Mastery Guide: Perimeter, Area, Parallelograms, Triangles & Commercial Applications
Instructional Blueprint: Unit 8 Perimeter and Area
- Distinguish between Perimeter (1D boundary length) and Area (2D enclosed surface).
- Recognize that shapes with the same area can have different perimeters and vice versa.
- Apply standard formulas for Perimeter & Area of Squares and Rectangles.
- Calculate the Area of Parallelograms ($A = \text{base} \times \text{height}$) and Triangles ($A = \frac{1}{2} \times \text{base} \times \text{height}$).
- Solve multi-step real-world problems involving fencing, wall construction, tiling, painting, and carpeting costs.
- 00–15m: Hook & Intuition: The Playground Fence vs Grass Lawn.
- 15–30m: Square & Rectangle Formulas & Same Area / Different Perimeter Investigation.
- 30–45m: Area of Parallelograms & Triangles ($A = b \times h$ and $A = \frac{1}{2} b h$).
- 45–60m: Commercial Costing, Tiling & Complete Review Exercise.
- The Ant on the Fence (Perimeter): Imagine an ant walking all the way along the boundary fence!
- The Paint Roller (Area): Imagine rolling paint across the floor tiles to cover every flat spot!
- The Parallelogram Slice-and-Slide: Cut a right triangle from one end of a slanted parallelogram and slide it to the other end — it turns into a neat rectangle!
🏞️ Real-Life Challenge: The Park Fence & Grass Lawn
Sara and Raza are playing in their school playground. Raza asks: "If a wooden fence is to be fixed all around our rectangular playground, what will be its total length?"
Sara smiles: "That is the Perimeter! But if we want to lay lush green grass over the whole playing ground, we need its Area!"
1. Perimeter & Area of Square and Rectangular Regions
- Perimeter ($P$): Total length of 4 equal sides:
$$\mathbf{P = 4 \times L} \quad \left(\text{Side } L = \frac{P}{4}\right)$$ - Area ($A$): Space covered by the square:
$$\mathbf{A = L \times L = L^2}$$
- Perimeter ($P$): Boundary length around 2 lengths + 2 widths:
$$\mathbf{P = 2(L + W)} \quad \left(L + W = \frac{P}{2}\right)$$ - Area ($A$): Space enclosed inside rectangle:
$$\mathbf{A = L \times W} \quad \left(L = \frac{A}{W}, \quad W = \frac{A}{L}\right)$$
2. Surprising Discovery: Same Area vs. Same Perimeter
Consider a Rectangle ($8\text{ cm} \times 2\text{ cm}$) and a Square ($4\text{ cm} \times 4\text{ cm}$):
• $\text{Area of Rectangle} = 8 \times 2 = \mathbf{16\text{ cm}^2}$
• $\text{Area of Square} = 4 \times 4 = \mathbf{16\text{ cm}^2}$ (Same Area!)
• $\text{Perimeter of Rectangle} = 2(8 + 2) = \mathbf{20\text{ cm}}$
• $\text{Perimeter of Square} = 4 \times 4 = \mathbf{16\text{ cm}}$
Conclusion: Shapes with identical area can have totally different perimeters!
Consider a Rectangle ($6\text{ cm} \times 2\text{ cm}$) and a Square ($4\text{ cm} \times 4\text{ cm}$):
• $\text{Perimeter of Rectangle} = 2(6 + 2) = \mathbf{16\text{ cm}}$
• $\text{Perimeter of Square} = 4 \times 4 = \mathbf{16\text{ cm}}$ (Same Perimeter!)
• $\text{Area of Rectangle} = 6 \times 2 = \mathbf{12\text{ cm}^2}$
• $\text{Area of Square} = 4 \times 4 = \mathbf{16\text{ cm}^2}$
Conclusion: Shapes with identical perimeter can enclose completely different areas!
3. Area of a Parallelogram ($A = \text{base} \times \text{height}$)
A parallelogram has pairs of parallel and equal opposite sides. If we draw a perpendicular line from one side to the opposite side, that vertical distance is called the altitude (height).
4. Area of a Triangle ($A = \frac{1}{2} \times \text{base} \times \text{height}$)
Any diagonal divides a parallelogram into two identical, equal triangles. Therefore, the area of a triangle is exactly half ($\frac{1}{2}$) of the area of a parallelogram on the same base and height!
