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Please help me. I think I have it figured out but I just wanted to double check.

Please help me. I think I have it figured out but I just wanted to double check.-example-1

2 Answers

4 votes

Answer:

Let's consider triangle ABCLength AB can be obtained using PythagorasSimilarly, we can consider triangle ACD, so that length AD will be obtained through PythagorasConsidering triangle ABD, with BD being the hypotenuseExpanding the parentheses Divide both sides by 2Option A is correct

Explanation:

User PlasmaHH
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4 votes

Let's consider triangle ABC

Length AB can be obtained using Pythagoras


\begin{gathered} AB^2=x^2+y^2 \\ AB\text{ = }\sqrt[]{x^2+y^2} \\ \end{gathered}

Similarly, we can consider triangle ACD, so that length AD will be obtained through Pythagoras


\begin{gathered} AD^2=x^2+z^2 \\ AD\text{ = }\sqrt[]{x^2+z^2} \end{gathered}

Considering triangle ABD, with BD being the hypotenuse


\begin{gathered} BD^2=AD^2+AB^2 \\ (y+z)^{2\text{ }}=(x^2+z^2)+(x^2+y^2\text{)} \end{gathered}

Expanding the parentheses


\begin{gathered} y^2+2yz+z^2=x^2+z^2+x^2+y^2 \\ \\ y^2-y^2+z^2-z^2+2yz=2x^2 \\ 2yz=2x^2 \\ \end{gathered}

Divide both sides by 2


\begin{gathered} (2yz)/(2)=\text{ }(2x^2)/(2) \\ yz=x^2 \\ \\ x^2\text{ =yz} \\ \\ x\text{ = }\sqrt[]{yz} \end{gathered}

Option A is correct

Please help me. I think I have it figured out but I just wanted to double check.-example-1
User Malovich
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3.8k points