Important Formulas, Rules & Key Points

• Heron's Formula: Area = √(s(s - a)(s - b)(s - c))
• Semi-perimeter (s): s = (a + b + c) / 2
• Area of Right Triangle: Area = 1/2 × Base × Height
• Pythagoras Theorem: Hypotenuse² = Base² + Perpendicular²
• Quadrilateral Area Rule: Divide non-standard quadrilaterals into two triangles using a diagonal, then find the sum of their individual areas.

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 to divide quadrilateral ABCD into two triangles: ΔBCD and ΔABD.

2. In right-angled triangle BCD (∠C = 90°):
Using Pythagoras Theorem: BD² = BC² + CD² = 12² + 5² = 144 + 25 = 169
⇒ BD = √169 = 13 m.
Area of ΔBCD = 1/2 × Base × Height = 1/2 × 12 × 5 = 30 m².

3. For ΔABD with sides a = 9 m, b = 8 m, c = 13 m:
Semi-perimeter, s = (9 + 8 + 13) / 2 = 30 / 2 = 15 m.
Area of ΔABD = √(15(15 - 9)(15 - 8)(15 - 13)) = √(15 × 6 × 7 × 2) = √1260 ≈ 35.5 m².

4. Total Area of Park 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. Consider ΔABC with sides AB = 3 cm, BC = 4 cm, AC = 5 cm.
Since 3² + 4² = 9 + 16 = 25 = 5², ΔABC is a right triangle at B.
Area of ΔABC = 1/2 × 3 × 4 = 6 cm².

2. Consider ΔADC with sides AD = 5 cm, CD = 4 cm, AC = 5 cm.
Semi-perimeter, s = (5 + 4 + 5) / 2 = 14 / 2 = 7 cm.
Area of ΔADC = √(7(7 - 5)(7 - 4)(7 - 5)) = √(7 × 2 × 3 × 2) = 2√21 ≈ 2 × 4.58 = 9.16 cm².

3. Total Area of Quadrilateral ABCD = Area(ΔABC) + Area(ΔADC) = 6 + 9.16 = 15.16 cm² (approx).

Question 3:

A rhombus-shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting?

Solution:

1. The diagonal divides the rhombus into two congruent triangles, each having sides a = 30 m, b = 30 m, and c = 48 m.

2. For one triangle:
Semi-perimeter, s = (30 + 30 + 48) / 2 = 108 / 2 = 54 m.
Area = √(54(54 - 30)(54 - 30)(54 - 48)) = √(54 × 24 × 24 × 6) = √(324 × 24²) = 18 × 24 = 432 m².

3. Total area of rhombus field = 2 × 432 = 864 m².

4. Area available for each cow = Total Area / 18 = 864 / 18 = 48 m².

Aptitude Corner: Quick Exam Tricks

Shortcut for Rhombus Area: If the diagonals d₁ and d₂ of a rhombus are known, the area is simply (1/2) × d₁ × d₂. If only one diagonal and side are given, use Pythagoras' theorem on the right-angled quarter triangles (since diagonals bisect at 90°) to find the second diagonal quickly!

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