Difference between revisions of "2024 AMC 8 Problems/Problem 11"
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<cmath>\dfrac{6h}{2}=3h=12.</cmath> | <cmath>\dfrac{6h}{2}=3h=12.</cmath> | ||
This means that <math>h=4, </math> so the answer is <math>7+4=\boxed{(D) 11}</math> | This means that <math>h=4, </math> so the answer is <math>7+4=\boxed{(D) 11}</math> | ||
+ | |||
+ | ==Solution 2== | ||
+ | <asy> | ||
+ | size(10cm); | ||
+ | draw((5,7)--(11,7)--(3,11)--cycle); | ||
+ | draw((3,11)--(3,7)--(5,7),red); | ||
+ | draw((3,7.5)--(3.5,7.5)--(3.5,7)); | ||
+ | label("$A(5,7)$", (5,7),S); | ||
+ | label("$B(11,7)$", (11,7),S); | ||
+ | label("$C(3,y)$", (3,11),W); | ||
+ | label("$D(3,7)$", (3,7),SW); | ||
+ | </asy> | ||
+ | Label point <math>D(3,7)</math> as the point at which <math>CD\perp DA</math>. We now have <math>[\triangle ABC] = [\triangle BCD] - [\triangle ACD]</math>, where the brackets denote areas. On the righthand side, both of these triangles are right, so we can just compute the two sides of each triangle. The two side lengths of <math>\triangle ACD</math> are <math>y-7</math> and <math>5-3=2</math>. The two side lengths of <math>\triangle BCD</math> are <math>y-7</math> and <math>11-3 = 8.</math> Now, | ||
+ | |||
+ | <cmath>[\triangle ABC] = 12 = \frac{1}{2}\cdot (y-7)\cdot 8 - \frac{1}{2}\cdot (y-7)\cdot 2 = 3(y-7)</cmath> | ||
+ | |||
+ | Dividing by <math>3</math> gives <math>y -7 = 4,</math> so <math>y = \boxed{\textbf{(D)\ 11}}.</math> | ||
+ | |||
+ | -Benedict T (countmath1) | ||
==Video Solution by Math-X (First understand the problem!!!)== | ==Video Solution by Math-X (First understand the problem!!!)== |
Revision as of 15:04, 26 January 2024
Contents
Problem
The coordinates of are
,
, and
, with
. The area of
is 12. What is the value of
?
Solution 1
The triangle has base which means its height satisfies
This means that
so the answer is
Solution 2
Label point
as the point at which
. We now have
, where the brackets denote areas. On the righthand side, both of these triangles are right, so we can just compute the two sides of each triangle. The two side lengths of
are
and
. The two side lengths of
are
and
Now,
Dividing by gives
so
-Benedict T (countmath1)
Video Solution by Math-X (First understand the problem!!!)
~Math-X
Video Solution (easy to digest) by Power Solve
https://www.youtube.com/watch?v=2UIVXOB4f0o