A plane figure (or plane shape) is a flat, two-dimensional shape that lies entirely on a flat surface (plane). It has length and breadth (width) but no thickness.
Examples of plane figures include:
Triangle, Square, Rectangle, Parallelogram, Rhombus, Trapezium, Kite, Circle.
Each of these shapes has its own properties, such as the number of sides, angles, and lines of symmetry.
A parallelogram is a quadrilateral (four-sided figure) whose opposite sides are parallel and equal in length.
Properties of a Parallelogram:
Examples of Parallelograms: Rectangle, Square, and Rhombus are all special types of parallelograms.
Diagram of a Parallelogram:
_________
/ /
/________/
A rectangle is a parallelogram with four right angles (90° each).
Properties of a Rectangle:
Example:
If a rectangle has length = 8 cm and breadth = 4 cm:
Perimeter = 2(l + b) = 2(8 + 4) = 24 cm
Area = l × b = 8 × 4 = 32 cm²
Diagram of a Rectangle: +----------+ | | breadth = 4 cm | | +----------+ length = 8 cm
A rhombus is a parallelogram with all sides equal in length.
Properties of a Rhombus:
Example:
If each side of a rhombus is 6 cm and one diagonal is 8 cm, find the other diagonal (d₂) if area = (½ × d₁ × d₂):
Area = ½ × 8 × d₂
Suppose area = 24 cm²,
24 = 4 × d₂ → d₂ = 6 cm
Diagram of a Rhombus: /\ / \ \ / \/ (Diagonals cross at 90°)
A square is a special type of rectangle and rhombus — all sides are equal and all angles are 90°.
Properties of a Square:
Example:
If a square has side 5 cm:
Perimeter = 4 × 5 = 20 cm
Area = 5² = 25 cm²
Diagram of a Square: +----+ | | side = 5 cm +----+ (All angles 90°)
A trapezium (US: trapezoid) is a quadrilateral with only one pair of parallel sides.
Properties of a Trapezium:
Example:
If the parallel sides of a trapezium are 8 cm and 4 cm, and the height is 3 cm:
Area = ½ × (sum of parallel sides) × height
= ½ × (8 + 4) × 3
= ½ × 12 × 3 = 18 cm²
Diagram of a Trapezium:
______8 cm______
/ \
/__________________\
4 cm
height = 3 cm
A kite is a quadrilateral with two pairs of adjacent sides equal.
Properties of a Kite:
Example:
If diagonals of a kite are 6 cm and 8 cm,
Area = ½ × d₁ × d₂ = ½ × 6 × 8 = 24 cm²
Diagram of a Kite: /\ / \ \ / \/ (Diagonals cross at 90°)
| Shape | Opposite Sides Parallel | All Sides Equal | All Angles 90° | Diagonals Equal | Diagonals Bisect at 90° |
|---|---|---|---|---|---|
| Parallelogram | ✔ | ✖ | ✖ | ✖ | ✖ |
| Rectangle | ✔ | ✖ | ✔ | ✔ | ✖ |
| Rhombus | ✔ | ✔ | ✖ | ✖ | ✔ |
| Square | ✔ | ✔ | ✔ | ✔ | ✔ |
| Trapezium | One pair | ✖ | ✖ | Sometimes | ✖ |
| Kite | ✖ | Two pairs (adjacent) | ✖ | ✖ | ✔ |
Question:
A parallelogram has base 10 cm and height 6 cm. Find its area and perimeter if the side length is 8 cm.
Solution:
Diagram of a Parallelogram:
_________
/ /
/________/
base = 10 cm, height = 6 cm
Problem: A rectangular vegetable plot is 12 m long and 8 m wide. How much fencing is needed to go all the way round the plot? Also find the area of the plot.
Solution:
Perimeter of rectangle = 2 × (length + width).
Step 1: length + width = 12 + 8 = 20.
Step 2: Perimeter = 2 × 20 = 40.
So 40 m of fencing is needed.
Area of rectangle = length × width = 12 × 8.
12 × 8 = 96.
So area = 96 m².
Diagram: 12 m ┌────────────────────┐ │ │ 8 m │ │ └────────────────────┘
Problem: A gardener makes a parallelogram bed whose base is 7 m and perpendicular height is 3 m. Find the area.
Solution:
Area of parallelogram = base × height.
Step: 7 × 3 = 21.
So area = 21 m².
