Question

which triangles unknown side length measures \\( \sqrt { 53 } \\) units?

Explanation:

Step1: Apply Pythagorean theorem

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

First triangle:

Let the unknown side be \(x\). Given \(c = 20\), \(b=\sqrt{73}\). Then \(x^{2}+(\sqrt{73})^{2}=20^{2}\), \(x^{2}+73 = 400\), \(x^{2}=400 - 73=327\), \(x=\sqrt{327}
eq\sqrt{53}\).

Second triangle:

Let the unknown side be \(y\). Given \(a = 6\), \(c=\sqrt{91}\). Then \(y^{2}+6^{2}=(\sqrt{91})^{2}\), \(y^{2}+36 = 91\), \(y^{2}=91 - 36 = 55\), \(y=\sqrt{55}
eq\sqrt{53}\).

Third triangle:

Let the unknown side be \(z\). Given \(a=\sqrt{19}\), \(c=\sqrt{34}\). Then \(z^{2}+(\sqrt{19})^{2}=(\sqrt{34})^{2}\), \(z^{2}+19 = 34\), \(z^{2}=34 - 19=15\), \(z=\sqrt{15}
eq\sqrt{53}\).

Fourth triangle:

Let the unknown side be \(m\). Given \(c = 10\), \(b=\sqrt{47}\). Then \(m^{2}+(\sqrt{47})^{2}=10^{2}\), \(m^{2}+47 = 100\), \(m^{2}=100 - 47 = 53\), \(m=\sqrt{53}\).

Answer:

The fourth triangle.