Area of General Quadrilateral
A quadrilateral is a polygon with four sides, four vertices, and four angles. While special quadrilaterals like rectangles, squares, parallelograms, and trapeziums have specific area formulas, the general quadrilateral does not have a single direct formula.
To find the area of any general quadrilateral, we divide it into two triangles by drawing one of its diagonals. The area of the quadrilateral then equals the sum of the areas of these two triangles.
This method works for every quadrilateral — regular, irregular, convex, or concave. It is one of the most versatile techniques in Class 8 mensuration.
The diagonal method is essential when dealing with land plots, irregular floor plans, and other real-world shapes that do not fit standard formulas.
What is Area of General Quadrilateral?
Definition: A quadrilateral is a closed figure bounded by four line segments.
The area of a general quadrilateral is calculated by dividing it into two triangles using a diagonal, and summing their areas.
Key terms:
- Diagonal — a line segment joining two non-adjacent vertices.
- Perpendicular heights (h₁ and h₂) — the perpendicular distances from the other two vertices to the chosen diagonal.
- A quadrilateral has two diagonals; either can be used.
Area of General Quadrilateral Formula
Formula:
Area = ½ × d × (h₁ + h₂)
Where:
- d = length of the diagonal
- h₁ = perpendicular distance from one vertex to the diagonal
- h₂ = perpendicular distance from the opposite vertex to the diagonal
Alternative: If the quadrilateral is divided into two triangles by diagonal AC, then:
Area = Area of △ABC + Area of △ACD
Derivation and Proof
Derivation:
Step 1: Let ABCD be a quadrilateral. Draw diagonal AC of length d.
Step 2: This diagonal divides ABCD into two triangles: △ABC and △ACD.
Step 3: Drop a perpendicular from B to AC. Let its length be h₁.
- Area of △ABC = ½ × AC × h₁ = ½ × d × h₁
Step 4: Drop a perpendicular from D to AC. Let its length be h₂.
- Area of △ACD = ½ × AC × h₂ = ½ × d × h₂
Step 5: Area of quadrilateral ABCD = Area of △ABC + Area of △ACD
- = ½ × d × h₁ + ½ × d × h₂
- = ½ × d × (h₁ + h₂)
Types and Properties
Problems on the area of a general quadrilateral can be classified as follows:
1. Direct computation:
- Diagonal and both perpendicular heights are given. Apply Area = ½ × d × (h₁ + h₂).
2. Finding a missing measurement:
- Area and diagonal are given; find h₁ + h₂ or one height when the other is known.
3. Splitting into triangles with given side lengths:
- Use Heron’s formula to find each triangle’s area when all sides and the diagonal are given.
4. Coordinate geometry method:
- When vertices are given as coordinates, use the Shoelace formula (studied in higher classes).
5. Word problems:
- Land plots, garden beds, and floor plans that form irregular quadrilaterals.
Solved Examples
Example 1: Example 1: Direct application
Problem: The diagonal of a quadrilateral is 20 cm. The perpendicular distances from the other two vertices to this diagonal are 6 cm and 8 cm. Find the area.
Solution:
Given:
- d = 20 cm, h₁ = 6 cm, h₂ = 8 cm
Using the formula:
- Area = ½ × d × (h₁ + h₂)
- Area = ½ × 20 × (6 + 8)
- Area = ½ × 20 × 14
- Area = 140 cm²
Answer: The area is 140 cm².
Example 2: Example 2: Finding a perpendicular height
Problem: The area of a quadrilateral is 252 cm². Its diagonal is 24 cm and one perpendicular height is 9 cm. Find the other height.
Solution:
Given:
- Area = 252 cm², d = 24 cm, h₁ = 9 cm
Using the formula:
- 252 = ½ × 24 × (9 + h₂)
- 252 = 12 × (9 + h₂)
- 9 + h₂ = 252 / 12 = 21
- h₂ = 21 − 9 = 12 cm
Answer: The other perpendicular height is 12 cm.
