Difference between revisions of "2008 AMC 10A Problems/Problem 20"
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==Problem== | ==Problem== | ||
− | [[Trapezoid]] <math>ABCD</math> has bases <math>\overline{AB}</math> and <math>\overline{CD}</math> and diagonals intersecting at <math>K</math> | + | [[Trapezoid]] <math>ABCD</math> has bases <math>\overline{AB}</math> and <math>\overline{CD}</math> and diagonals intersecting at <math>K.</math> Suppose that <math>AB = 9</math>, <math>DC = 12</math>, and the area of <math>\triangle AKD</math> is <math>24.</math> What is the area of trapezoid <math>ABCD</math>? |
<math>\mathrm{(A)}\ 92\qquad\mathrm{(B)}\ 94\qquad\mathrm{(C)}\ 96\qquad\mathrm{(D)}\ 98 \qquad\mathrm{(E)}\ 100</math> | <math>\mathrm{(A)}\ 92\qquad\mathrm{(B)}\ 94\qquad\mathrm{(C)}\ 96\qquad\mathrm{(D)}\ 98 \qquad\mathrm{(E)}\ 100</math> | ||
==Solution== | ==Solution== | ||
+ | ==Solution 1== | ||
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Therefore, the area of <math>ABCD = [AKD] + [AKB] + [BKC] + [CKD] = 24 + 18 + 24 + 32 = 98\ \mathrm{(D)}</math>. | Therefore, the area of <math>ABCD = [AKD] + [AKB] + [BKC] + [CKD] = 24 + 18 + 24 + 32 = 98\ \mathrm{(D)}</math>. | ||
+ | ==Solution 2== | ||
+ | We denote <math>KA</math> with length <math>x</math> and <math>KD</math> with length <math>\frac{4x}{3}</math> (which follows from similar triangles), and we denote <math>\angle{AKD}=\theta</math>. Note that <math>\frac{4x^2}{3}\cdot \sin\theta=48\implies 4x^2\cdot \sin\theta=36</math>. The areas of triangles <math>ABK</math> and <math>CDK</math> combined are <math>\frac{x^2\cdot\sin\theta+\frac{16x^2}{9}\cdot\sin\theta}{2}=\frac{25x^2}{18}\cdot\sin\theta=36\cdot\frac{25}{18}=50</math>. Thus, <math>[ABCD]=[ABK]+[BCK]+[CDK]+[ADK]=48+50=98\ \mathrm{(D)}</math>, as desired. | ||
+ | -mop | ||
==See also== | ==See also== |
Revision as of 02:53, 27 January 2022
Problem
Trapezoid has bases and and diagonals intersecting at Suppose that , , and the area of is What is the area of trapezoid ?
Solution
Solution 1
Since it follows that . Thus .
We now introduce the concept of area ratios: given two triangles that share the same height, the ratio of the areas is equal to the ratio of their bases. Since share a common altitude to , it follows that (we let denote the area of the triangle) , so . Similarly, we find and .
Therefore, the area of .
Solution 2
We denote with length and with length (which follows from similar triangles), and we denote . Note that . The areas of triangles and combined are . Thus, , as desired. -mop
See also
2008 AMC 10A (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 10 Problems and Solutions |
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