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Difference between revisions of "2018 AMC 10A Problems/Problem 10"

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Substituting into the equation <math>\sqrt{49-x^2}+\sqrt{25-x^2}</math>, we get <math>\sqrt{49-\frac{75}{4}}+\sqrt{25-\frac{75}{4}}</math>. Immediately, we simplify into <math>\sqrt{\frac{121}{4}}+\sqrt{\frac{25}{4}}</math>. The two numbers inside the square roots are simplified to be <math>\frac{11}{2}</math> and <math>\frac{5}{2}</math>, so you add them up: <math>\frac{11}{2}+\frac{5}{2}=\boxed{\textbf{(A) 8}}</math>
 
Substituting into the equation <math>\sqrt{49-x^2}+\sqrt{25-x^2}</math>, we get <math>\sqrt{49-\frac{75}{4}}+\sqrt{25-\frac{75}{4}}</math>. Immediately, we simplify into <math>\sqrt{\frac{121}{4}}+\sqrt{\frac{25}{4}}</math>. The two numbers inside the square roots are simplified to be <math>\frac{11}{2}</math> and <math>\frac{5}{2}</math>, so you add them up: <math>\frac{11}{2}+\frac{5}{2}=\boxed{\textbf{(A) 8}}</math>
  
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===Solution 4 (Geometric Interpretation)===
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 +
Draw a right triangle <math>ABC</math> with a hypotenuse <math>AC</math> of length <math>7</math> and leg <math>AB</math> of length <math>x</math>. Draw <math>D</math> on <math>BC</math> such that <math>AD=5</math>. Note that <math>BC=\sqrt{49-x^2}</math> and <math>BD=\sqrt{25-x^2}</math>. Thus, from the given equation, <math>BC-BD=DC=3</math>. Using Law of Cosines on triangle <math>ADC</math>, we see that <math>\angle{ADC}=120^{\circ}</math> so <math>\angle{ADB}=60^{\circ}</math>. Since <math>ADB</math> is a <math>30-60-90</math> triangle, <math>\sqrt{25-x^2}=BD=\frac{5}{2}</math> and <math>\sqrt{49-x^2}=\frac{5}{2}+3=\frac{11}{2}</math>. Finally, <math>\sqrt{49-x^2}+\sqrt{25-x^2}=\frac{5}{2}+\frac{11}{2}=\boxed{\textbf{(A)~8}}</math> ~anser
  
 
{{AMC10 box|year=2018|ab=A|num-b=9|after=Problem 11}}
 
{{AMC10 box|year=2018|ab=A|num-b=9|after=Problem 11}}

Revision as of 18:22, 11 February 2018

Problem

Suppose that real number $x$ satisfies \[\sqrt{49-x^2}-\sqrt{25-x^2}=3\]What is the value of $\sqrt{49-x^2}+\sqrt{25-x^2}$?

$\textbf{(A) }8\qquad \textbf{(B) }\sqrt{33}+8\qquad \textbf{(C) }9\qquad \textbf{(D) }2\sqrt{10}+4\qquad \textbf{(E) }12\qquad$

Solutions

Solution 1

In order to get rid of the square roots, we multiply by the conjugate. Its value is the solution.The $x^2$ terms cancel nicely. $(\sqrt {49-x^2} + \sqrt {25-x^2}) * (\sqrt {49-x^2} - \sqrt {25-x^2}) = 49-x^2 - 25 +x^2 = 24$

Given that $(\sqrt {49-x^2} - \sqrt {25-x^2})$ = 3, $(\sqrt {49-x^2} + \sqrt {25-x^2}) = \frac {24} {3} = \boxed{\textbf{(A) } 8}$

Solution by PancakeMonster2004, explanations added by a1b2.

Solution 2

Let $u=\sqrt{49-x^2}$, and let $v=\sqrt{25-x^2}$. Then $v=\sqrt{u^2-24}$. Substituting, we get $u-\sqrt{u^2-24}=3$. Rearranging, we get $u-3=\sqrt{u^2-24}$. Squaring both sides and solving, we get $u=\frac{11}{2}$ and $v=\frac{11}{2}-3=\frac{5}{2}$. Adding, we get that the answer is $\boxed{\textbf{(A) } 8}$

Solution 3

Put the equations to one side. $\sqrt{49-x^2}-\sqrt{25-x^2}=3$ can be changed into $\sqrt{49-x^2}=\sqrt{25-x^2}+3$.

We can square both sides, getting us $49-x^2=(25-x^2)+(3^2)+ 2\cdot 3 \cdot \sqrt{25-x^2}.$

That simplifies out to $15=6 \sqrt{25-x^2}.$ Dividing both sides by 6 gets us $\frac{5}{2}=\sqrt{25-x^2}$.

Following that, we can square both sides again, resulting in the equation $\frac{25}{4}=25-x^2$. Simplifying that, we get $x^2 = \frac{75}{4}$.

Substituting into the equation $\sqrt{49-x^2}+\sqrt{25-x^2}$, we get $\sqrt{49-\frac{75}{4}}+\sqrt{25-\frac{75}{4}}$. Immediately, we simplify into $\sqrt{\frac{121}{4}}+\sqrt{\frac{25}{4}}$. The two numbers inside the square roots are simplified to be $\frac{11}{2}$ and $\frac{5}{2}$, so you add them up: $\frac{11}{2}+\frac{5}{2}=\boxed{\textbf{(A) 8}}$

Solution 4 (Geometric Interpretation)

Draw a right triangle $ABC$ with a hypotenuse $AC$ of length $7$ and leg $AB$ of length $x$. Draw $D$ on $BC$ such that $AD=5$. Note that $BC=\sqrt{49-x^2}$ and $BD=\sqrt{25-x^2}$. Thus, from the given equation, $BC-BD=DC=3$. Using Law of Cosines on triangle $ADC$, we see that $\angle{ADC}=120^{\circ}$ so $\angle{ADB}=60^{\circ}$. Since $ADB$ is a $30-60-90$ triangle, $\sqrt{25-x^2}=BD=\frac{5}{2}$ and $\sqrt{49-x^2}=\frac{5}{2}+3=\frac{11}{2}$. Finally, $\sqrt{49-x^2}+\sqrt{25-x^2}=\frac{5}{2}+\frac{11}{2}=\boxed{\textbf{(A)~8}}$ ~anser

2018 AMC 10A (Problems • Answer Key • Resources)
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Problem 9 		Followed by
Problem 11
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All AMC 10 Problems and Solutions
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