📝 Unit 8 Solved Exercises (100% Complete & Exhaustive Textbook Bank)
(a) Find the required length of the ribbon:
$$\text{Length of ribbon} = \text{Perimeter} = 2(12 + 8) = 2(20) = \mathbf{40\text{ cm}} = \mathbf{0.4\text{ metres}}$$
(b) What will be the total cost of the ribbon if $1\text{ metre}$ of it costs $\text{Rs. } 5$?
$$\text{Total Cost} = 0.4\text{ m} \times \text{Rs. } 5 = \mathbf{\text{Rs. } 2}$$
(a) Find its perimeter:
$$\text{Perimeter} = 2(L + W) = 2(128 + 96.5) = 2(224.5) = \mathbf{449\text{ metres}}$$
(b) Find the total cost required for the construction of a boundary wall around this building if the rate of construction of a wall is $\text{Rs. } 470\text{ per metre}$:
$$\text{Total Cost} = 449\text{ m} \times \text{Rs. } 470 = \mathbf{\text{Rs. } 211,030}$$
(a) The width of the gate is $2.3\text{ metres}$. Find its length:
$$\text{Length} = \frac{\text{Area}}{\text{Width}} = \frac{19.55}{2.3} = \mathbf{8.5\text{ metres}}$$
(b) Find the cost of painting the gate, if the rate of painting is $\text{Rs. } 275\text{ per m}^2$:
$$\text{Cost of painting} = 19.55 \times \text{Rs. } 275 = \mathbf{\text{Rs. } 5,376.25}$$
(a) The perimeter of the Masjid:
$$\text{Width} = \frac{\text{Area}}{\text{Length}} = \frac{27,540}{255} = 108\text{ metres}$$
$$\text{Perimeter} = 2(L + W) = 2(255 + 108) = 2(363) = \mathbf{726\text{ metres}}$$
(b) The cost of carpeting the Masjid, if the rate of carpeting is $\text{Rs. } 275\text{ per m}^2$:
$$\text{Cost of carpeting} = 27,540\text{ m}^2 \times \text{Rs. } 275 = \mathbf{\text{Rs. } 7,573,500}$$
(i) Take a rectangle of length $9\text{ cm}$, width $4\text{ cm}$ and find its area:
$$\text{Area of rectangle} = 9 \times 4 = \mathbf{36\text{ cm}^2}$$
(ii) Take a square of length $6\text{ cm}$ and find its area:
$$\text{Area of square} = 6 \times 6 = \mathbf{36\text{ cm}^2}$$
(iii) Show that shapes with the same area can have different perimeter:
$$\text{Perimeter of rectangle} = 2(9 + 4) = 2(13) = \mathbf{26\text{ cm}}$$
$$\text{Perimeter of square} = 4 \times 6 = \mathbf{24\text{ cm}}$$
Since $36\text{ cm}^2 = 36\text{ cm}^2$ but $26\text{ cm} \neq 24\text{ cm}$, shapes with the same area can indeed have different perimeters.
(i) Take a rectangle of length $10\text{ cm}$, width $2\text{ cm}$ and find its perimeter:
$$\text{Perimeter of rectangle} = 2(10 + 2) = 2(12) = \mathbf{24\text{ cm}}$$
(ii) Take a square of length $6\text{ cm}$ and find its perimeter:
$$\text{Perimeter of square} = 4 \times 6 = \mathbf{24\text{ cm}}$$
(iii) Show that shapes with the same perimeter can have different area:
$$\text{Area of rectangle} = 10 \times 2 = \mathbf{20\text{ cm}^2}$$
$$\text{Area of square} = 6 \times 6 = \mathbf{36\text{ cm}^2}$$
Since $24\text{ cm} = 24\text{ cm}$ but $20\text{ cm}^2 \neq 36\text{ cm}^2$, shapes with the same perimeter can have different areas.
$A = 5 \times 2 = \mathbf{10\text{ cm}^2}$
$A = 8 \times 3.5 = \mathbf{28\text{ m}^2}$
$A = 50 \times 22 = \mathbf{1,100\text{ mm}^2}$
$A = 8.6 \times 7 = \mathbf{60.2\text{ cm}^2}$
$A = 5.5 \times 2.5 = \mathbf{13.75\text{ cm}^2}$
$A = 8.5 \times 3.2 = \mathbf{27.2\text{ cm}^2}$
$A = \frac{1}{2} \times 7 \times 4 = \mathbf{14\text{ cm}^2}$
$A = \frac{1}{2} \times 8 \times 2.5 = \mathbf{10\text{ m}^2}$
$A = \frac{1}{2} \times 30 \times 9 = \mathbf{135\text{ mm}^2}$
$A = \frac{1}{2} \times 7 \times 2.2 = \mathbf{7.7\text{ cm}^2}$
$A = \frac{1}{2} \times 12 \times 11 = \mathbf{66\text{ m}^2}$
$A = \frac{1}{2} \times 6.8 \times 5 = \mathbf{17\text{ cm}^2}$
| S.No. | Base | Altitude | Area of Parallelogram | Area of Triangle |
|---|---|---|---|---|
