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I was helping my sophomore friend with her geometry homework and came to problem 13a:

Visual

My first instinct was to use similar triangles and say $\triangle BKC \sim \triangle CMD$, and from there we can easily find the sidelengths.

However, in the class they have not learned triangle similarity (nor trigonometry), so they are not allowed to use it to solve problems.

What they have learned are triangle congruency (SSS, SAS, etc.), vertical angles, parallel lines with transversals (alternate interior angles, etc.), and Pythagoras' theorem. How would one solve this problem only using this knowledge? I'm at a loss.

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    $\begingroup$ The easiest thing is good old Pythagoras. $\endgroup$ – user122283 Apr 4 '14 at 1:33
  • $\begingroup$ Can you elaborate? I figured maybe if I could find the length BD or the length AK I could find the others, but couldn't find it without resorting to Law of Cosines $\endgroup$ – MCT Apr 4 '14 at 1:35
  • $\begingroup$ Okay, I think I have something. Not sure if you did the computation, but I get $CM = 5 \sqrt 7$, $CD = 20$? $\endgroup$ – MCT Apr 4 '14 at 1:45
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$\overline{BC}$ and $\overline{DM}$ are a base and altitude for the parallelogram (bases are not always horizontal!), so the product of their lengths gives the figure's area. Likewise with $\overline{DC}$ and $\overline{KC}$. So, $$|\overline{BC}| |\overline{DM}| = |\overline{DC}||\overline{KC}| \quad \to \quad 12 \cdot 15 = |\overline{DC}|\cdot 9 \quad \to\quad |\overline{DC}| = 20$$

Then, because $\triangle CDM$ is a right triangle with right angle at $M$: $$|\overline{DM}|^2 + |\overline{CM}|^2 = |\overline{DC}|^2 \quad \to \quad 15^2 + |\overline{CM}|^2 = 20^2 \quad\to\quad |\overline{CM}| = \sqrt{175} = 5\;\sqrt{7}$$

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Let $CM = x$, $CD = y$, $BD = z$

By Pythagoras' theorem, we can set up the following three equations:

$$x^2 + 15^2 = y^2$$ $$(y + 3 \sqrt 7)^2 + 9^2 = z^2$$ $$(x + 12)^2 + 15^2 = z^2$$

We have three variables and three equations, so we can solve:

Equate the second and third equations to get $(y + 3 \sqrt 7)^2 - (x + 12)^2 = 144$. We have $y^2 - x^2 = 225$, and after some effort we get $x = \frac{\sqrt 7}{4}y$.

We can now plug back into original equations to get $x$ and $y$, coming to $x = 5 \sqrt 7, y = 20$.

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