Question

name: maraley winboul geometry part b assessment 2

1. find the length of the third side. if necessary, round to the nearest tenth.
2. find the length of the third side. if necessary, round to the nearest tenth.
3. solve for x to the nearest tenth.
4. a 50 - foot ladder is set against the side of a house that it reaches up 48 feet. if mila grabs the ladder at its base and pulls it 6 feet farther from the house, how far up the side of the house will the ladder reach now? (the answer is not 42 ft.) round to the nearest tenth of a foot.
5. which segment is adjacent to ∠j? a. (overline{re}) b. (overline{jr}) c. (overline{je})

Explanation:

Step1: Recall Pythagorean theorem

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

Step2: Solve problem 1

Let the two given sides of the right - triangle be \(a = 14\) and \(b = 18\). Then \(x^{2}=14^{2}+18^{2}=196 + 324=520\), so \(x=\sqrt{520}\approx22.8\).

Step3: Solve problem 2

Let the two given sides of the right - triangle be \(a = 12\) and \(b = 9\). Then \(x^{2}=12^{2}+9^{2}=144 + 81 = 225\), so \(x = 15\).

Step4: Solve problem 3

First, find the length of the hypotenuse of the small right - triangle with sides \(2\) and \(4\): \(h_{1}=\sqrt{2^{2}+4^{2}}=\sqrt{4 + 16}=\sqrt{20}\). Then, for the large right - triangle with one side \(10\) and the hypotenuse \(h_{2}\) and the other side \(x\), using similar triangles or the Pythagorean theorem. The two right - triangles are similar. The ratio of the sides of the similar triangles gives us the relationship. Another way, using the Pythagorean theorem for the large triangle: Let the hypotenuse of the large triangle be \(H=\sqrt{( \sqrt{20})^{2}+10^{2}}=\sqrt{20 + 100}=\sqrt{120}\). If we consider the proportion of similar triangles, we can also calculate. Let's use the Pythagorean theorem for the large triangle with sides \(x\), \(10\) and hypotenuse \(h=\sqrt{(2^{2}+4^{2})+10^{2}}=\sqrt{20 + 100}=\sqrt{120}\). Then \(x=\sqrt{120 - 100}=\sqrt{20}\approx4.5\).

Step5: Solve problem 4

The length of the ladder is the hypotenuse \(c = 50\) feet. Initially, the height on the house is \(a = 48\) feet. Using the Pythagorean theorem, the initial distance from the base of the ladder to the house is \(b_{1}=\sqrt{50^{2}-48^{2}}=\sqrt{(50 + 48)(50 - 48)}=\sqrt{98\times2}=\sqrt{196}=14\) feet. After pulling the base \(6\) feet farther from the house, the new distance from the base of the ladder to the house is \(b_{2}=14 + 6=20\) feet. Let the new height on the house be \(a_{2}\). Then \(a_{2}=\sqrt{50^{2}-20^{2}}=\sqrt{2500 - 400}=\sqrt{2100}\approx45.8\) feet.

Step6: Solve problem 5

In a right - triangle \(\triangle JRE\) with right - angle at \(R\), the side adjacent to \(\angle J\) is \(JR\).

Answer:

1. \(22.8\)
2. \(15\)
3. \(4.5\)
4. \(45.8\)
5. B. \(JR\)