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The triangles are drawn to scale. The first triangle has side lengths of 17, 17, 16 while the second triangle has side lengths of 17,17,$k$. The triangles have the same area.

Find the value of $k$ algebraically.

So for the first triangle, I know that the height of the triangle splits the base into two parts of 8 each. sO then using pythagorean theorem, I get the height to be 15 and then the area is $\frac{1}{2} 16 *15 = 120$

For the second triangle, I'm not sure what to do and I don't know what solving for $k$ algebraically means.

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  • $\begingroup$ If we assume the "half" triangles are identical in both cases, only arranged differently, then it's obvious that $k/2=15$ leads to some solution. It's not obvious there are no other solutions though (where our assumption doesn't hold). And this is not "algebraically". Still the simplicity of finding this particular solution deserves a comment. $\endgroup$ – Kamil Maciorowski Nov 6 '18 at 23:02
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HINT

Exactly the same idea. Split in two parts using the height, and in the half-triangle you have the hypotenuse of $17$ and one of the legs is $k/2$.

  1. By the Pythagorean theorem, height $h$ satisfies $h^2 + (k/2)^2 = 17^2$, can you find $h(k)$?
  2. Now the area of the big triangle is $k \cdot h(k) /2$, but you already know this is $120$, can you solve for $k$?

Remark It's obvious one of the answers will be $k=16$ because then the triangles are identical. Are there other values?

Update

You have $$k = \sqrt{4\left(17^2 - h^2\right)} = 2\sqrt{17^2 - h^2},$$ hence the final equation is $$ 120 = k(h) \cdot h /2 = \frac{h}{2} \cdot 2\sqrt{17^2 - h^2} = h \sqrt{17^2 - h^2} $$

To solve this, square both sides to get $$ 120^2 = h^2 \left(17^2 - h^2\right) $$ and let $z = h^2$ to get a quadratic in $z$.

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  • $\begingroup$ well I have that $\frac{k^{2}h}{2} = 120$ so $k^{2}h = 240$ but now I don't know what to do $\endgroup$ – user8290579 Nov 6 '18 at 18:40
  • $\begingroup$ I know that $k^{2} = \frac{240}{h}$ and plugged that into the first equation but that doesn't help at all $\endgroup$ – user8290579 Nov 6 '18 at 18:40
  • $\begingroup$ i don't know how to find h*k $\endgroup$ – user8290579 Nov 6 '18 at 18:41
  • $\begingroup$ @user8290579 see update $\endgroup$ – gt6989b Nov 6 '18 at 19:00
  • $\begingroup$ I'm sorry maybe I'm just not understanding but with your update I get that $k^{2} = 1156-4h^{2}$ I can't take the square root of both sides and get a simple answer because square root doesn't distribute over addition or subtraction right? $\endgroup$ – user130306 Nov 6 '18 at 19:02
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Don’t calculate, cut the triangle along the height instead to get two triangles with sides $8-17-15$. Now glue them together at their common sides of length $8$ to get the wanted triangle; its sides are $17-17-30$.

Alternatively call the angle on top $\phi$. Then the triangle’s area equals $0.5\cdot 17^2\cdot\sin(\phi)$. Firstly, this shows that there’s exactly one other solution, provided that $\phi\neq\pi/2$. This other solution has an angle of $\pi-\phi$ on top. For this solution holds that $0.5k/17=\sin(\pi/2-\phi/2)=\cos(\phi/2)$. As $\sin(\phi/2)=8/17$ we have $(8/17)^2+(0.5k/17)^2=1$, from which $k=30$ follows immediately.

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