Question

the pythagorean theorem and its converse
find x.
1.
2.
3.
4.
5.
6.

Explanation:

Step1: Apply Pythagorean theorem

For a right - triangle with legs \(a\) and \(b\) and hypotenuse \(c\), \(a^{2}+b^{2}=c^{2}\).

Step2: Solve for \(x\) in problem 1

In problem 1, \(a = 13\), \(b = 23\), \(c=x\).

$$x=\sqrt{13^{2}+23^{2}}=\sqrt{169 + 529}=\sqrt{698}\approx26.42$$

Step3: Solve for \(x\) in problem 2

In problem 2, \(a=x\), \(b = 21\), \(c = 34\).

$$x=\sqrt{34^{2}-21^{2}}=\sqrt{(34 + 21)(34 - 21)}=\sqrt{55\times13}=\sqrt{715}\approx26.74$$

Step4: Solve for \(x\) in problem 3

First, find the base of the large right - triangle. Let the base of the large right - triangle be \(y\). Using the Pythagorean theorem for the large right - triangle with hypotenuse \(26\) and one leg \(18\), \(y=\sqrt{26^{2}-18^{2}}=\sqrt{(26 + 18)(26 - 18)}=\sqrt{44\times8}=\sqrt{352}\).
Since the two right - triangles (the large and the small with hypotenuse \(x\)) are similar (by AA similarity, as they share an angle and both have a right angle), or we can use the property that in an isosceles triangle (two sides are \(26\)) and using the Pythagorean theorem for the small right - triangle: \(x=\sqrt{26^{2}-18^{2}}=\sqrt{676-324}=\sqrt{352}\approx18.76\)

Step5: Solve for \(x\) in problem 4

In problem 4, using the Pythagorean theorem for the rectangle (where the diagonal \(x\) is the hypotenuse of a right - triangle with legs \(34\) and \(22\))

$$x=\sqrt{34^{2}+22^{2}}=\sqrt{1156 + 484}=\sqrt{1640}\approx40.5$$

Step6: Solve for \(x\) in problem 5

Using the Pythagorean theorem, if we consider the right - triangle with sides \(16\) and \(14\) and hypotenuse \(x\) (assuming the figure is composed of right - triangles in a way that this formula applies)

$$x=\sqrt{16^{2}-14^{2}}=\sqrt{(16 + 14)(16 - 14)}=\sqrt{30\times2}=\sqrt{60}\approx7.75$$

Step7: Solve for \(x\) in problem 6

Let the base of each of the two congruent right - triangles (formed by the altitude \(x\) in the isosceles triangle) be \(z\). Then \(2z=42\), so \(z = 21\). Using the Pythagorean theorem for the right - triangle with hypotenuse \(24\) and base \(21\)

$$x=\sqrt{24^{2}-21^{2}}=\sqrt{(24 + 21)(24 - 21)}=\sqrt{45\times3}=\sqrt{135}\approx11.62$$

Answer:

1. \(x\approx26.42\)
2. \(x\approx26.74\)
3. \(x\approx18.76\)
4. \(x\approx40.5\)
5. \(x\approx7.75\)
6. \(x\approx11.62\)