Example 3: Example 3: Equal perpendicular heights
Problem: A quadrilateral has a diagonal of 30 cm. Both perpendicular heights from the opposite vertices are equal at 10 cm each. Find the area.
Solution:
- Area = ½ × 30 × (10 + 10)
- Area = ½ × 30 × 20
- Area = 300 cm²
Answer: The area is 300 cm².
Example 4: Example 4: Finding the diagonal
Problem: The area of a quadrilateral is 180 cm². The sum of the perpendicular heights from the two vertices to a diagonal is 18 cm. Find the diagonal.
Solution:
- 180 = ½ × d × 18
- 180 = 9d
- d = 180 / 9 = 20 cm
Answer: The diagonal is 20 cm.
Example 5: Example 5: Quadrilateral with two triangles (Heron’s formula)
Problem: Quadrilateral ABCD has diagonal AC = 13 cm. In △ABC, the sides are AB = 5 cm, BC = 12 cm, AC = 13 cm. In △ACD, the sides are AC = 13 cm, CD = 14 cm, AD = 15 cm. Find the area of ABCD.
Solution:
△ABC:
- Check: 5² + 12² = 25 + 144 = 169 = 13². Right-angled triangle.
- Area = ½ × 5 × 12 = 30 cm²
△ACD (Heron’s formula):
- s = (13 + 14 + 15) / 2 = 21
- Area = √[21 × (21−13) × (21−14) × (21−15)]
- = √[21 × 8 × 7 × 6]
- = √[7056]
- = 84 cm²
Total area = 30 + 84 = 114 cm²
Answer: The area of ABCD is 114 cm².
Example 6: Example 6: Land plot problem
Problem: A farmer’s plot is shaped like a quadrilateral. A surveyor measures one diagonal as 50 m and the perpendicular distances from the other two corners as 18 m and 22 m. Find the area of the plot.
Solution:
- Area = ½ × 50 × (18 + 22)
- Area = ½ × 50 × 40
- Area = 1,000 m²
Answer: The area of the plot is 1,000 m².
Example 7: Example 7: Converting units
Problem: A quadrilateral field has diagonal 80 m and perpendicular heights 25 m and 35 m. Find the area in hectares. (1 hectare = 10,000 m²)
Solution:
- Area = ½ × 80 × (25 + 35)
- Area = ½ × 80 × 60 = 2,400 m²
- Area in hectares = 2,400 / 10,000 = 0.24 hectares
Answer: The area is 0.24 hectares.
Example 8: Example 8: Quadrilateral with one height zero
Problem: In quadrilateral ABCD, vertex D lies on diagonal AC. If AC = 16 cm and the perpendicular from B to AC is 10 cm, find the area.
Solution:
- Since D lies on AC, h₂ = 0.
- Area = ½ × 16 × (10 + 0)
- Area = ½ × 16 × 10 = 80 cm²
Note: The quadrilateral degenerates to a triangle when one vertex lies on the diagonal.
Answer: The area is 80 cm².
Example 9: Example 9: Cost of turfing
Problem: A garden in the shape of a quadrilateral has diagonal 40 m and perpendicular heights 12 m and 18 m. If the cost of turfing is ₹5 per m², find the total cost.
Solution:
- Area = ½ × 40 × (12 + 18) = ½ × 40 × 30 = 600 m²
- Cost = 600 × 5 = ₹3,000
Answer: The total cost of turfing is ₹3,000.
Example 10: Example 10: Comparing two methods
Problem: Quadrilateral PQRS has diagonal PR = 26 cm. The perpendicular from Q is 10 cm and from S is 14 cm. Verify by computing areas of the two triangles separately.
Solution:
Method 1 (Direct formula):
- Area = ½ × 26 × (10 + 14) = ½ × 26 × 24 = 312 cm²
Method 2 (Two triangles):
- Area of △PQR = ½ × 26 × 10 = 130 cm²
- Area of △PRS = ½ × 26 × 14 = 182 cm²
- Total = 130 + 182 = 312 cm² ✓
Answer: Both methods give 312 cm².
