2003 AMC 10B Problem 20

Below is the professionally curated solution for Problem 20 of the 2003 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2003 AMC 10B solutions, or check the answer key.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Difficulty rating: 1480

20.

In rectangle $ABCD,$ $AB=5$ and $BC=3.$ Points $F$ and $G$ are on $\overline{CD}$ so that $DF=1$ and $GC=2.$ Lines $AF$ and $BG$ intersect at $E.$ Find the area of $\triangle AEB.$

$10$

$\dfrac{21}{2}$

$12$

$\dfrac{25}{2}$

$15$

Solution:

Here $FG = CD - DF - GC = 5 - 1 - 2 = 2.$ Let $h$ be the distance from $E$ down to line $CD.$ Since $\triangle FEG \sim \triangle AEB$ with ratio $\dfrac{FG}{AB}=\dfrac25,$ we have $\dfrac{h}{h+3}=\dfrac25,$ so $h=2.$

The height of $\triangle AEB$ from $E$ to $AB$ is $h+3=5,$ giving area $\dfrac12 \cdot 5 \cdot 5 = \dfrac{25}{2}.$

Thus, the correct answer is D.

← Problem 19 Full Exam Problem 21 →

Problem 20 in Other Years

2000 AMC 10 · 2001 AMC 10 · 2002 AMC 10A · 2002 AMC 10B · 2003 AMC 10A · 2004 AMC 10A · 2004 AMC 10B · 2005 AMC 10A · 2005 AMC 10B · 2006 AMC 10A · 2006 AMC 10B · 2007 AMC 10A · 2007 AMC 10B · 2008 AMC 10A · 2008 AMC 10B · 2009 AMC 10A · 2009 AMC 10B · 2010 AMC 10A · 2010 AMC 10B · 2011 AMC 10A · 2011 AMC 10B · 2012 AMC 10A · 2012 AMC 10B · 2013 AMC 10A · 2013 AMC 10B · 2014 AMC 10A · 2014 AMC 10B · 2015 AMC 10A · 2015 AMC 10B · 2016 AMC 10A · 2016 AMC 10B · 2017 AMC 10A · 2017 AMC 10B · 2018 AMC 10A · 2018 AMC 10B · 2019 AMC 10A · 2019 AMC 10B · 2020 AMC 10A · 2020 AMC 10B · 2021 AMC 10A Spring · 2021 AMC 10B Spring · 2021 AMC 10A Fall · 2021 AMC 10B Fall · 2022 AMC 10A · 2022 AMC 10B · 2023 AMC 10A · 2023 AMC 10B · 2024 AMC 10A · 2024 AMC 10B · 2025 AMC 10A · 2025 AMC 10B