Question

special right triangles version d - criterion b: pattern investigation. using pythagoreans theorem and exact answers, determine the value of x and sometimes y in each tri
1)
2)
3)
4)

Explanation:

Step1: Recall Pythagorean theorem

For a right - triangle with legs \(a\) and \(b\) and hypotenuse \(c\), \(a^{2}+b^{2}=c^{2}\). For isosceles right - triangles (with angles \(45^{\circ}-45^{\circ}-90^{\circ}\)), the legs are of equal length, say \(a = b\), and the hypotenuse \(c=a\sqrt{2}\).

Step2: Solve for \(x\) in the third triangle

In the third triangle, it is a \(45^{\circ}-45^{\circ}-90^{\circ}\) triangle. The hypotenuse is \(5\sqrt{2}\) and the legs are of length \(x\). Using the relationship \(c = a\sqrt{2}\) (where \(c\) is the hypotenuse and \(a\) is the length of a leg), if \(c = 5\sqrt{2}\) and \(c=a\sqrt{2}\), then \(x = 5\).

Step3: Solve for \(x\) in the fourth triangle

In the fourth triangle, it is a \(45^{\circ}-45^{\circ}-90^{\circ}\) triangle. The legs are of length \(11\). Using the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\), where \(a = 11\) and \(b = 11\), we have \(x^{2}=11^{2}+11^{2}=121 + 121=242\), so \(x=\sqrt{242}=11\sqrt{2}\).

Answer:

For the third triangle, \(x = 5\).
For the fourth triangle, \(x = 11\sqrt{2}\).