Difference between revisions of "2012 AMC 8 Problems/Problem 25"
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<math> \textbf{(A)}\hspace{.05in}\frac{1}5\qquad\textbf{(B)}\hspace{.05in}\frac{2}5\qquad\textbf{(C)}\hspace{.05in}\frac{1}2\qquad\textbf{(D)}\hspace{.05in}1\qquad\textbf{(E)}\hspace{.05in}4 </math> | <math> \textbf{(A)}\hspace{.05in}\frac{1}5\qquad\textbf{(B)}\hspace{.05in}\frac{2}5\qquad\textbf{(C)}\hspace{.05in}\frac{1}2\qquad\textbf{(D)}\hspace{.05in}1\qquad\textbf{(E)}\hspace{.05in}4 </math> | ||
+ | |||
+ | ==Solution 1== | ||
+ | The total area of the four congruent triangles formed by the squares is <math> 1 </math>. Therefore, the area of one of these triangles is <math> \frac{1}{4} </math>. The height of one of these triangles is <math> a </math> and the base is <math> b </math>. Using the formula for area of the triangle, we have <math> \frac{ab}{2} = \frac{1}{4} </math>. Multiply by <math> 2 </math> on both sides to find that the value of <math> ab </math> is <math> \boxed{\textbf{(C)}\ \frac{1}2} </math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | To solve this problem you could also use algebraic manipulation. | ||
+ | |||
+ | Since the area of the large square is <math> 5 </math>, the sidelength is <math> \sqrt{5} </math>. | ||
+ | |||
+ | We then have the equation <math> a + b = \sqrt{5} </math>. | ||
+ | |||
+ | We also know that the sidelength of the smaller square is <math> 2 </math>, since its area is <math> 4 </math>. Then, the segment of length <math> a </math> and segment of length <math> b </math> form a right triangle whose hypotenuse would have length <math> 2 </math>. | ||
+ | |||
+ | So our second equation is <math> \sqrt{{a^2}+{b^2}} = 2 </math>. | ||
+ | |||
+ | Square both equations. | ||
+ | |||
+ | <math> a^2 + 2ab + b^2 = 5 </math> | ||
+ | |||
+ | <math> a^2 + b^2 = 4 </math> | ||
+ | |||
+ | Now, subtract, and obtain the equation <math> 2ab = 1 </math>. We can deduce that the value of <math> ab </math> is <math> \boxed{\textbf{(C)}\ \frac{1}2} </math>. | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2012|num-b=24|after=Last Problem}} | {{AMC8 box|year=2012|num-b=24|after=Last Problem}} |
Revision as of 12:38, 24 November 2012
A square with area 4 is inscribed in a square with area 5, with one vertex of the smaller square on each side of the larger square. A vertex of the smaller square divides a side of the larger square into two segments, one of length , and the other of length . What is the value of ?
Solution 1
The total area of the four congruent triangles formed by the squares is . Therefore, the area of one of these triangles is . The height of one of these triangles is and the base is . Using the formula for area of the triangle, we have . Multiply by on both sides to find that the value of is .
Solution 2
To solve this problem you could also use algebraic manipulation.
Since the area of the large square is , the sidelength is .
We then have the equation .
We also know that the sidelength of the smaller square is , since its area is . Then, the segment of length and segment of length form a right triangle whose hypotenuse would have length .
So our second equation is .
Square both equations.
Now, subtract, and obtain the equation . We can deduce that the value of is .
See Also
2012 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
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 AJHSME/AMC 8 Problems and Solutions |