Question

1. a baby bird left the nest that was 14 feet above the base of a tree. if the bird flew diagonally 21 feet to the ground, how far from the base of the tree did it land? round answer to the nearest tenth.

a. 25.2 feet
b. 15.7 feet
c. 9.7 feet
d. 5.9 feet

1. find the length of the diagonal line that runs through the box from the top left corner to the bottom right corner. round answer to the nearest tenth.

a. 4.9 cm
b. 2.6 cm
c. 5.7 cm
d. 6.4 cm

Explanation:

Step1: Use Pythagorean theorem for problem 4

Let the height of the tree be \(a = 14\) feet (one - leg of the right - triangle), the diagonal distance the bird flew be \(c=21\) feet (hypotenuse of the right - triangle). Let the distance from the base of the tree be \(b\). According to the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\), so \(b=\sqrt{c^{2}-a^{2}}\). Substitute \(a = 14\) and \(c = 21\) into the formula: \(b=\sqrt{21^{2}-14^{2}}=\sqrt{(21 + 14)(21 - 14)}=\sqrt{35\times7}=\sqrt{245}\approx15.7\) feet.

Step2: Use three - dimensional Pythagorean theorem for problem 5

For a rectangular box with length \(l = 4\) cm, width \(w = 3\) cm, and height \(h = 4\) cm. The formula for the space diagonal \(d\) of a rectangular box is \(d=\sqrt{l^{2}+w^{2}+h^{2}}\). Substitute \(l = 4\), \(w = 3\), and \(h = 4\) into the formula: \(d=\sqrt{4^{2}+3^{2}+4^{2}}=\sqrt{16 + 9+16}=\sqrt{41}\approx6.4\) cm.

Answer:

1. B. 15.7 feet
2. D. 6.4 cm