| (i) | $12\text{ cm}$ | $10\text{ cm}$ | $12 \times 10 = \mathbf{120\text{ cm}^2}$ | $\frac{1}{2} \times 120 = \mathbf{60\text{ cm}^2}$ |
| (ii) | $8\text{ m}$ | $6\text{ m}$ | $8 \times 6 = \mathbf{48\text{ m}^2}$ | $\frac{1}{2} \times 48 = \mathbf{24\text{ m}^2}$ |
| (iii) | $11\text{ m}$ | $4\text{ m}$ | $11 \times 4 = \mathbf{44\text{ m}^2}$ | $\frac{1}{2} \times 44 = \mathbf{22\text{ m}^2}$ |
| (iv) | $25\text{ mm}$ | $12\text{ mm}$ | $25 \times 12 = \mathbf{300\text{ mm}^2}$ | $\frac{1}{2} \times 300 = \mathbf{150\text{ mm}^2}$ |
(i) Right-angled Triangle: Base $= 6\text{ cm}$, Altitude $= 8\text{ cm} \implies \text{Area} = \frac{1}{2} \times 6 \times 8 = \mathbf{24\text{ cm}^2}$
(ii) Inverted Triangle: Base $= 6\text{ cm}$, Altitude $= 6\text{ cm} \implies \text{Area} = \frac{1}{2} \times 6 \times 6 = \mathbf{18\text{ cm}^2}$
(iii) Obtuse Triangle: Base $= 9\text{ cm}$, Altitude $= 4\text{ cm} \implies \text{Area} = \frac{1}{2} \times 9 \times 4 = \mathbf{18\text{ cm}^2}$
(iv) Acute Triangle: Base $= 6\text{ cm}$, Altitude $= 8\text{ cm} \implies \text{Area} = \frac{1}{2} \times 6 \times 8 = \mathbf{24\text{ cm}^2}$
(v) Parallelogram: Base $= 8\text{ cm}$, Altitude $= 4\text{ cm} \implies \text{Area} = 8 \times 4 = \mathbf{32\text{ cm}^2}$
(vi) Parallelogram: Base $= 12\text{ cm}$, Altitude $= 8\text{ cm} \implies \text{Area} = 12 \times 8 = \mathbf{96\text{ cm}^2}$
(a) If length of rectangle is $4\text{ cm}$ and width is $3.4\text{ cm}$, perimeter is: (iii) $14.8\text{ cm}$ [$2(4 + 3.4) = 14.8$]
(b) Formula to find perimeter of square: (iii) $4L$
(c) Formula to find perimeter of rectangle: (i) $2(L + W)$
(d) Area of rectangle is $45\text{ m}^2$, length is $15\text{ m}$, width is: (ii) $3\text{ m}$ [$45 \div 15 = 3$]
(e) Formula to find area of square: (i) $L \times L$
(f) Formula to find area of rectangle: (i) $L \times W$
(g) Perimeter of rectangle is $34\text{ cm}$, if length increases by $2\text{ cm}$, difference in perimeter is: (ii) $4\text{ cm}$ [$2 \times 2\text{ cm} = 4\text{ cm}$]
(h) If length of side of square is $14\text{ cm}$, its perimeter is: (ii) $56\text{ cm}$ [$4 \times 14 = 56$]
(i) Base and height of parallelogram are $6\text{ cm}$ and $5\text{ cm}$. Area is: (iii) $30\text{ cm}^2$ [$6 \times 5 = 30$]
(j) Base and height of triangle are $10\text{ cm}$ and $8\text{ cm}$. Area is: (iv) $40\text{ cm}^2$ [$\frac{1}{2} \times 10 \times 8 = 40$]
(a) Calculate the area of the kitchen:
$$\text{Area of kitchen} = 12\text{ m} \times 10\text{ m} = \mathbf{120\text{ m}^2}$$
(b) How many tiles will be required if the length of side of a square tile is $2\text{ metres}$?
$$\text{Area of 1 tile} = 2\text{ m} \times 2\text{ m} = 4\text{ m}^2$$
$$\text{Number of tiles} = \frac{\text{Area of kitchen}}{\text{Area of 1 tile}} = \frac{120}{4} = \mathbf{30\text{ tiles}}$$
🎯 Unit 8 Synthesis Summary
Perimeter measures 1-dimensional boundary length around a closed shape ($P = 4L$ for squares, $P = 2(L+W)$ for rectangles) in linear units ($\text{m}, \text{cm}$). Area measures the enclosed 2-dimensional surface ($A = L^2$ for squares, $A = L \times W$ for rectangles, $A = b \times h$ for parallelograms, and $A = \frac{1}{2}bh$ for triangles) in square units ($\text{m}^2, \text{cm}^2$). Fencing and boundary costs depend on perimeter, whereas tiling, painting, and carpeting costs depend on area.
More Chapter Notes for Class 5 (FBISE)
MathematicsTest Your Knowledge on Chapter 8: Mastery Guide: Perimeter, Area, Parallelograms, Triangles & Commercial Applications
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Class 5 Mathematics - Ch 8: Perimeter and Area Chapter Mock Test
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