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

Revision as of 20:49, 29 April 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, $(\sqrt {49-x^2} + \sqrt {25-x^2}) = \frac {24} {3} = \boxed{\textbf{(A) } 8}$ . - 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: $\frac{11}{2}+\frac{5}{2}=\boxed{\textbf{(A) 8}}$ .

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}$", (0,1.25), E); label("3", (0,4), E); label("5", (-5*s/4, 5/4), N); [/asy]$

2018 AMC 10A (Problems • Answer Key • Resources)
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

@@ Line 25: / Line 25: @@
 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>.
-e can square both sides, getting us <math>49-x^2=(25-x^2)+(3^2)+ 2\cdot 3 \cdot \sqrt{25-x^2}.</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>
 That simplifies out to <math>15=6 \sqrt{25-x^2}.</math> Dividing both sides by 6  gets us <math>\frac{5}{2}=\sqrt{25-x^2}</math>.

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