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| | \textbf{(E) }12\qquad | | \textbf{(E) }12\qquad |
| | </math> | | </math> |
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| − | ==Solutions==
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| − | ===Solution 1===
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| − | In order to eliminate the square roots, we multiply by the conjugate. Its value is the solution. The <math>x^2</math> terms cancel nicely. <math>(\sqrt {49-x^2} + \sqrt {25-x^2})(\sqrt {49-x^2} - \sqrt {25-x^2}) = 49-x^2 - 25 +x^2 = 24</math>
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| − | Given that <math>(\sqrt {49-x^2} - \sqrt {25-x^2}) = 3, (\sqrt {49-x^2} + \sqrt {25-x^2}) = \frac {24} {3} = \boxed{\textbf{(A) } 8}</math>. - cookiemonster2004
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| − | ===Solution 2===
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| − | Let <math>u=\sqrt{49-x^2}</math>, and let <math>v=\sqrt{25-x^2}</math>. Then <math>v=\sqrt{u^2-24}</math>. Substituting, we get <math>u-\sqrt{u^2-24}=3</math>. Rearranging, we get <math>u-3=\sqrt{u^2-24}</math>. Squaring both sides and solving, we get <math>u=\frac{11}{2}</math> and <math>v=\frac{11}{2}-3=\frac{5}{2}</math>. Adding, we get that the answer is <math>\boxed{\textbf{(A) } 8}</math>.
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| − | ===Solution 3===
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| − | Put the equations to one side. <math>\sqrt{49-x^2}-\sqrt{25-x^2}=3</math> can be changed into <math>\sqrt{49-x^2}=\sqrt{25-x^2}+3</math>.
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| − | We can square both sides, getting us <math>49-x^2=(25-x^2)+(3^2)+ 2\cdot 3 \cdot \sqrt{25-x^2}.</math>
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| − | That simplifies out to <math>15=6 \sqrt{25-x^2}.</math> Dividing both sides by <math>6</math> gets us <math>\frac{5}{2}=\sqrt{25-x^2}</math>.
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| − | Following that, we can square both sides again, resulting in the equation <math>\frac{25}{4}=25-x^2</math>. Simplifying that, we get <math>x^2 = \frac{75}{4}</math>.
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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>.
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| − | ~kevinmathz
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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>.
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| − | <asy>
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| − | var s = sqrt(3);
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| − | pair A = (-5*s/2, 0);
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| − | pair B = (0,0);
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| − | pair C = (0,5.5);
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| − | pair D = (0,2.5);
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| − | draw(A--B--C--A--D);
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| − | rightanglemark(A, B, D);
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| − | label("A", A, SW);
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| − | label("B", B, SE);
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| − | label("C", C, NE);
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| − | label("D", D, E);
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| − | label("7", (-5*s/4, 5.5/2), NW);
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| − | label("120$^\circ$", D, NW);
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| − | label("60$^\circ$", (0,2), SW);
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| − | label("$x$", 0.5*A, S);
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| − | draw(rightanglemark(A, B, C));
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| − | draw(anglemark(A, D, B));
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| − | markscalefactor = 0.04;
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| − | draw(anglemark(C, D, A));
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| − | label("$\frac{5}{2}$", (0,1.25), E);
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| − | label("3", (0,4), E);
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| − | label("5", (-5*s/4, 5/4), N);
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| − | </asy>
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| − | ===Solution 6 (Symmetric Substitution)===
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| − | Since <math>\frac{25+49}{2}=37</math>, let <math>37-x^2 = y</math>. Then we have <math>\sqrt{y+12}-\sqrt{y-12}=3</math>. Squaring both sides gives us <math>2y-2\sqrt{y^2-144}=9</math>. Isolating the term with the square root, and squaring again, we get <math>4y^2-36y+81=4y^2-576 \implies y=\frac{73}{4}</math>. Then <math>\sqrt{y+12}+\sqrt{y-12} = \sqrt{\frac{121}{4}}+\sqrt{\frac{25}{4}} = \frac{16}{2}=\boxed{\textbf{(A)}\ 8}</math>.
