I already know how to solve a hypotenuse when both leg and adjacent are given. But my instructor gave as an assignment which is to find the hypotenuse. The problem is this:
One leg of a right triangle is 24 inches ($a$) long and the hypotenuse is $6$ inches shorter than twice the length of the other leg ($b$). Find the length of the hypotenuse ($c$).
So, I have $a=24$, $b=?$, and $c=2b-6$. The formula I am using is Pythagorean Theorem which is $c^2=a^2 + b^2$. so applying the given I came up with $2b-6=24^2 + b^2$ now my problem is how to solve them T.T can someone help me?
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Hagen von Eitzen
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iAthena
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1Isn't there an (...)^2 missing in the last equation? – Hagen von Eitzen Sep 15 '15 at 14:50
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You forgot to square $2b-6$ in your equation. You will have to solve for $b$ using the quadratic equation – CSCFCEM Sep 15 '15 at 14:50
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When reading the title, I was feeling prophetic: http://meta.math.stackexchange.com/questions/21406/what-is-should-be-the-right-way-for-closing-nonsensical-questions?cb=1. – Jack D'Aurizio Sep 15 '15 at 16:15
2 Answers
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Notice, in a right triangle with legs $a$ & $b$ & hypotenuse $c$, we have
$$a^2+b^2=c^2$$
Setting the corresponding values, we get $$(24)^2+b^2=(2b-6)^2$$ $$b^2-8b-180=0$$ Solving the above quadratic equation $$b=\frac{-(-8)\pm\sqrt{(-8)^2-4(1)(-180)}}{2(1)}$$ $$b=\frac{8\pm28}{2}\iff b=18, -10$$
But, $b>0$ Hence, we get $ {\color{red}{b=18}}\iff \color{red}{c}=2\times 18-6=\color{red}{30}$
Harish Chandra Rajpoot
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Alright, notice we can simplify the equation as follows $$(24)^2+b^2=(2b-6)^2$$ $$576+b^2=4b^2+36-24b$$ $$3b^2-24b-540=0$$ Dividing by $3$, we get $$b^2-8b-180=0$$ – Harish Chandra Rajpoot Sep 15 '15 at 16:23
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This is a quadratic equation in $b$, after slight rearrangement of $$ (2b-6)^2=24^2+b^2$$ we obtain a quadratic in standard form: $$ b^2-8b-180=0$$ This happens to have two integer solutions.
Hagen von Eitzen
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