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Geometry Difficulty 2.6 Junior Find the answer Canada

In the diagram, $\triangle ABC$ is a right-angled isosceles triangle. $D$ is the midpoint of $BC$ and $E$ is the midpoint of $AB$ .If $AB=BC=24$ cm, what is the area of $\triangle AED$ ?

Pick one

Solution

Solution 1

The area of $\triangle AED$ is equal to one-half its base times its height. Suppose the base of $\triangle AED$ is $AE$ , then its height is $BD$ ( $AE$ is perpendicular to $BD$ ). Since $AB=BC=24$ cm and $E$ and $D$ are the midpoints of their respective sides, then $AE=12$ cm and $BD=12$ cm. Thus, the area of $\triangle AED$ is $\$\frac12\times12\text{ cm}\times12\text{}$
cm}=72 $\text{}$ cm}^2 $. Solution 2 The area of$ $\triangle$ AED $is equal to the area of$ $\triangle$ ABD $minus the area of$ $\triangle$
EBD $. Suppose the base of$ $\triangle$ EBD $is$ BD $, then its height is$ EB $. Since$ AB=BC=24 $cm and$ E $and$ D are the midpoints of their respective sides, then

Figure for this problem EB=12 $cm and$ BD=12 $cm. Thus, the area of$ $\triangle$ EBD $is$ $\frac12\times12\text{ cm}\times12\text{}$
cm}=72 $\text{}$ cm}^2 $. The area of$ $\triangle$ ABD $is equal to$ $\frac12\times BD\times AB=\frac12\times 12\text{}$ cm} $\times 24\text{ cm}=144\text{}$
cm}^2 $. Thus, the area of$ $\triangle$ AED $is$ $144\text{ cm}^2-72\text{ cm}^2=72\text{}$
cm}^2$.

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