PNG Examination Questions Solutions : Perimeter and Area
Perimeter and Area are core topics in the Papua New Guinea mathematics curriculum and are commonly tested in Grade 9, Grade 10, and Grade 12 examinations. Questions involving rectangles, algebraic expressions, and problem-solving form a regular part of internal assessments and national exams.
PNG students must master perimeter and area questions because these problems test more than just formulas. They assess a student’s ability to interpret diagrams, substitute algebraic expressions, simplify equations, and solve real-life measurement problems. These skills are essential for success in examinations, especially in structured and long-answer questions where clear working is required.
Understanding perimeter and area also helps students apply mathematics to practical situations such as land measurement, fencing, building design, and planning spaces — all of which are relevant in everyday life in Papua New Guinea. By confidently solving perimeter and area questions, students improve their problem-solving skills, boost exam confidence, and increase their chances of achieving higher marks.
This worked example explains how to find the perimeter and area of a rectangle using algebra. It is suitable for Grade 9–10 mathematics and examination revision.
Given
- Length of the rectangle = \(2x\)
- Width of the rectangle = \(x - 1\)
1. Perimeter in terms of \(x\)
The formula for the perimeter of a rectangle is:
\[ P = 2(\text{length} + \text{width}) \]
Substituting the given values:
\[ P = 2(2x + (x - 1)) \]
Simplifying:
\[ P = 2(3x - 1) \]
\[ \boxed{P = 6x - 2} \]
2. Perimeter when \(x = 2\)
Substitute \(x = 2\) into the perimeter expression:
\[ P = 6x - 2 \]
\[ P = 6(2) - 2 \]
\[ P = 12 - 2 \]
Answer:
\[ \boxed{P = 10 \text{ meters}} \]
3. Area in terms of \(x\)
The formula for the area of a rectangle is:
\[ A = \text{length} \times \text{width} \]
Substituting the expressions:
\[ A = 2x(x - 1) \]
Expanding:
\[ \boxed{A = 2x^2 - 2x} \]
4. Find the value of \(x\) if the area is 24 m²
Set the area equal to 24:
\[ 2x^2 - 2x = 24 \]
Bring all terms to one side:
\[ 2x^2 - 2x - 24 = 0 \]
Divide through by 2:
\[ x^2 - x - 12 = 0 \]
Factorise:
\[ (x - 4)(x + 3) = 0 \]
Solving:
\[ x = 4 \quad \text{or} \quad x = -3 \]
Since a length cannot be negative, reject \(x = -3\).
Final Answer:
\[ \boxed{x = 4} \]
Final Answers Summary
- Perimeter in terms of \(x\): \(6x - 2\)
- Perimeter when \(x = 2\): 10 m
- Area in terms of \(x\): \(2x^2 - 2x\)
- Value of \(x\) when area is 24 m²: \(x = 4\)
This example helps students understand algebraic expressions involving perimeter and area.
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