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

(Problem: Someone asked for an asymptote diagram??)
(Undo revision 187800 by Themathguyd (talk))
(Tag: Undo)
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Isosceles <math>\triangle ABC</math> has equal side lengths <math>AB</math> and <math>BC</math>. In the figure below, segments are drawn parallel to <math>\overline{AC}</math> so that the shaded portions of <math>\triangle ABC</math> have the same area. The heights of the two unshaded portions are 11 and 5 units, respectively. What is the height of <math>h</math> of <math>\triangle ABC</math>?
 
Isosceles <math>\triangle ABC</math> has equal side lengths <math>AB</math> and <math>BC</math>. In the figure below, segments are drawn parallel to <math>\overline{AC}</math> so that the shaded portions of <math>\triangle ABC</math> have the same area. The heights of the two unshaded portions are 11 and 5 units, respectively. What is the height of <math>h</math> of <math>\triangle ABC</math>?
  
<asy>
+
*Add asymptote diagram*
// Diagram by TheMathGuyd
 
size(9cm);
 
real h = 2.5; // height
 
real g=4; //c2c space
 
real s = 0.65; //Xcord of Hline
 
 
 
pair A,B,C;
 
B=(0,h);
 
C=(1,0);
 
A=-conj(C);
 
pair P=(s,h*(1-s)); //Endpoint of Hline
 
path L=P--(-conj(P)); //Hline
 
path T=A--B--C--cycle; //Triangle
 
 
 
 
 
draw(L);
 
draw(T);
 
label("$A$",A,SW);
 
label("$B$",B,N);
 
label("$C$",C,SE);
 
 
 
draw(shift(g,0)*L);
 
draw(shift(g,0)*T);
 
label("$A$",shift(g,0)*A,SW);
 
label("$B$",shift(g,0)*B,N);
 
label("$C$",shift(g,0)*C,SE);
 
 
 
draw(B--shift(g,0)*B,dashed);
 
draw(C--shift(g,0)*A,dashed);
 
draw((g/2,0)--(g/2,h),dashed);
 
draw((0,h*(1-s))--B,dashed);
 
draw((g,h*(1-s))--(g,0),dashed);
 
label("$5$", midpoint((g,h*(1-s))--(g,0)),UnFill);
 
label("$h$", midpoint((g/2,0)--(g/2,h)),UnFill);
 
label("$11$", midpoint((0,h*(1-s))--B),UnFill);
 
</asy>
 
 
 
 
(note: diagrams are not necessarily drawn to scale)
 
(note: diagrams are not necessarily drawn to scale)
  

Revision as of 03:17, 25 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)

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

Solution 1

First, we notice that the smaller isosceles triangles are similar to the larger isosceles triangles. We can find that the area of the gray area in the first triangle is $[\text{ABC}]\cdot\left(1-\left(\tfrac{11}{h}\right)^2\right)$. Similarly, we can find that the area of the gray part in the second triangle is $[\text{ABC}]\cdot\left(\tfrac{h-5}{h}\right)^2$. These areas are equal, so $1-\left(\frac{11}{h}\right)^2=\left(\frac{h-5}{h}\right)^2$. Simplifying yields $10h=146$ so $h=\boxed{\textbf{(A) }14.6}$.

~MathFun1000 (~edits apex304)

Video Solution 1 by OmegaLearn (Using Similarity)

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

Video Solution 2 by SpreadTheMathLove(Using Area-Similarity Relaitionship)

https://www.youtube.com/watch?v=GTlkTwxSxgo

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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