 cm 25 cm² 4 cm c 9 cm 21 cm 144 cm²

Explanation:

Step1: Recall Pythagorean theorem

The sum of the areas of the two smaller squares (sides \(a\) and \(b\)) should equal the area of the largest square (side \(c\)), i.e., \(a^{2}+b^{2}=c^{2}\).

Step2: Analyze option A

For side lengths \(3\) cm and \(4\) cm, the areas of the two - smaller squares are \(3^{2}=9\) \(cm^{2}\) and \(4^{2}=16\) \(cm^{2}\). The sum is \(9 + 16=25\) \(cm^{2}\), which is the area of the largest square. So option A supports the statement.

Step3: Analyze option C

The area of the first small square with side \(9\) cm is \(9^{2}=81\) \(cm^{2}\), the area of the second small square is \(144\) \(cm^{2}\), and the area of the large square with side \(21\) cm is \(21^{2}=441\) \(cm^{2}\). Calculate \(81 + 144=225
eq441\). So option C does not support the statement.

Answer:

C