Diagram (height shown): base = 7 m ┌─────────┐ \ | \ | height = 3 m (perpendicular) \______|
Problem: A parallelogram has base 9 cm and the other side (adjacent side) is 6 cm. Find its perimeter.
Solution:
Perimeter of parallelogram = 2 × (sum of adjacent sides) = 2 × (base + side).
Step 1: base + side = 9 + 6 = 15.
Step 2: Perimeter = 2 × 15 = 30.
So perimeter = 30 cm.
Problem: A triangular-shaped wall panel is formed by splitting a parallelogram in half along a diagonal. The parallelogram has base 10 m and area 60 m². Find the perpendicular height of the parallelogram and hence the area of one triangular half.
Solution:
Area of parallelogram = base × height → height = area ÷ base.
Step 1: height = 60 ÷ 10 = 6.
So height = 6 m.
Area of one triangle = half the parallelogram area = 60 ÷ 2 = 30 m².
Problem: A decorative sign is a rhombus whose diagonals measure 8 cm and 6 cm. Find the area of the sign.
Solution:
Area of a rhombus = ½ × (product of the diagonals).
Step 1: product = 8 × 6 = 48.
Step 2: area = ½ × 48 = 24.
So area = 24 cm².
Diagram (diagonals):
\ | /
\ | /
\|/
/|\
/ | \
/ | \
d1=8 d2=6
Problem: A square courtyard has side 9 m. Tiles cost ₦800 per m². How much will it cost to tile the courtyard?
Solution:
Area of square = side² = 9 × 9 = 81 m².
Cost = area × price per m² = 81 × 800.
Step 1: 81 × 800 = 81 × (8 × 100) = (81 × 8) × 100.
Compute 81 × 8: 80×8 = 640, plus 1×8 = 8 → 640 + 8 = 648.
Then ×100 → 648 × 100 = 64,800.
So cost = ₦64,800.
Problem: A farmer makes a trapezium bed with parallel sides 12 m and 8 m, and height 4 m. Find area of the bed.
Solution:
Area of trapezium = ½ × (sum of parallel sides) × height.
Step 1: sum of parallels = 12 + 8 = 20.
Step 2: ½ × 20 = 10.
Step 3: area = 10 × height 4 = 40.
So area = 40 m².
Diagram:
______12 m______
/ \
/ \
/_____8 m___________\
height = 4 m
Problem: A rectangular pool is 6 m by 4 m. A uniform grass border 2 m wide is built all around the pool forming a larger rectangle. Find the area of the border (grass).
Solution:
Outer rectangle dimensions = (length + 2×border) and (width + 2×border).
Step 1: outer length = 6 + 2×2 = 6 + 4 = 10.
Step 2: outer width = 4 + 2×2 = 4 + 4 = 8.
Step 3: area outer = 10 × 8 = 80.
Step 4: area pool = 6 × 4 = 24.
Step 5: area border = outer area − pool area = 80 − 24 = 56.
So the grass border area = 56 m².
Diagram: Outer 10×8 ┌────────────────────┐ │ border 2m │ │ ┌───────6×4────┐ │ │ │ pool │ │ │ └─────────────┘ │ └────────────────────┘
Problem: A family wants a square garden. They have 48 m of fence available. What should be the length of each side?
Solution:
Perimeter of square = 4 × side → side = perimeter ÷ 4.
Step: side = 48 ÷ 4 = 12.
So each side = 12 m.
Area would be 12 × 12 = 144 m² (extra info).
Problem: A parallelogram roof panel has base length 15 m and area 75 m². A contractor says one side (adjacent side) must be 11 m so that the panel fits the supports. Is the contractor correct? (Find the perpendicular height, then compare with side length using right-triangle idea.)
Solution:
First find the perpendicular height from area: height = area ÷ base = 75 ÷ 15 = 5 m. So perpendicular height is 5 m.
If the side length along the slant is 11 m, is that inconsistent? The slanted side is not the perpendicular height; a slanted side can be larger than the height. To check plausibility, one can imagine the slanted side and height form a right triangle only if we know the horizontal offset (which we do not). So nothing in the area/height data contradicts a slanted side of 11 m.
But if the contractor meant the side length should equal the perpendicular height (i.e., 11 m = 5 m), that would be wrong. The correct perpendicular height is 5 m, and the adjacent (slanted) side can be any length ≥ height depending on the shape. So contractor is not correct if he meant height = 11 m; the height is 5 m.
