Difference between revisions of "2017 AMC 8 Problems/Problem 18"
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− | ==Problem | + | ==Problem== |
− | In the non-convex quadrilateral <math>ABCD</math> shown below, <math>\angle BCD</math> is a right angle, <math>AB=12</math>, <math>BC=4</math>, <math>CD=3</math>, and <math>AD=13</math>. | + | In the non-convex quadrilateral <math>ABCD</math> shown below, <math>\angle BCD</math> is a right angle, <math>AB=12</math>, <math>BC=4</math>, <math>CD=3</math>, and <math>AD=13</math>. What is the area of quadrilateral <math>ABCD</math>? |
− | < | + | <asy>draw((0,0)--(2.4,3.6)--(0,5)--(12,0)--(0,0)); label("$B$", (0, 0), SW); label("$A$", (12, 0), ESE); label("$C$", (2.4, 3.6), SE); label("$D$", (0, 5), N);</asy> |
− | ==Solution== | + | <math>\textbf{(A) }12 \qquad \textbf{(B) }24 \qquad \textbf{(C) }26 \qquad \textbf{(D) }30 \qquad \textbf{(E) }36</math> |
+ | |||
+ | ==Solution 1== | ||
We first connect point <math>B</math> with point <math>D</math>. | We first connect point <math>B</math> with point <math>D</math>. | ||
− | + | <asy>draw((0,0)--(2.4,3.6)--(0,5)--(12,0)--(0,0)); draw((0,0)--(0,5)); label("$B$", (0, 0), SW); label("$A$", (12, 0), ESE); label("$C$", (2.4, 3.6), SE); label("$D$", (0, 5), N);</asy> | |
+ | |||
+ | We can see that <math>\triangle BCD</math> is a 3-4-5 right triangle. We can also see that <math>\triangle BDA</math> is a right triangle, by the 5-12-13 Pythagorean triple. With these lengths, we can solve the problem. The area of <math>\triangle BDA</math> is <math>\frac{5\cdot 12}{2}</math>, and the area of <math>\triangle BCD</math> is <math>\frac{3\cdot 4}{2}</math>. Thus, the area of quadrilateral <math>ABCD</math> is <math>30-6 = \boxed{\textbf{(B)}\ 24}.</math> | ||
+ | |||
+ | ~CHECKMATE2021 | ||
+ | |||
+ | ==Solution 2== | ||
+ | <math>\triangle BCD</math> is a 3-4-5 right triangle. So the area of <math>\triangle BCD</math> is 6. Then we can use Heron's formula to compute the area of <math>\triangle ABD</math> whose sides have lengths 5,12,and 13. The area of <math>\triangle ABD</math> = <math>\sqrt{s(s-5)(s-12)(s-13)}</math> , where s is the semi-perimeter of the triangle, that is <math>s=(5+12+13)/2=15.</math> Thus, the area of <math>\triangle ABD</math> is 30, so the area of <math>ABCD</math> is <math>30-6 = \boxed{\textbf{(B)}\ 24}.</math> ---LarryFlora | ||
+ | |||
+ | ==Video Solution (CREATIVE THINKING + ANALYSIS!!!)== | ||
+ | https://youtu.be/ZYmCHlMBuIQ | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=DU95-maui9U&ab_channel=WhyMath (http://youtube/DU95-maui9U) | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | ==Video Solution by OmegaLearn== | ||
+ | https://youtu.be/51K3uCzntWs?t=2010 | ||
− | + | ~ pi_is_3.14 | |
==See Also== | ==See Also== |
Latest revision as of 17:37, 21 January 2024
Contents
Problem
In the non-convex quadrilateral shown below,
is a right angle,
,
,
, and
. What is the area of quadrilateral
?
Solution 1
We first connect point with point
.
We can see that is a 3-4-5 right triangle. We can also see that
is a right triangle, by the 5-12-13 Pythagorean triple. With these lengths, we can solve the problem. The area of
is
, and the area of
is
. Thus, the area of quadrilateral
is
~CHECKMATE2021
Solution 2
is a 3-4-5 right triangle. So the area of
is 6. Then we can use Heron's formula to compute the area of
whose sides have lengths 5,12,and 13. The area of
=
, where s is the semi-perimeter of the triangle, that is
Thus, the area of
is 30, so the area of
is
---LarryFlora
Video Solution (CREATIVE THINKING + ANALYSIS!!!)
~Education, the Study of Everything
Video Solution
https://www.youtube.com/watch?v=DU95-maui9U&ab_channel=WhyMath (http://youtube/DU95-maui9U)
~savannahsolver
Video Solution by OmegaLearn
https://youtu.be/51K3uCzntWs?t=2010
~ pi_is_3.14
See Also
2017 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.