Library / /124 of 213

, 2023

Algebra Difficulty 1.6 Junior Find the answer Canada

For 30 consecutive days, the daily high temperature was recorded.
On each of the first 25 days, the temperature recorded was 21℃. On each of the remaining 5
days, the temperature recorded was 15℃. For the 30 days, the mean
(average) of the temperatures recorded was

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$.

Want a route through all this instead of an archive? The track puts 2,444 problems in a working order, from Junior Challenge level to the IMO shortlist.

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