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| − | ===Solution 7 (Difference of Squares)===
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| − | Let <math>\sqrt{49-x^2}=a</math> and <math>\sqrt{25-x^2}=b</math>. Then by difference of squares:
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| − | <math>(a+b)(a-b)=a^2-b^2</math>.
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| − | We can simplify this expression to get our answer. <math>a^2-b^2=(49-x^2)-(25-x^2)=24</math> and from the given statement, <math>a-b=3</math>. Now we have:
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| − | <math>(a+b)(3)=24</math>.
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| − | Hence, <math>a+b=\sqrt{49-x^2}-\sqrt{25-x^2}=8</math> so our answer is <math>\boxed{\textbf{(A) } 8}</math>.
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| − | ~BakedPotato66
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| − | ===Solution 8 (Analytic Geometry)===
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| − | [[File:2018 AMC10 A P10.PNG|500px]]
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| − | The problem can be represented by the above diagram. The large circle with center <math>O</math> has a radius of 7, the small circle with center <math>O</math> has a radius of 5. Point <math>C</math>'s X coordinate is <math>x</math>. <math>AC=CD=\sqrt{49-x^2}</math>, <math>BC=\sqrt{25-x^2}</math>, <math>AB=AC-BC=\sqrt{49-x^2} - \sqrt{25-x^2} = 3</math>, <math>BD=CD+BC=\sqrt{49-x^2} + \sqrt{25-x^2}</math>.
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| − | By Power of a Point, <math>AB \cdot BD=BE \cdot BF=(7-5) \cdot (7+5)=24</math>, <math>BD=\boxed{\textbf{(A) } 8}</math>
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| − | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
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| − | ===Solution 9 (Pythagorean Theorem)===
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| − | Notice that <math>\sqrt{49-x^2} = \sqrt{7^2-x^2}</math> and <math>\sqrt{25-x^2} = \sqrt{5^2-x^2}</math> This is also the equation of finding a leg of a right triangle given the hypotenuse and the other leg using the [[Pythagorean Theorem]].
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| − | Now, <math>7</math> and <math>5</math> are the hypotenuses of the two triangles, and <math>x</math> is the one leg from each of the triangles. So, <math>\sqrt{7^2-x^2}</math> is the other leg of the 1st one, and <math>\sqrt{5^2-x^2}</math> is the other leg of the 2nd one.
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| − | For convenience, we name the other leg of the 1st triangle <math>a</math> (the one that's not <math>5</math> or <math>x</math>), and the other leg of the 2nd one <math>b</math> (the one that's not <math>7</math> or <math>x</math>). Using the Pythagorean Theorem, we set up 2 equations.
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| − | <cmath>\begin{align*}
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| − | a^2 + x^2 &= 7^2 \\
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| − | b^2 + x^2 &= 5^2
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| − | \end{align*}</cmath>
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| − | Subtracting the two equations and canceling out <math>x^2</math>, we have <math>a^2 - b^2 = 49-25</math>, which simplifies to <math>(a-b)(a+b)=24</math>.
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| − | We already know that <math>a-b</math> (or <math>\sqrt{49-x^2}-\sqrt{25-x^2}</math>) is equal to <math>3</math>, so plugging it in, we have <math>3(a+b)=24</math>, and dividing by <math>3</math> gives <math>a+b = \boxed{\textbf{(A)}\ 8}</math>
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| − | ~MrThinker
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| | ==Solution 10 (Solution 1 but alternate)== | | ==Solution 10 (Solution 1 but alternate)== |
satisfies ![\[\sqrt{49-x^2}-\sqrt{25-x^2}=3\]](http://latex.artofproblemsolving.com/6/0/2/60260471de484d4c7c3af264a757d4ff5de2ad3d.png)
What is the value of 
?
; in other words, we want to find 
. We know that 
Thus, 
