Area of Quadrilateral Using Diagonals
A general quadrilateral does not have a single simple area formula like a rectangle or parallelogram. However, we can find its area by dividing it into two triangles using one of its diagonals.
The diagonal splits the quadrilateral into two triangles. The area of each triangle is ½ × base × height, where the base is the diagonal and the heights are the perpendicular distances from the opposite vertices to the diagonal.
This method works for any quadrilateral — regular or irregular.
What is Area of Quadrilateral Using Diagonals?
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
Area of Quadrilateral Using Diagonals Formula
Derivation:
- Let ABCD be a quadrilateral with diagonal AC = d.
- Drop perpendiculars from B and D to AC. Let their lengths be h₁ and h₂.
- Area of △ABC = ½ × d × h₁
- Area of △ACD = ½ × d × h₂
- Total area = ½ × d × h₁ + ½ × d × h₂ = ½ × d × (h₁ + h₂)
Special cases:
- Rhombus: Diagonals bisect each other at right angles. Area = ½ × d₁ × d₂.
- Kite: One diagonal bisects the other. Area = ½ × d₁ × d₂ (if diagonals are perpendicular).
Solved Examples
Example 1: Example 1: Basic quadrilateral
Problem: A quadrilateral has a diagonal of 16 cm. The perpendicular distances from the two opposite vertices to this diagonal are 5 cm and 7 cm. Find the area.
Solution:
- Area = ½ × d × (h₁ + h₂) = ½ × 16 × (5 + 7) = ½ × 16 × 12 = 96 cm²
Answer: Area = 96 cm².
Example 2: Example 2: Finding the diagonal
Problem: The area of a quadrilateral is 150 cm². The perpendicular distances from the vertices to a diagonal are 10 cm and 5 cm. Find the diagonal.
Solution:
- 150 = ½ × d × (10 + 5) = ½ × d × 15
- 150 = 7.5d
- d = 20 cm
Answer: Diagonal = 20 cm.
Example 3: Example 3: Quadrilateral-shaped field
Problem: A field shaped as a quadrilateral ABCD has diagonal AC = 40 m. The perpendicular from B to AC is 12 m and from D to AC is 18 m. Find the area of the field.
Solution:
- Area = ½ × 40 × (12 + 18) = ½ × 40 × 30 = 600 m²
Answer: Area = 600 m².
Example 4: Example 4: Cost of levelling
Problem: A plot is shaped like a quadrilateral with diagonal 50 m and perpendicular heights 15 m and 20 m. Find the cost of levelling at ₹8 per m².
Solution:
- Area = ½ × 50 × (15 + 20) = ½ × 50 × 35 = 875 m²
- Cost = 875 × 8 = ₹7,000
Answer: Cost = ₹7,000.
Example 5: Example 5: Both heights equal
Problem: A quadrilateral has diagonal 24 cm and both perpendicular distances are 9 cm. Find the area. What special quadrilateral could this be?
Solution:
- Area = ½ × 24 × (9 + 9) = ½ × 24 × 18 = 216 cm²
When both heights are equal, the diagonal bisects the quadrilateral into two equal-area triangles. This could be a parallelogram (where the diagonal divides it into two congruent triangles).
Answer: Area = 216 cm².
Example 6: Example 6: Finding a perpendicular height
Problem: A quadrilateral has area 180 cm², diagonal 20 cm, and one perpendicular height 8 cm. Find the other height.
Solution:
- 180 = ½ × 20 × (8 + h₂)
- 180 = 10 × (8 + h₂)
- 18 = 8 + h₂
- h₂ = 10 cm
Answer: The other height = 10 cm.
Example 7: Example 7: Rhombus using diagonals
Problem: A rhombus has diagonals 14 cm and 20 cm. Find its area.
Solution:
- Area of rhombus = ½ × d₁ × d₂ = ½ × 14 × 20 = 140 cm²
(This is a special case of the general formula where the diagonals are perpendicular bisectors.)
Answer: Area = 140 cm².
Example 8: Example 8: Irregular quadrilateral with coordinates
Problem: In quadrilateral PQRS, diagonal PR = 30 cm. Q is at 8 cm from PR and S is at 12 cm from PR (on the same side). Can we use the formula directly?
