PRELIMINARY STAGE, ROUND 3 PROBLEM OF THE DAY
2026
St. Hubert Seminary SHS: 51 points
Bright SHS: 48 points
Akwamuman SHS: 29 points
QUESTION
Find the area of triangle ABC, which is right-angled at B. Angle C measures $22.5^\circ$, and the length of AC is 20 cm.
ANSWER: $50\sqrt2$ cm$^2$
FORMULA
$\text{Area} = \dfrac{1}{4} h^2 \sin(2\theta)$
(where $h$ is the hypotenuse $AC$, and $\theta$ is the acute angle $C$)
SOLUTION
Identify the given parameters:
$h = 20\text{ cm}$
$\theta = 22.5^\circ$
Double the angle for the sine function:
$2\theta = 2 \times 22.5^\circ = 45^\circ$
Recall the exact trigonometric value:
$\sin(45^\circ) = \dfrac{\sqrt{2}}{2}$
Substitute the parameters vertically into the area formula:
$\text{Area} = \dfrac{1}{4} \times 20^2 \times \sin(45^\circ)$
$\text{Area} = \dfrac{1}{4} \times 400 \times \dfrac{\sqrt{2}}{2}$
$\text{Area} = 100 \times \dfrac{\sqrt{2}}{2}$
$\text{Area} = 50\sqrt{2}\text{ cm}^2$
ANSWER: $50\sqrt2$ cm$^2$
PRACTICE QUESTIONS
1. Find the area of triangle $ABC$, right-angled at $B$. Angle $C$ measures $22.5^\circ$, and the length of $AC$ is $20$ cm.
ANSWER: $100\sqrt2\text{ cm}^2$
SOLUTION
Use $\text{Area}=\dfrac{1}{4}h^2\sin(2\theta)$, where $h$ is the hypotenuse $AC$ and $\theta$ is angle $C$.
$2\theta=2\times22.5^\circ=45^\circ$
$\sin(45^\circ)=\dfrac{\sqrt2}{2}$
$\text{Area}=\dfrac{1}{4}\times20^2\times\dfrac{\sqrt2}{2}$
$\text{Area}=100\times\dfrac{\sqrt2}{2}=100\sqrt2\text{ cm}^2$
2. Find the area of triangle $ABC$, right-angled at $B$. Angle $C$ measures $22.5^\circ$, and the length of $AC$ is $24$ cm.
ANSWER: $144\sqrt2\text{ cm}^2$
SOLUTION
Use $\text{Area}=\dfrac{1}{4}h^2\sin(2\theta)$, where $h$ is the hypotenuse $AC$ and $\theta$ is angle $C$.
$2\theta=2\times22.5^\circ=45^\circ$
$\sin(45^\circ)=\dfrac{\sqrt2}{2}$
$\text{Area}=\dfrac{1}{4}\times24^2\times\dfrac{\sqrt2}{2}$
$\text{Area}=144\times\dfrac{\sqrt2}{2}=144\sqrt2\text{ cm}^2$
3. Find the area of triangle $ABC$, right-angled at $B$. Angle $C$ measures $67.5^\circ$, and the length of $AC$ is $20$ cm.
ANSWER: $100\sqrt2\text{ cm}^2$
SOLUTION
Use $\text{Area}=\dfrac{1}{4}h^2\sin(2\theta)$, where $h$ is the hypotenuse $AC$ and $\theta$ is angle $C$.
$2\theta=2\times67.5^\circ=135^\circ$
$\sin(135^\circ)=\dfrac{\sqrt2}{2}$
$\text{Area}=\dfrac{1}{4}\times20^2\times\dfrac{\sqrt2}{2}$
$\text{Area}=100\times\dfrac{\sqrt2}{2}=100\sqrt2\text{ cm}^2$
4. Find the area of triangle $ABC$, right-angled at $B$. Angle $C$ measures $67.5^\circ$, and the length of $AC$ is $12$ cm.
ANSWER: $36\sqrt2\text{ cm}^2$
SOLUTION
Use $\text{Area}=\dfrac{1}{4}h^2\sin(2\theta)$, where $h$ is the hypotenuse $AC$ and $\theta$ is angle $C$.
$2\theta=2\times67.5^\circ=135^\circ$
$\sin(135^\circ)=\dfrac{\sqrt2}{2}$
$\text{Area}=\dfrac{1}{4}\times12^2\times\dfrac{\sqrt2}{2}$
$\text{Area}=36\times\dfrac{\sqrt2}{2}=36\sqrt2\text{ cm}^2$
5. Find the area of triangle $ABC$, right-angled at $B$. Angle $C$ measures $45^\circ$, and the length of $AC$ is $16$ cm.
ANSWER: $64\text{ cm}^2$
SOLUTION
Use $\text{Area}=\dfrac{1}{4}h^2\sin(2\theta)$, where $h$ is the hypotenuse $AC$ and $\theta$ is angle $C$.
$2\theta=2\times45^\circ=90^\circ$
$\sin(90^\circ)=1$
$\text{Area}=\dfrac{1}{4}\times16^2\times1$
$\text{Area}=64\times1=64\text{ cm}^2$
6. Find the area of triangle $ABC$, right-angled at $B$. Angle $C$ measures $45^\circ$, and the length of $AC$ is $24$ cm.
ANSWER: $144\text{ cm}^2$
SOLUTION
Use $\text{Area}=\dfrac{1}{4}h^2\sin(2\theta)$, where $h$ is the hypotenuse $AC$ and $\theta$ is angle $C$.
$2\theta=2\times45^\circ=90^\circ$
$\sin(90^\circ)=1$
$\text{Area}=\dfrac{1}{4}\times24^2\times1$
$\text{Area}=144\times1=144\text{ cm}^2$