2018 AMC 10A Problems/Problem 10

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 eliminate 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 6 (Symmetric Substitution)

Since $\frac{25+49}{2}=37$ , let $37-x^2 = y$ . Then we have $\sqrt{y+12}-\sqrt{y-12}=3$ . Squaring both sides gives us $2y-2\sqrt{y^2-144}=9$ . Isolating the term with the square root, and squaring again, we get $4y^2-36y+81=4y^2-576 \implies y=\frac{73}{4}$ . Then $\sqrt{y+12}+\sqrt{y-12} = \sqrt{\frac{121}{4}}+\sqrt{\frac{25}{4}} = \frac{16}{2}=\boxed{\textbf{(A)}\ 8}$ .

Solution 7 (Difference of Squares)

Let $\sqrt{49-x^2}=a$ and $\sqrt{25-x^2}=b$ . Then by difference of squares:

$(a+b)(a-b)=a^2-b^2$ .

We can simplify this expression to get our answer. $a^2-b^2=(49-x^2)-(25-x^2)=24$ and from the given statement, $a-b=3$ . Now we have:

$(a+b)(3)=24$ .

Hence, $a+b=\sqrt{49-x^2}-\sqrt{25-x^2}=8$ so our answer is $\boxed{\textbf{(A) } 8}$ .

~BakedPotato66

Solution 8 (Analytic Geometry)

The problem can be represented by the above diagram. The large circle with center $O$ has a radius of 7, the small circle with center $O$ has a radius of 5. Point $C$ 's X coordinate is $x$ . $AC=CD=\sqrt{49-x^2}$ , $BC=\sqrt{25-x^2}$ , $AB=AC-BC=\sqrt{49-x^2} - \sqrt{25-x^2} = 3$ , $BD=CD+BC=\sqrt{49-x^2} + \sqrt{25-x^2}$ .

By Power of a Point, $AB \cdot BD=BE \cdot BF=(7-5) \cdot (7+5)=24$ , $BD=\boxed{\textbf{(A) } 8}$

~isabelchen

Solution 9 (Pythagorean Theorem)

Notice that $\sqrt{49-x^2} = \sqrt{7^2-x^2}$ and $\sqrt{25-x^2} = \sqrt{5^2-x^2}$ This is also the equation of finding a leg of a right triangle given the hypotenuse and the other leg using the Pythagorean Theorem.

Now, $7$ and $5$ are the hypotenuses of the two triangles, and $x$ is the one leg from each of the triangles. So, $\sqrt{7^2-x^2}$ is the other leg of the 1st one, and $\sqrt{5^2-x^2}$ is the other leg of the 2nd one.

For convenience, we name the other leg of the 1st triangle $a$ (the one that's not $5$ or $x$ ), and the other leg of the 2nd one $b$ (the one that's not $7$ or $x$ ). Using the Pythagorean Theorem, we set up 2 equations. $\begin{align*} a^2 + x^2 &= 7^2 \\ b^2 + x^2 &= 5^2 \end{align*}$

Subtracting the two equations and canceling out $x^2$ , we have $a^2 - b^2 = 49-25$ , which simplifies to $(a-b)(a+b)=24$ .

We already know that $a-b$ (or $\sqrt{49-x^2}-\sqrt{25-x^2}$ ) is equal to $3$ , so plugging it in, we have $3(a+b)=24$ , and dividing by $3$ gives $a+b = \boxed{\textbf{(A)}\ 8}$

~MrThinker

Video Solution (HOW TO THINK CREATIVELY!)

https://youtu.be/P-atxiiTw2I

~Education, the Study of Everything

Video Solutions

Video Solution 1

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

~ pi_is_3.14

Video Solution 2

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

Video Solution 3

https://youtu.be/ZiZVIMmo260

Video Solution 4

https://youtu.be/5cA87rbzFdw

~savannahsolver

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

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