Difference between revisions of "2023 AMC 8 Problems/Problem 24"

(Video Explanation (along with thought process) Using Lots of Similarity)
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*Add asymptote diagram*
 
*Add asymptote diagram*
(note: diagrams are not necessarily draw to scale)
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(note: diagrams are not necessarily drawn to scale)
  
<cmath>\textbf{(A) } 14.6 \qquad \textbf{(B) } 14.8 \qquad \textbf{(C) } 15 \qquad \textbf{(D) } 15.2 \qquad \textbf{(E) } ~15.4</cmath>
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<math>\textbf{(A) } 14.6 \qquad \textbf{(B) } 14.8 \qquad \textbf{(C) } 15 \qquad \textbf{(D) } 15.2 \qquad \textbf{(E) } 15.4</math>
  
 
==Video Solution 1 by OmegaLearn (Using Similarity)==
 
==Video Solution 1 by OmegaLearn (Using Similarity)==
 
https://youtu.be/almtw4n-92A
 
https://youtu.be/almtw4n-92A
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==See Also==
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{{AMC8 box|year=2023|num-b=23|num-a=25}}
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{{MAA Notice}}

Revision as of 21:45, 24 January 2023

Problem

Isosceles $\triangle ABC$ has equal side lengths $AB$ and $BC$. In the figure below, segments are drawn parallel to $\overline{AC}$ so that the shaded portions of $\triangle ABC$ have the same area. The heights of the two unshaded portions are 11 and 5 units, respectively. What is the height of $h$ of $\triangle ABC$?

  • Add asymptote diagram*

(note: diagrams are not necessarily drawn to scale)

$\textbf{(A) } 14.6 \qquad \textbf{(B) } 14.8 \qquad \textbf{(C) } 15 \qquad \textbf{(D) } 15.2 \qquad \textbf{(E) } 15.4$

Video Solution 1 by OmegaLearn (Using Similarity)

https://youtu.be/almtw4n-92A


See Also

2023 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
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

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