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

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Suppose that real number <math>x</math> satisfies <cmath>\sqrt{49-x^2}-\sqrt{25-x^2}=3</cmath>. What is the value of <math>\sqrt{49-x^2}+\sqrt{25-x^2}</math>?
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==Problem==
 +
 
 +
Suppose that real number <math>x</math> satisfies <cmath>\sqrt{49-x^2}-\sqrt{25-x^2}=3</cmath>What is the value of <math>\sqrt{49-x^2}+\sqrt{25-x^2}</math>?
  
 
<math>
 
<math>
\textbf{(A) }8 \qquad
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\textbf{(A) }8\qquad
 
\textbf{(B) }\sqrt{33}+8\qquad
 
\textbf{(B) }\sqrt{33}+8\qquad
\textbf{(C) }9 \qquad
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\textbf{(C) }9\qquad
\textbf{(D) }2\sqrt{10}+4 \qquad
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\textbf{(D) }2\sqrt{10}+4\qquad
\textbf{(E) }12 \qquad
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\textbf{(E) }12\qquad
 
</math>
 
</math>
  
== Solution ==
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==Solutions==
 +
 
 +
===Solution 1===
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In order to get rid of 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>.
 +
 
 +
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>.
 +
 
 +
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>.
  
<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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~kevinmathz
  
Given that <math>(\sqrt {49-x^2} - \sqrt {25-x^2})</math> = 3, <math>(\sqrt {49-x^2} + \sqrt {25-x^2}) = \frac {24} {3} = \boxed{A) 8}</math>
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===Solution 4 (Geometric Interpretation)===
  
== See Also ==
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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 5 (No Square Roots, Fastest)===
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We notice that the two expressions are conjugates, and therefore we can write them in a "difference-of-squares" format.
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Namely, we can write is as <math>((49-x^2) - (25 - x^2)) = (49 - x^2 - 25 + x^2) = 24</math>. Given the <math>3</math> in the problem, we can divide <math>24 / 3 = \boxed{\textbf{(A) } 8}</math>.
 +
 
 +
-aze.10
 +
 
 +
== Video Solution ==
 +
https://youtu.be/ba6w1OhXqOQ?t=1403
 +
 
 +
~ pi_is_3.14
 +
 
 +
== Video Solution ==
 +
https://youtu.be/zQG70XKAdeA
 +
~ North America Math Contest Go Go Go
 +
 
 +
==Video Solution==
 +
https://youtu.be/ZiZVIMmo260
 +
 
 +
https://youtu.be/5cA87rbzFdw
 +
 
 +
~savannahsolver
 +
 
 +
==See Also==
  
 
{{AMC10 box|year=2018|ab=A|num-b=9|num-a=11}}
 
{{AMC10 box|year=2018|ab=A|num-b=9|num-a=11}}
{{MAA Notice}}
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[[Category:Introductory Algebra Problems]]

Revision as of 16:11, 31 January 2021

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 $(\sqrt {49-x^2} - \sqrt {25-x^2}) = 3, (\sqrt {49-x^2} + \sqrt {25-x^2}) = \frac {24} {3} = \boxed{\textbf{(A) } 8}$. - cookiemonster2004

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: $\frac{11}{2}+\frac{5}{2}=\boxed{\textbf{(A) 8}}$.

~kevinmathz

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, $\sqrt{49-x^2}+\sqrt{25-x^2}=\frac{5}{2}+\frac{11}{2}=\boxed{\textbf{(A)~8}}$. [asy] var s = sqrt(3); pair A = (-5*s/2, 0); pair B = (0,0); pair C = (0,5.5); pair D = (0,2.5);  draw(A--B--C--A--D); rightanglemark(A, B, D); label("A", A, SW); label("B", B, SE); label("C", C, NE); label("D", D, E); label("7", (-5*s/4, 5.5/2), NW); label("120$^\circ$", D, NW); label("60$^\circ$", (0,2), SW); label("$x$", 0.5*A, S); draw(rightanglemark(A, B, C));  draw(anglemark(A, D, B)); markscalefactor = 0.04; draw(anglemark(C, D, A));  label("$\frac{5}{2}$", (0,1.25), E); label("3", (0,4), E); label("5", (-5*s/4, 5/4), N); [/asy]

Solution 5 (No Square Roots, Fastest)

We notice that the two expressions are conjugates, and therefore we can write them in a "difference-of-squares" format. Namely, we can write is as $((49-x^2) - (25 - x^2)) = (49 - x^2 - 25 + x^2) = 24$. Given the $3$ in the problem, we can divide $24 / 3 = \boxed{\textbf{(A) } 8}$.

-aze.10

Video Solution

https://youtu.be/ba6w1OhXqOQ?t=1403

~ pi_is_3.14

Video Solution

https://youtu.be/zQG70XKAdeA ~ North America Math Contest Go Go Go

Video Solution

https://youtu.be/ZiZVIMmo260

https://youtu.be/5cA87rbzFdw

~savannahsolver

See Also

2018 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions