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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 $EB=12$ 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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Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.