Solution:
If Q and S are on the same side of PR, the quadrilateral is concave. The formula becomes:
- Area = ½ × d × |h₁ − h₂| = ½ × 30 × |12 − 8| = ½ × 30 × 4 = 60 cm²
Note: When vertices are on the same side, use the difference, not the sum.
Answer: Area = 60 cm².
Example 9: Example 9: Combined with Heron's formula
Problem: Quadrilateral ABCD has diagonal AC = 25 cm. Triangle ABC has sides 15, 20, 25 and triangle ACD has sides 7, 24, 25. Find the total area.
Solution:
△ABC: sides 15, 20, 25 → check: 15² + 20² = 225 + 400 = 625 = 25². Right triangle!
- Area = ½ × 15 × 20 = 150 cm²
△ACD: sides 7, 24, 25 → check: 7² + 24² = 49 + 576 = 625 = 25². Right triangle!
- Area = ½ × 7 × 24 = 84 cm²
Total area = 150 + 84 = 234 cm²
Answer: Area = 234 cm².
Example 10: Example 10: Finding area of a park
Problem: A park is in the shape of a quadrilateral. Its diagonal is 100 m and the two perpendicular offsets from the diagonal are 35 m and 45 m. Find the area in hectares (1 hectare = 10,000 m²).
Solution:
- Area = ½ × 100 × (35 + 45) = ½ × 100 × 80 = 4000 m²
- In hectares = 4000/10000 = 0.4 hectares
Answer: Area = 4000 m² = 0.4 hectares.
Real-World Applications
Applications:
- Land surveying: Measuring irregular plots by taking diagonal and offset measurements.
- Agriculture: Calculating field areas for irrigation and fertiliser planning.
- Construction: Estimating floor area of irregularly shaped rooms.
- Mapping: Dividing map regions into triangles for area calculation.
Key Points to Remember
- Area of any quadrilateral = ½ × diagonal × (sum of perpendicular heights from opposite vertices).
- This works because the diagonal divides the quadrilateral into two triangles.
- For rhombus: Area = ½ × d₁ × d₂ (diagonals are perpendicular).
- If both vertices are on the same side of the diagonal, use the difference of heights instead of the sum.
- The formula can be combined with Heron's formula for more complex problems.
- Always identify the diagonal and the two perpendicular distances clearly.
- Units must be consistent — convert before calculating.
- This method is used in land surveying with chain and offset measurements.
Practice Problems
- A quadrilateral has diagonal 28 cm with perpendicular heights 9 cm and 11 cm. Find the area.
- Find the diagonal of a quadrilateral with area 270 m² and perpendicular heights 12 m and 15 m.
- A rhombus has diagonals 16 cm and 30 cm. Find its area.
- A field shaped as a quadrilateral has diagonal 80 m. The offsets from the diagonal are 25 m and 35 m. Find the cost of fencing at ₹10/m if perimeter is 240 m, and cost of levelling at ₹5/m².
- The area of a quadrilateral is 420 cm². One perpendicular height is 14 cm, and the diagonal is 30 cm. Find the other height.
- A park has two quadrilateral sections sharing a common diagonal of 50 m. One section has offsets 20 and 30 m, the other has offsets 15 and 25 m. Find total area.
Frequently Asked Questions
Q1. What is the formula for area of a quadrilateral using its diagonal?
Area = ½ × d × (h₁ + h₂), where d is the diagonal length and h₁, h₂ are the perpendicular distances from the other two vertices to this diagonal.
Q2. Does this formula work for all quadrilaterals?
Yes, for convex quadrilaterals use the sum of heights. For concave (where both vertices are on the same side), use the difference.
Q3. How is this different from Heron's formula method?
The diagonal method needs the diagonal length and two perpendicular heights. Heron's formula needs the side lengths of the resulting triangles. Both give the same answer.
Q4. What are offsets in land surveying?
Offsets are perpendicular distances measured from a base line (the diagonal or chain line) to the boundary points of the land.
Q5. Can we use either diagonal?
Yes. You can use either diagonal. The area will be the same regardless of which diagonal you choose.
Q6. How is the rhombus area formula related?
In a rhombus, the diagonals bisect each other at right angles. Using one diagonal as d, the heights h₁ and h₂ are each half of the other diagonal. So Area = ½ × d₁ × (d₂/2 + d₂/2) = ½ × d₁ × d₂.
