Triangle in a square

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fortenforge

200 posts

fortenforge

#1 h Feb 6, 2013, 5:41 PM • 7 Y

Y by samrocksnature, IdkHowToAddNumbers, icematrix2, megarnie, Adventure10, Mango247, Rounak_iitr

Square $ABCD$ has side length $10$ . Point $E$ is on $\overline{BC}$ , and the area of $\bigtriangleup ABE$ is $40$ . What is $BE$ ?

$\textbf{(A)} \ 4 \qquad \textbf{(B)} \ 5 \qquad \textbf{(C)} \ 6 \qquad \textbf{(D)} \ 7 \qquad \textbf{(E)} \ 8 \qquad$

$[asy] pair A,B,C,D,E; A=(0,0); B=(0,50); C=(50,50); D=(50,0); E = (30,50); draw(A--B); draw(B--E); draw(E--C); draw(C--D); draw(D--A); draw(A--E); dot(A); dot(B); dot(C); dot(D); dot(E); label("A",A,SW); label("B",B,NW); label("C",C,NE); label("D",D,SE); label("E",E,N); [/asy]$

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DiscipulusBonus

241 posts

DiscipulusBonus

#2 h Feb 6, 2013, 5:46 PM • 6 Y

Y by icematrix2, samrocksnature, IdkHowToAddNumbers, Adventure10, Mango247, and 1 other user

This one's pretty easy: we know the area will be 1/2*BE*BA = 40, which is the same as 5BE = 40, which makes BE = 8 (E).

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mcdonalds106_7

1138 posts

mcdonalds106_7

#3 h Feb 6, 2013, 5:58 PM • 4 Y

Y by samrocksnature, IdkHowToAddNumbers, icematrix2, Adventure10

Also AMC10A #3.

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Eagle_Student7

27 posts

Eagle_Student7

#5 h Nov 3, 2020, 4:41 PM • 6 Y

Y by samrocksnature, IdkHowToAddNumbers, tenebrine, judgefan99, icematrix2, Pengu14

should have been AMC10 #1
Solution

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permute_xy16

90 posts

permute_xy16

#6 h Nov 3, 2020, 7:50 PM • 3 Y

Y by samrocksnature, IdkHowToAddNumbers, icematrix2

Eagle_Student7 wrote:

should have been AMC10 #1
Solution

The formula is bh/2 not bh.

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sugar_rush

1341 posts

sugar_rush

#7 h May 14, 2021, 9:10 PM • 2 Y

Y by samrocksnature, IdkHowToAddNumbers

solution

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samrocksnature

8791 posts

samrocksnature

#8 h May 14, 2021, 9:21 PM • 2 Y

Y by IdkHowToAddNumbers, icematrix2

Unique solution

This post has been edited 1 time. Last edited by samrocksnature, May 14, 2021, 9:22 PM

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mathMagicOPS

836 posts

mathMagicOPS

#9 h May 14, 2021, 9:34 PM • 4 Y

Y by samrocksnature, IdkHowToAddNumbers, tigerzhang, icematrix2

samrocksnature wrote:

Unique solution

what did you do to problem 1
aaaaaaaa
why in the world did you bother to use shoelace theorem? ~~DO PICK'S THEOREM!~~

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AOPqghj

546 posts

AOPqghj

#10 h May 14, 2021, 9:37 PM • 4 Y

Y by samrocksnature, IdkHowToAddNumbers, icematrix2, AbhayAttarde01

Troll Solution

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mathMagicOPS

836 posts

mathMagicOPS

#11 h May 14, 2021, 10:22 PM • 3 Y

Y by IdkHowToAddNumbers, samrocksnature, icematrix2

AOPqghj wrote:

Troll Solution

did you.....seriously...do....an...uncomplicated answer....

Jokes aside, did you use complementary area or did you just do a random function that worked by coincidence

This post has been edited 1 time. Last edited by mathMagicOPS, May 14, 2021, 10:23 PM
Reason: edit

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AOPqghj

546 posts

AOPqghj

#12 h May 14, 2021, 10:35 PM • 3 Y

Y by IdkHowToAddNumbers, samrocksnature, icematrix2

I used the trapezoid formula (10-BE) as one of the bases, and 10 was the other. Then the height was obviously 10

This post has been edited 1 time. Last edited by AOPqghj, May 14, 2021, 10:35 PM

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mathMagicOPS

836 posts

mathMagicOPS

#13 h May 14, 2021, 11:05 PM • 3 Y

Y by IdkHowToAddNumbers, samrocksnature, icematrix2

AOPqghj wrote:

I used the trapezoid formula (10-BE) as one of the bases, and 10 was the other. Then the height was obviously 10

cool
also this is too easy for amc12

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judgefan99

2905 posts

judgefan99

#14 h May 14, 2021, 11:10 PM • 3 Y

Y by samrocksnature, icematrix2, Mango247

mathMagicOPS wrote:

AOPqghj wrote:

I used the trapezoid formula (10-BE) as one of the bases, and 10 was the other. Then the height was obviously 10

cool
also this is too easy for amc12

It is problem #1 so I see how they put it in.

