Question

1. a rectangular picture frame has a diagonal length of 15 inches and a height of 6 inches. what is the width of the frame?
2. jack swam from the dock east 26 meters. he turned and swam another 10 meters, as shown in the diagram. what is the value of x, the distance jack swam to return to the dock?
3. a drone rises 100 feet vertically and then flies due east 65 feet. how many feet is the drone from its starting point?

Explanation:

Step1: Apply Pythagorean theorem

For a right - triangle (a rectangle's diagonal forms a right - triangle with its sides), the Pythagorean theorem is \(a^{2}+b^{2}=c^{2}\), where \(c\) is the hypotenuse (diagonal of the rectangle or the distance Jack swam back to the dock or the drone's distance from the starting point) and \(a\) and \(b\) are the other two sides.

For problem 7:

Let the width be \(w\), height \(h = 6\) inches and diagonal \(d=15\) inches.
By the Pythagorean theorem \(w^{2}+h^{2}=d^{2}\), so \(w^{2}=d^{2}-h^{2}\).
Substitute \(d = 15\) and \(h = 6\):
\(w^{2}=15^{2}-6^{2}=225 - 36=189\)
\(w=\sqrt{189}\approx13.7\) inches

For problem 9:

Let \(a = 26\) meters, \(b = 10\) meters and \(x\) be the hypotenuse.
By the Pythagorean theorem \(x^{2}=26^{2}+10^{2}\)
\(x^{2}=676+100 = 776\)
\(x=\sqrt{776}\approx27.9\) meters

For problem 11:

Let \(a = 100\) feet, \(b = 65\) feet and \(y\) be the hypotenuse (distance of the drone from the starting point).
By the Pythagorean theorem \(y^{2}=100^{2}+65^{2}\)
\(y^{2}=10000 + 4225=14225\)
\(y=\sqrt{14225}\approx119.3\) feet

Answer:

1. The width of the frame is approximately \(13.7\) inches.
2. The value of \(x\) is approximately \(27.9\) meters.
3. The drone is approximately \(119.3\) feet from its starting point.