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# geometry

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Let triangle PQR be a triangle in the plane, and let S be a point outside the plane of PQR, so that SPQR is a pyramid whose faces are all triangles. Suppose that every edge of SPQR has length 18 or 41, but no face of SPQR is equilateral. Then what is the surface area of SPQR?

Guest Jun 9, 2018

#1
+2248
+2

I think this is one of those times where simplifying the problem may make it easier to understand. Mathematicians utilize this strategy all the time, and it amazes me how many times it works. Instead of considering a three-dimensional figure, let's solely focus on the base of the tetrahedron, figure PQR.

A few rules govern the existence of this triangle. Let's consider the rules one at a time. If the sidelengths of $$\triangle PQR$$ can only consist of lengths of 18 or 41, then this narrows the possibilities to 4 options. The images below are not drawn to scale. The central label indicates how I will differentiate and refer to the triangles:

The next rule is that no face is equilateral; in other words, the triangle cannot have all sides with equal length. Since both triangle A and triangle B are equilateral in nature, these two cannot be the base of the tetrahedron.

There are no more rules given in the problem, yet there are still two triangles left as possibilities. Can we eliminate one? Yes, absolutely! Triangle C cannot be a triangle because it cannot exist. By the Triangle Inequality Theorem, the sum of the lengths of any two sides of a triangle must be greater than the length of the remaining side. The triangle fails this theorem. $$18+18\ngtr 41$$. If a triangle fails the theorem, then the figure cannot be a triangle.

This leaves us with one and only one possibility: the 18-41-41 triangle. Since all the faces of this figure, known as a tetrahedron, is a triangle with the rules aforementioned, then all other faces must be an 18-41-41 triangle, too. A tetrahedron has 4 faces, so I can find the area of one base and multiply it by 4. The area of the base can be obtained with heron's formula. Doing this will get the surface area of the three-dimensional figure.

 $$SA_{\text{tetrahedron}}=4\sqrt{s(s-a)(s-b)(s-c)}$$ In heron's formula, a, b, and c are side lengths of any triangle. s is the semiperimeter. Let's find the semiperimeter first. $$s=\frac{18+41+41}{2}=50\\ a=18\\ b=41\\ c=41$$ Now, substitute into heron's formula and solve. $$SA_{\text{tetrahedron}}=4\sqrt{50(50-18)(50-41)(50-41)}$$ Simplify the expressions within parentheses. $$SA_{\text{tetrahedron}}=4\sqrt{\textcolor{red}{50}*\textcolor{blue}{32}*\textcolor{green}{9*9}}$$ Let's try to extract all the perfect squares from this. $$SA_{\text{tetrahedron}}=4\sqrt{\textcolor{red}{25*2}*\textcolor{blue}{16*2}*\textcolor{green}{9^2}}$$ We can combine the red 2 with the blue 2 to make another perfect square. $$SA_{\text{tetrahedron}}=4\sqrt{5^2*4^2*9^2*2^2}$$ Let's take out all the perfect squares now! $$SA_{\text{tetrahedron}}=4*\textcolor{red}{5*4}*\textcolor{blue}{9*2}$$ Now, multiply in any order you choose. I chose to multiply this way because it felt easiest. $$SA_{\text{tetrahedron}}=\textcolor{red}{4}*20*\textcolor{red}{18}$$ $$SA_{\text{tetrahedron}}=20*72$$ $$SA_{\text{tetrahedron}}=1440\text{units}^2$$ Do not forget about those units!
TheXSquaredFactor  Jun 9, 2018
#1
+2248
+2

I think this is one of those times where simplifying the problem may make it easier to understand. Mathematicians utilize this strategy all the time, and it amazes me how many times it works. Instead of considering a three-dimensional figure, let's solely focus on the base of the tetrahedron, figure PQR.

A few rules govern the existence of this triangle. Let's consider the rules one at a time. If the sidelengths of $$\triangle PQR$$ can only consist of lengths of 18 or 41, then this narrows the possibilities to 4 options. The images below are not drawn to scale. The central label indicates how I will differentiate and refer to the triangles:

The next rule is that no face is equilateral; in other words, the triangle cannot have all sides with equal length. Since both triangle A and triangle B are equilateral in nature, these two cannot be the base of the tetrahedron.

There are no more rules given in the problem, yet there are still two triangles left as possibilities. Can we eliminate one? Yes, absolutely! Triangle C cannot be a triangle because it cannot exist. By the Triangle Inequality Theorem, the sum of the lengths of any two sides of a triangle must be greater than the length of the remaining side. The triangle fails this theorem. $$18+18\ngtr 41$$. If a triangle fails the theorem, then the figure cannot be a triangle.

This leaves us with one and only one possibility: the 18-41-41 triangle. Since all the faces of this figure, known as a tetrahedron, is a triangle with the rules aforementioned, then all other faces must be an 18-41-41 triangle, too. A tetrahedron has 4 faces, so I can find the area of one base and multiply it by 4. The area of the base can be obtained with heron's formula. Doing this will get the surface area of the three-dimensional figure.

 $$SA_{\text{tetrahedron}}=4\sqrt{s(s-a)(s-b)(s-c)}$$ In heron's formula, a, b, and c are side lengths of any triangle. s is the semiperimeter. Let's find the semiperimeter first. $$s=\frac{18+41+41}{2}=50\\ a=18\\ b=41\\ c=41$$ Now, substitute into heron's formula and solve. $$SA_{\text{tetrahedron}}=4\sqrt{50(50-18)(50-41)(50-41)}$$ Simplify the expressions within parentheses. $$SA_{\text{tetrahedron}}=4\sqrt{\textcolor{red}{50}*\textcolor{blue}{32}*\textcolor{green}{9*9}}$$ Let's try to extract all the perfect squares from this. $$SA_{\text{tetrahedron}}=4\sqrt{\textcolor{red}{25*2}*\textcolor{blue}{16*2}*\textcolor{green}{9^2}}$$ We can combine the red 2 with the blue 2 to make another perfect square. $$SA_{\text{tetrahedron}}=4\sqrt{5^2*4^2*9^2*2^2}$$ Let's take out all the perfect squares now! $$SA_{\text{tetrahedron}}=4*\textcolor{red}{5*4}*\textcolor{blue}{9*2}$$ Now, multiply in any order you choose. I chose to multiply this way because it felt easiest. $$SA_{\text{tetrahedron}}=\textcolor{red}{4}*20*\textcolor{red}{18}$$ $$SA_{\text{tetrahedron}}=20*72$$ $$SA_{\text{tetrahedron}}=1440\text{units}^2$$ Do not forget about those units!
TheXSquaredFactor  Jun 9, 2018