Question:
In the above diagram, ABCD is a rectangle with $AE = EF = FB$. What is the ratio of the areas of $\triangle CEF$ and that of the rectangle?
In the above diagram, ABCD is a rectangle with $AE = EF = FB$. What is the ratio of the areas of $\triangle CEF$ and that of the rectangle?
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When a base is divided equally, use the fraction directly in the triangle area formula to find the ratio.
Updated On: Aug 4, 2025
- $\frac{1}{6}$
- $\frac{1}{8}$
- $\frac{1}{9}$
- None of these
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The Correct Option is A
Solution and Explanation
Let rectangle $ABCD$ have length $l$ and height $h$. Since $AE = EF = FB$, the base $AB$ is divided into three equal parts, each of length $\frac{l}{3}$.
Triangle $\triangle CEF$ has base $EF = \frac{l}{3}$ and height $h$. Area of $\triangle CEF = \frac{1}{2} \times \frac{l}{3} \times h = \frac{lh}{6}$.
Area of rectangle $ABCD = l \times h$. Ratio = $\frac{\frac{lh}{6}}{lh} = \frac{1}{6}$.
Triangle $\triangle CEF$ has base $EF = \frac{l}{3}$ and height $h$. Area of $\triangle CEF = \frac{1}{2} \times \frac{l}{3} \times h = \frac{lh}{6}$.
Area of rectangle $ABCD = l \times h$. Ratio = $\frac{\frac{lh}{6}}{lh} = \frac{1}{6}$.
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