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aaja3427

1918 posts

aaja3427

#15 h May 15, 2021, 12:27 AM • 3 Y

Y by samrocksnature, icematrix2, Mango247

permute_xy16 wrote:

Eagle_Student7 wrote:

should have been AMC10 #1
Solution

The formula is bh/2 not bh.

Well that's funny that the person who said it should be #1 got it wrong

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Alex-131

5306 posts

Alex-131

#16 h May 15, 2021, 1:36 AM • 2 Y

Y by samrocksnature, icematrix2

This is too easy

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DaBobWhoLikeMath1234

397 posts

DaBobWhoLikeMath1234

#17 h May 15, 2021, 2:17 AM • 2 Y

Y by samrocksnature, icematrix2

Am I missing out on something with people bumping old threads for AMC problems?

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Ladka13

2467 posts

Ladka13

#18 h May 15, 2021, 3:06 AM • 6 Y

Y by smartguy888, samrocksnature, icematrix2, Mango247, Mango247, Mango247

DaBobWhoLikeMath1234 wrote:

Am I missing out on something with people bumping old threads for AMC problems?

no it's mainly sugar rush and rft trolling no one knows what's really going now

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Spakian

304 posts

Spakian

#19 h May 15, 2021, 2:15 PM • 5 Y

Y by samrocksnature, icematrix2, Mango247, Mango247, Mango247

fortenforge wrote:

Square $ABCD$ has side length $10$ . Point $E$ is on $\overline{BC}$ , and the area of $\bigtriangleup ABE$ is $40$ . What is $BE$ ?

$\textbf{(A)} \ 4 \qquad \textbf{(B)} \ 5 \qquad \textbf{(C)} \ 6 \qquad \textbf{(D)} \ 7 \qquad \textbf{(E)} \ 8 \qquad$

$[asy] pair A,B,C,D,E; A=(0,0); B=(0,50); C=(50,50); D=(50,0); E = (30,50); draw(A--B); draw(B--E); draw(E--C); draw(C--D); draw(D--A); draw(A--E); dot(A); dot(B); dot(C); dot(D); dot(E); label("A",A,SW); label("B",B,NW); label("C",C,NE); label("D",D,SE); label("E",E,N); [/asy]$

Well this one is pretty simple

Solution

This post has been edited 1 time. Last edited by Spakian, May 15, 2021, 2:16 PM
Reason: adding some stuff

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fidgetboss_4000

3472 posts

fidgetboss_4000

#20 h May 15, 2021, 5:11 PM • 5 Y

Y by AOPqghj, samrocksnature, icematrix2, centslordm, megarnie

sugar_rush wrote:

solution

nice necroposting

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Pyramix

419 posts

Pyramix

#21 h Jul 18, 2022, 11:46 AM • 2 Y

Y by Mango247, Mango247

Solution

This post has been edited 1 time. Last edited by Pyramix, Jul 18, 2022, 11:47 AM
Reason: Small typo

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ihatemath123

3440 posts

ihatemath123

#22 h Jul 18, 2022, 12:31 PM

Y by

Imagine if this year's AMCs all start with geometry instead of computation

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MathophileSAR1

14 posts

MathophileSAR1

#23 h Aug 15, 2023, 9:16 AM

Y by

MathophileSAR1 wrote:

Square $ABCD$ has side length $10$ . Point $E$ is on $\overline{BC}$ , and the area of $\bigtriangleup ABE$ is $40$ . What is $BE$ ?

$\textbf{(A)} \ 4 \qquad \textbf{(B)} \ 5 \qquad \textbf{(C)} \ 6 \qquad \textbf{(D)} \ 7 \qquad \textbf{(E)} \ 8 \qquad$

$[asy] pair A,B,C,D,E; A=(0,0); B=(0,50); C=(50,50); D=(50,0); E = (30,50); draw(A--B); draw(B--E); draw(E--C); draw(C--D); draw(D--A); draw(A--E); dot(A); dot(B); dot(C); dot(D); dot(E); label("A",A,SW); label("B",B,NW); label("C",C,NE); label("D",D,SE); label("E",E,N); [/asy]$

Area of square is 1/2 * base * height
1*x*10=80
x=8
BE = 8

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JetFire008

115 posts

JetFire008

#24 h Mar 23, 2025, 1:52 PM

Y by

Area of a triangle is $\frac{1}{2} \cdot bh$ where $b$ is the base of the triangle and $h$ is the height.
In $\triangle ABE$ ,
$40=\frac{1}{2}*BE*10$ $\implies \frac{1}{2}BE=\frac{40}{10}$ $\implies BE=4 \cdot 2=8$ Hence the answer is $\boxed{(E)8}$ .

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