## Calculator, Q 17 - what it getting at

What is this question going for. I can identify that OTA is a right angled triangle..... Where do I go from
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 Quote by thomas49th What is this question going for. I can identify that OTA is a right angled triangle..... Where do I go from
Pythagoras?
 $$(x+8)(x+8) = x^{2} + (x+5)(x+5)$$ $$x ^ {2} + 16x + 64 = x^4+10x+25$$ take LHS from RHS $$x^{2} - 6x - 39 = 0$$ but how did you know to use pythagerous?

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## Calculator, Q 17 - what it getting at

What relationships DO you know hold for right-angled triangles that might come in handy?

Mentor
 Quote by thomas49th $$(x+8)(x+8) = x^{2} + (x+5)(x+5)$$ $$x ^ {2} + 16x + 64 = x^4+10x+25$$ take LHS from RHS $$x^{2} - 6x - 39 = 0$$
Sorry, one correction to your second line.

$$x ^ {2} + 16x + 64 = 2x^2+10x+25$$

 but how did you know to use pythagerous?
Well, you did the hard part; spotting that it was a right angled triangle. Pythagoras' theorem holds for right angled triangles, and is a relationship relating the squares of the sides. Since the solution contains an x^2, this is quite a big hint as to what you should use.

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