PRELIMINARY CONTEST 2024
Contest 40
Chemu SHS – 43 pts (Winner)
Our Lady of Mt. Carmel Girls’ SHS – 33 pts
Manya Krobo SHS –28 pts
Contest 41
Adiembra SHS – 34 pts (Winner)
St. Paul’s Senior High & Minor Seminary – 24 pts
Biakoye Community SHS – 07 pts
Contest 42
Swedru Secondary School – 49 pts (Winner)
Edinaman SHS – 29 pts
Techiman SHS – 23 pts
Contest 37
Winneba SHS – 53 pts (Winner)
Armed Forces SHTS – 43 pts
Vakpo SHS – 22 pts
Contest 38
St. Joseph’s Seminary SHS – 26 pts (Winner)
Suhum Senior High Technical School – 22 pts
Breman Asikuma SHS – 18 pts
Contest 39
Ghana National College – 66 pts (Winner)
Nkwatia Presby SHS – 46 pts
Nafana Presby SHS – 15 pts
ROUND 1
SECOND SET OF QUESTIONS
PREAMBLE
Find the length of the sides of a rhombus if the diagonals have length:
FORMULA
$\text{Side}=\sqrt{\left(\dfrac{d_1}{2}\right)^2+\left(\dfrac{d_2}{2}\right)^2}$
$d_1$ and $d_2$ are the diagonals of the rhombus.
1. $20$ cm and $30$ cm
ANSWER: $5\sqrt{13}$ cm
SOLUTION
$\text{Side}=\sqrt{\left(\dfrac{20}{2}\right)^2+\left(\dfrac{30}{2}\right)^2}$
$\text{Side}=\sqrt{10^2+15^2}$
$\text{Side}=\sqrt{100+225}$
$\text{Side}=\sqrt{325}$
$\text{Side}=\sqrt{25\times13}$
$\text{Side}=5\sqrt{13}\text{ cm}$
2. $10$ cm and $16$ cm
ANSWER: $\sqrt{89}$ cm
SOLUTION
$\text{Side}=\sqrt{\left(\dfrac{10}{2}\right)^2+\left(\dfrac{16}{2}\right)^2}$
$\text{Side}=\sqrt{5^2+8^2}$
$\text{Side}=\sqrt{25+64}$
$\text{Side}=\sqrt{89}\text{ cm}$
3. $30$ cm and $24$ cm
ANSWER: $3\sqrt{41}$ cm
SOLUTION
$\text{Side}=\sqrt{\left(\dfrac{30}{2}\right)^2+\left(\dfrac{24}{2}\right)^2}$
$\text{Side}=\sqrt{15^2+12^2}$
$\text{Side}=\sqrt{225+144}$
$\text{Side}=\sqrt{369}$
$\text{Side}=\sqrt{9\times41}$
$\text{Side}=3\sqrt{41}\text{ cm}$
PRACTICE QUESTIONS
Find the length of the sides of a rhombus if the diagonals have length:
1. $20$ cm and $30$ cm
ANSWER: $5\sqrt{13}$ cm
SOLUTION
$\text{Side}=\sqrt{\left(\dfrac{20}{2}\right)^2+\left(\dfrac{30}{2}\right)^2}=\sqrt{10^2+15^2}$
$\text{Side}=\sqrt{100+225}=\sqrt{325}$
$\text{Side}=5\sqrt{13}\text{ cm}$
2. $10$ cm and $16$ cm
ANSWER: $\sqrt{89}$ cm
SOLUTION
$\text{Side}=\sqrt{\left(\dfrac{10}{2}\right)^2+\left(\dfrac{16}{2}\right)^2}=\sqrt{5^2+8^2}$
$\text{Side}=\sqrt{25+64}=\sqrt{89}$
$\text{Side}=\sqrt{89}\text{ cm}$
3. $30$ cm and $24$ cm
ANSWER: $3\sqrt{41}$ cm
SOLUTION
$\text{Side}=\sqrt{\left(\dfrac{30}{2}\right)^2+\left(\dfrac{24}{2}\right)^2}=\sqrt{15^2+12^2}$
$\text{Side}=\sqrt{225+144}=\sqrt{369}$
$\text{Side}=3\sqrt{41}\text{ cm}$
4. $12$ cm and $16$ cm
ANSWER: $10$ cm
SOLUTION
$\text{Side}=\sqrt{\left(\dfrac{12}{2}\right)^2+\left(\dfrac{16}{2}\right)^2}=\sqrt{6^2+8^2}$
$\text{Side}=\sqrt{36+64}=\sqrt{100}$
$\text{Side}=10\text{ cm}$
5. $18$ cm and $24$ cm
ANSWER: $15$ cm
SOLUTION
$\text{Side}=\sqrt{\left(\dfrac{18}{2}\right)^2+\left(\dfrac{24}{2}\right)^2}=\sqrt{9^2+12^2}$
$\text{Side}=\sqrt{81+144}=\sqrt{225}$
$\text{Side}=15\text{ cm}$
6. $14$ cm and $48$ cm
ANSWER: $25$ cm
SOLUTION
$\text{Side}=\sqrt{\left(\dfrac{14}{2}\right)^2+\left(\dfrac{48}{2}\right)^2}=\sqrt{7^2+24^2}$
$\text{Side}=\sqrt{49+576}=\sqrt{625}$
$\text{Side}=25\text{ cm}$