Real-World Applications
Land Surveying: Most land plots are irregular quadrilaterals. Surveyors measure a diagonal and drop perpendiculars to calculate the plot area for legal records and sale deeds.
Architecture: Irregular room shapes in old buildings or custom-designed homes require the diagonal method to calculate floor area for tiling and carpeting.
Agriculture: Farmers with irregularly shaped fields use this method to calculate area for crop planning, irrigation, and fertiliser estimation.
Map Reading: When a region on a map is bounded by four landmarks, the diagonal method helps estimate the enclosed area.
Construction: Engineers split irregular plot shapes into triangles for foundation layout and material estimation.
Key Points to Remember
- A general quadrilateral’s area is found by dividing it into two triangles using a diagonal.
- Formula: Area = ½ × d × (h₁ + h₂).
- d is the diagonal; h₁ and h₂ are perpendicular distances from the other two vertices to the diagonal.
- Either diagonal can be used — the result is the same.
- For special quadrilaterals, this formula reduces to their standard formulas.
- When all sides and one diagonal are known, use Heron’s formula for each triangle.
- Area is always in square units.
- This method works for both convex and concave quadrilaterals.
- If a vertex lies on the diagonal, h = 0 and the quadrilateral reduces to a triangle.
- The diagonal method is the foundation for computing areas of all polygons by splitting into triangles.
Practice Problems
- A quadrilateral has a diagonal of 28 cm. The perpendicular heights from the other two vertices are 7 cm and 11 cm. Find the area.
- The area of a quadrilateral is 390 cm². Its diagonal is 26 cm. If one perpendicular height is 12 cm, find the other.
- A plot of land shaped like a quadrilateral has diagonal 60 m and perpendicular heights 15 m and 25 m. Find the area in m².
- Find the cost of painting a quadrilateral-shaped wall with diagonal 3 m and perpendicular heights 1.2 m and 1.8 m, at ₹120 per m².
- A quadrilateral ABCD has AC = 10 cm, AB = 6 cm, BC = 8 cm, CD = 7 cm, AD = 9 cm. Find its area using Heron’s formula for each triangle.
- The diagonals of a quadrilateral are 16 cm and 12 cm and they intersect at right angles. Find the area.
Frequently Asked Questions
Q1. What is the formula for the area of a general quadrilateral?
Area = ½ × d × (h₁ + h₂), where d is the length of a diagonal and h₁, h₂ are the perpendicular distances from the other two vertices to that diagonal.
Q2. Why do we divide the quadrilateral into triangles?
Because there is no single direct formula for the area of an arbitrary quadrilateral. Dividing into triangles using a diagonal lets us use the well-known triangle area formula.
Q3. Does it matter which diagonal we choose?
No. Using either diagonal will give the same total area, because the quadrilateral is the same shape regardless of how it is divided.
Q4. Can this method be used for concave quadrilaterals?
Yes. For concave quadrilaterals, one perpendicular height may fall outside the quadrilateral, but the formula still works if the correct signed heights are used.
Q5. How is this different from the trapezium area formula?
The trapezium formula Area = ½ × (a + b) × h uses the two parallel sides and the height between them. The general quadrilateral formula uses a diagonal and perpendicular heights from the remaining vertices.
Q6. What if both diagonals and their intersection angle are known?
If the diagonals intersect at right angles, Area = ½ × d₁ × d₂ (same as the rhombus formula). For other angles, Area = ½ × d₁ × d₂ × sinθ, where θ is the angle between the diagonals.
Q7. What is Heron’s formula?
For a triangle with sides a, b, c: s = (a+b+c)/2, Area = √[s(s−a)(s−b)(s−c)]. It is used when the perpendicular height is not directly known but all sides are given.
Q8. Can this formula be extended to polygons with more than 4 sides?
Yes. Any polygon can be divided into triangles by drawing diagonals from one vertex. The total area is the sum of all the triangle areas.
