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$AD=3cm$, $DC=4cm$, $BC=5cm$. The diagonals of this quad figure must also be perpendicular. Find $AB$.

Darth Geek
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Pawsome
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    Why did you tag it by [tag:general-relativity]? – Hakim Aug 04 '14 at 10:08
  • Hi and welcome to the site! Since this is a site that encourages learning, you will get much more help if you show us what you have already done. Could you edit your question with your thoughts and ideas? – 5xum Aug 04 '14 at 10:13

2 Answers2

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Let the intersection of two diagonals be $P$. Then we have: $$\begin{multline}AD^2+BC^2=(AP^2+DP^2)+(BP^2+CP^2)\\=(AP^2+BP^2)+(CP^2+DP^2)=AB^2+CD^2,\end{multline}$$ since the diagonals are perpendicular.

Jaehyeon Seo
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Let $O$ be the intersecting point between the two diagonals $AC$ and $BD$.

Then by Pythagoras' Theorem we have the following identities:

$$\begin{cases} AO^2 + DO^2 = AD^2 = 9 & (1)\\ DO^2 + CO^2 = CD^2 = 16 & (2)\\ CO^2 + BO^2 = BC^2 = 25 & (3)\\ BO^2 + AO^2 = AB^2 & (4) \end{cases}$$

Note that $(1) + (3) - (2) = (4)$ thus $AB^2 = 9+25-16 = 18\Rightarrow AB = 3\sqrt{2}$

Darth Geek
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