# What are we learning ?

[/et_pb_text][et_pb_text _builder_version="4.1" text_font="Raleway||||||||" text_font_size="20px" text_letter_spacing="1px" text_line_height="1.5em" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" hover_enabled="0" box_shadow_style="preset2"]Competency in Focus: Area of triangles This problem from American Mathematics contest (AMC 10A, 2013) is based on calculation of area of triangles . [/et_pb_text][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

# Next understand the problem

[/et_pb_text][et_pb_text _builder_version="4.1" text_font="Raleway||||||||" text_font_size="20px" text_letter_spacing="1px" text_line_height="1.5em" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" hover_enabled="0" box_shadow_style="preset2"]Square $ABCD$ has side length $10$. Point $E$ is on $\overline{BC}$, and the area of $\triangle ABE$ is $40$. What is $BE$?$[asy] pair A,B,C,D,E; A=(0,0); B=(0,50); C=(50,50); D=(50,0); E = (30,50); draw(A--B); draw(B--E); draw(E--C); draw(C--D); draw(D--A); draw(A--E); dot(A); dot(B); dot(C); dot(D); dot(E); label("A",A,SW); label("B",B,NW); label("C",C,NE); label("D",D,SE); label("E",E,N); [/asy]$ $\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$ [/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="4.0"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="4.1" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" hover_enabled="0"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="4.1" hover_enabled="0"]American Mathematical Contest 2013, AMC 10A  Problem 3 [/et_pb_accordion_item][et_pb_accordion_item title="Key Competency" _builder_version="4.1" hover_enabled="0" open="off"]Area of triangles [/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.1" hover_enabled="0" open="off"]4/10 [/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.0.9" open="off"]Challenges and Thrills in Pre College Mathematics Excursion Of Mathematics

[/et_pb_text][et_pb_tabs _builder_version="4.1" hover_enabled="0"][et_pb_tab title="HINT 0" _builder_version="4.0.9"]Do you really need a hint? Try it first![/et_pb_tab][et_pb_tab title="HINT 1" _builder_version="4.1" hover_enabled="0"]Given Square $ABCD$ has side length $10$.  So,  $AB=10$. [/et_pb_tab][et_pb_tab title="HINT 2" _builder_version="4.1" hover_enabled="0"]Now, we know area of a triangle =$\frac{(height).(base)}{2}$. Try to use this here . [/et_pb_tab][et_pb_tab title="HINT 3" _builder_version="4.1" hover_enabled="0"]So , we have the area of $\triangle ABE$ is equal to $\frac{AB(BE)}{2}$Plugging in $AB=10$ , what we get ? [/et_pb_tab][et_pb_tab title="HINT 4" _builder_version="4.1" hover_enabled="0"]we get $80 = 10BE$. Dividing, we find that $BE=\boxed{\textbf{(E) }8}$ [/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" min_height="12px" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]