Difference between revisions of "2018 AMC 10A Problems/Problem 10"
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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>. | 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>. | ||
+ | |||
+ | ~kevinmathz | ||
===Solution 4 (Geometric Interpretation)=== | ===Solution 4 (Geometric Interpretation)=== |
Revision as of 17:30, 18 February 2019
Contents
Problem
Suppose that real number satisfies
What is the value of
?
Solutions
Solution 1
In order to get rid of the square roots, we multiply by the conjugate. Its value is the solution.The terms cancel nicely.
Given that . - cookiemonster2004
Solution 2
Let , and let
. Then
. Substituting, we get
. Rearranging, we get
. Squaring both sides and solving, we get
and
. Adding, we get that the answer is
.
Solution 3
Put the equations to one side. can be changed into
.
We can square both sides, getting us
That simplifies out to Dividing both sides by 6 gets us
.
Following that, we can square both sides again, resulting in the equation . Simplifying that, we get
.
Substituting into the equation , we get
. Immediately, we simplify into
. The two numbers inside the square roots are simplified to be
and
, so you add them up:
.
~kevinmathz
Solution 4 (Geometric Interpretation)
Draw a right triangle with a hypotenuse
of length
and leg
of length
. Draw
on
such that
. Note that
and
. Thus, from the given equation,
. Using Law of Cosines on triangle
, we see that
so
. Since
is a
triangle,
and
. Finally,
.
See Also
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 |