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CEMC Gauss (Grade 7), 2023

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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