Difference between revisions of "2015 AMC 12A Problems/Problem 20"
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The area of <math>T</math> is <math>\dfrac{1}{2} \cdot 8 \cdot 3 = 12</math> and the perimeter is 18. | The area of <math>T</math> is <math>\dfrac{1}{2} \cdot 8 \cdot 3 = 12</math> and the perimeter is 18. | ||
− | The area of <math>T</math> is <math>\dfrac{1}{2} b \sqrt{a^2 - (\dfrac{b}{2})^2}</math> and the perimeter is <math>2a + b</math>. | + | The area of <math>T'</math> is <math>\dfrac{1}{2} b \sqrt{a^2 - (\dfrac{b}{2})^2}</math> and the perimeter is <math>2a + b</math>. |
− | Thus <math>2a + b = 18</math>, so <math>2a = 18 - | + | Thus <math>2a + b = 18</math>, so <math>2a = 18 - b</math>. |
Thus <math>12 = \dfrac{1}{2} b \sqrt{a^2 - (\dfrac{b}{2})^2}</math>, so <math>48 = b \sqrt{4a^2 - b^2} = b \sqrt{(18 - b)^2 - b^2} = b \sqrt{324 - 36b}</math>. | Thus <math>12 = \dfrac{1}{2} b \sqrt{a^2 - (\dfrac{b}{2})^2}</math>, so <math>48 = b \sqrt{4a^2 - b^2} = b \sqrt{(18 - b)^2 - b^2} = b \sqrt{324 - 36b}</math>. |
Revision as of 02:34, 5 February 2015
Problem
Isosceles triangles and are not congruent but have the same area and the same perimeter. The sides of have lengths , , and , while those of have lengths , , and . Which of the following numbers is closest to ?
Solution
Solution 1
The area of is and the perimeter is 18.
The area of is and the perimeter is .
Thus , so .
Thus , so .
We square and divide 36 from both sides to obtain , so . This factors as . Because clearly but , we have The answer is .
Solution 2
Triangle , being isosceles, has an area of and a perimeter of . Triangle similarly has an area of and .
Now we apply our computational fortitude.
Plug in to obtain Plug in to obtain We know that is a valid solution by . Factoring out , we obtain Utilizing the quadratic formula gives We clearly must pick the positive solution. Note that , and so , which clearly gives an answer of , as desired.
See Also
2015 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
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 12 Problems and Solutions |