Question

a triangular pyramid is formed from three right triangles as shown below. use the information given in the figure to find the length su. if applicable, round your answer to the nearest whole number. the lengths on the figure are not drawn accurately.

Explanation:

Step1: Use the Pythagorean theorem in \(\triangle STU\)

The Pythagorean theorem states that for a right - triangle \(a^{2}+b^{2}=c^{2}\), where \(c\) is the hypotenuse. In \(\triangle STU\), if \(SU = x\), \(ST = 105\), and \(TU\) can be found using the Pythagorean theorem in \(\triangle RTU\) first. In \(\triangle RTU\), \(RU = 16\), \(RT=65\), then \(TU=\sqrt{RT^{2}-RU^{2}}\) (by \(a^{2}+b^{2}=c^{2}\Rightarrow b=\sqrt{c^{2}-a^{2}}\)).

$$TU=\sqrt{65^{2}-16^{2}}=\sqrt{(65 + 16)(65 - 16)}=\sqrt{81\times49}=\sqrt{81}\times\sqrt{49}=9\times7 = 63$$

Step2: Use the Pythagorean theorem in \(\triangle STU\) to find \(SU\)

In \(\triangle STU\), \(ST = 105\) (hypotenuse), \(TU = 63\), and let \(SU=x\). By the Pythagorean theorem \(x=\sqrt{ST^{2}-TU^{2}}\)

$$x=\sqrt{105^{2}-63^{2}}=\sqrt{(105 + 63)(105 - 63)}=\sqrt{168\times42}=\sqrt{42\times4\times42}=\sqrt{42^{2}\times4}=42\times2=84$$

Answer:

\(84\)