Important Formulas, Rules & Key Points

• Heron's Formula: Area of a triangle = √(s(s - a)(s - b)(s - c))
• Semi-Perimeter (s): s = (a + b + c) / 2
• Area of Right-angled Triangle: Area = 1/2 × Base × Height
• Quadrilateral Area Strategy: Divide the quadrilateral into two triangles using a diagonal, calculate the area of each triangle separately, and add them together.

Question 1:

A park, in the shape of a quadrilateral ABCD, has ∠C = 90°, AB = 9 m, BC = 12 m, CD = 5 m, and AD = 8 m. How much area does it occupy?

Solution:

1. Join diagonal BD. Triangle BCD forms a right-angled triangle at C.
2. Using Pythagoras theorem in ΔBCD: BD² = BC² + CD² = 12² + 5² = 144 + 25 = 169 ⇒ BD = 13 m.
3. Area of ΔBCD: 1/2 × Base × Height = 1/2 × 12 × 5 = 30 m².
4. For ΔABD: sides are a = 9 m, b = 8 m, c = 13 m.
5. Semi-perimeter s = (9 + 8 + 13)/2 = 30/2 = 15 m.
6. Area of ΔABD: √(15(15-9)(15-8)(15-13)) = √(15 × 6 × 7 × 2) = √(1260) ≈ 35.5 m².
7. Total Area of Quadrilateral ABCD: Area(ΔBCD) + Area(ΔABD) = 30 + 35.5 = 65.5 m² (approx).

Question 2:

Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm, and AC = 5 cm.

Solution:

1. Quadrilateral ABCD is divided into two triangles: ΔABC and ΔACD by diagonal AC.
2. In ΔABC, sides are 3 cm, 4 cm, and 5 cm. Since 3² + 4² = 5² (9 + 16 = 25), ΔABC is a right triangle at B.
3. Area of ΔABC: 1/2 × 3 × 4 = 6 cm².
4. In ΔACD, sides are a = 5 cm, b = 4 cm, c = 5 cm.
5. s = (5 + 4 + 5)/2 = 14/2 = 7 cm.
6. Area of ΔACD: √(7(7-5)(7-4)(7-5)) = √(7 × 2 × 3 × 2) = 2√21 ≈ 2 × 4.58 = 9.16 cm².
7. Total Area of Quadrilateral ABCD: 6 + 9.16 = 15.16 cm² (approx).

Aptitude Corner (Competitive Exam Shortcut)

When dealing with standard Pythagorean triplets like (3, 4, 5) or (5, 12, 13), directly apply Area = 1/2 × base × height to save crucial time during competitive exams instead of using Heron's Formula!

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