Important Formulas, Rules & Key Points

• Semi-perimeter (s): s = (a + b + c) / 2
• Heron's Formula: Area = √[s(s - a)(s - b)(s - c)]
• Area of Right Triangle: Area = 1/2 × Base × Height
• Quadrilateral Triangulation: A quadrilateral can be split into two triangles by a diagonal to find its total area.

Question 1

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:

The diagonal AC divides quadrilateral ABCD into two triangles: ΔABC and ΔADC.

For ΔABC: Sides are a = 3 cm, b = 4 cm, c = 5 cm.
Since 3² + 4² = 9 + 16 = 25 = 5², ΔABC is a right-angled triangle at B.
Area(ΔABC) = 1/2 × Base × Height = 1/2 × 3 × 4 = 6 cm².

For ΔADC: Sides are a = 5 cm, b = 5 cm, c = 4 cm.
Semi-perimeter (s) = (5 + 5 + 4) / 2 = 7 cm.
Area(ΔADC) = √[7(7 - 5)(7 - 5)(7 - 4)] = √[7 × 2 × 2 × 3] = 2√21 cm² ≈ 2 × 4.58 = 9.16 cm².

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

Question 2

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:

Join diagonal BD. In right-angled ΔBCD (∠C = 90°):
By Pythagoras theorem, BD² = BC² + CD² = 12² + 5² = 144 + 25 = 169
BD = √169 = 13 m.
Area(ΔBCD) = 1/2 × Base × Height = 1/2 × 12 × 5 = 30 m².

Now for ΔABD, sides are a = 9 m, b = 8 m, c = 13 m.
Semi-perimeter (s) = (9 + 8 + 13) / 2 = 15 m.
Area(ΔABD) = √[15(15 - 9)(15 - 8)(15 - 13)] = √[15 × 6 × 7 × 2] = √1260 ≈ 35.5 m².

Total Area of Park ABCD = Area(ΔBCD) + Area(ΔABD) = 30 + 35.5 = 65.5 m² (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:

A diagonal divides the rhombus into two congruent triangles.
Sides of one triangle are a = 30 m, b = 30 m, c = 48 m.
Semi-perimeter (s) = (30 + 30 + 48) / 2 = 54 m.
Area of one triangle = √[54(54 - 30)(54 - 30)(54 - 48)] = √[54 × 24 × 24 × 6] = √[9 × 6 × 24 × 24 × 6] = 3 × 6 × 24 = 432 m².

Total Area of Rhombus Field = 2 × 432 = 864 m².
Area available for 18 cows = 864 m².
Area each cow gets = 864 / 18 = 48 m².

Aptitude Corner & Shortcuts

Pythagorean Triples Shortcut: When solving quadrilateral problems, always check if a triangle formed by a diagonal is a right-angled triangle using Pythagorean triples like (3, 4, 5) or (5, 12, 13). If it is, use Area = 1/2 × base × height instead of Heron's formula to save significant time during exams!

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