 # What are we learning ?

Competency in Focus: Area of triangles This problem from American Mathematics contest (AMC 10A, 2013) is based on calculation of area of triangles .

# First look at the knowledge graph. # Next understand the problem

Square $ABCD$ has side length $10$. Point $E$ is on $\overline{BC}$, and the area of $\triangle ABE$ is $40$. What is $BE$? $[asy] pair A,B,C,D,E; A=(0,0); B=(0,50); C=(50,50); D=(50,0); E = (30,50); draw(A--B); draw(B--E); draw(E--C); draw(C--D); draw(D--A); draw(A--E); dot(A); dot(B); dot(C); dot(D); dot(E); label("A",A,SW); label("B",B,NW); label("C",C,NE); label("D",D,SE); label("E",E,N); [/asy]$ $\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$
##### Source of the problem
American Mathematical Contest 2013, AMC 10A  Problem 3
##### Key Competency
Area of triangles
4/10
##### Suggested Book
Challenges and Thrills in Pre College Mathematics Excursion Of Mathematics

Do you really need a hint? Try it first!
Given Square $ABCD$ has side length $10$.  So, $AB=10$.
Now, we know area of a triangle =$\frac{(height).(base)}{2}$. Try to use this here .
So , we have the area of $\triangle ABE$ is equal to $\frac{AB(BE)}{2}$Plugging in $AB=10$ , what we get ?
we get $80 = 10BE$. Dividing, we find that $BE=\boxed{\textbf{(E) }8}$

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