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The Task
Consider the rectangle
ABCD. Point
E is the midpoint of
bar(DC) and point
F is the midpoint of
bar(BC). If points
A,E, and
F are joined to form a triangle, what is the ratio of the area of triangle
AEF to the area of triangle
ABF.
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11:26 AM

4//27//2024

Consider the rectangle $ABCD$ . Point $E$ is the midpoint of $\overline{DC}$ and point $F$ is the midpoint of $\overline{BC}$ . If points $A$ , $E$ , and $F$ are joined to form a triangle, what is the ratio of the area of triangle $AEF$ to the area of triangle $ABF$ .

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Area of quadrilaterals and triangles: word problems

Full solution

Q. Consider the rectangle

ABCD

. Point

E

is the midpoint of

\overline{DC}

and point

F

is the midpoint of

\overline{BC}

. If points

A

,

E

, and

F

are joined to form a triangle, what is the ratio of the area of triangle

AEF

to the area of triangle

ABF

.

Identify Properties: Identify the properties of the points $E$ and $F$ on rectangle $ABCD$ . Since $E$ and $F$ are midpoints of $DC$ and $BC$ respectively, $DE = EC$ and $BF = FC$ .
Consider Triangles: Consider the triangles $AEF$ and $ABF$ . Triangle $AEF$ is formed by joining points $A$ , $E$ , and $F$ , and triangle $ABF$ is formed by joining points $A$ , $B$ , and $F$ .
Observe Common Height: Observe that triangles $AEF$ and $ABF$ share a common height from point $A$ perpendicular to line $BF$ . The base of triangle $AEF$ is $EF$ , and the base of triangle $ABF$ is $BF$ . Since $F$ is the midpoint of $BC$ , $BF$ is half of $BC$ .
Calculate Length of EF: Calculate the length of EF. Since $E$ is the midpoint of $DC$ , and $DC$ is equal to $BC$ in a rectangle, $EF$ is half of $BC$ . Therefore, $EF = \frac{1}{2} BC$ .
Compare Areas of Triangles: Compare the areas of triangles $AEF$ and $ABF$ . The area of a triangle is $\frac{1}{2} \times \text{base} \times \text{height}$ . Both triangles share the same height, so the ratio of their areas will be the ratio of their bases.
Calculate Ratio of Bases: Calculate the ratio of the bases $EF$ and $BF$ . Since $EF = \frac{1}{2} BC$ and $BF = \frac{1}{2} BC$ , the ratio $\frac{EF}{BF} = \frac{\frac{1}{2} BC}{\frac{1}{2} BC} = 1$ .

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