2018 AMC 10A Problems/Problem 10


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

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


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


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:


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:


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


Solution 8 (Analytic Geometry)

2018 AMC10 A P10.PNG

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


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 $a+b = \boxed{\textbf{(A)}\ 8}$


Solution 10 (Solution 1 but alternate)

We let $a=\sqrt{49-x^2}+\sqrt{25-x^2}$; in other words, we want to find $a$. We know that $a\cdot3=\left(\sqrt{49-x^2}+\sqrt{25-x^2}\right)\cdot\left(\sqrt{49-x^2}-\sqrt{25-x^2}\right)=\left(\sqrt{49-x^2}\right)^2-\left(\sqrt{25-x^2}\right)^2=\left(49-x^2\right)-\left(25-x^2\right)=24.$ Thus, $a=8.$




~Education, the Study of Everything

Video Solutions

Video Solution 1


~ pi_is_3.14

Video Solution 2

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

Video Solution 3


Video Solution 4



See Also

2018 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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All AMC 10 Problems and Solutions