CBSE Class 9 Mathematics Heron's Formula Notes Set 03

Area of a Triangle

With base b and height h, the area of a triangle is ½ × b × h, so h = 2A ÷ b and b = 2A ÷ h.
Equilateral triangle of side a: height (√3/2)a and area (√3/4)a². Isosceles triangle with equal sides a and base b: area ¼ b √(4a² − b²).
Pythagoras: in a right triangle the square on the hypotenuse equals the sum of the squares on the other two sides.

Heron's Formula

When the three sides a, b, c are known but the height is not, let s = (a + b + c) ÷ 2 be the semi-perimeter. Then the area is √(s(s − a)(s − b)(s − c)).
Example: sides 13, 14, 15 give s = 21 and area √(21 × 8 × 7 × 6) = 84.
Combine it with the altitude formula to find a height: h = 2A ÷ b, with the smallest altitude on the longest side.

Area of Quadrilaterals

General quadrilateral with a diagonal d and perpendiculars h1, h2 to it from the other vertices: area = ½ d (h1 + h2).
Perpendicular diagonals d1, d2 (kite or rhombus): area = ½ d1 d2.
Four sides and a diagonal known: split into two triangles and use Heron's formula for each. Cyclic quadrilateral with sides a, b, c, d: area = √((s − a)(s − b)(s − c)(s − d)), where s is the semi-perimeter.

Trapezium, Parallelogram and Rhombus

Trapezium: area = ½ × (sum of parallel sides) × distance between them.
Parallelogram: area = base × corresponding altitude, so side × altitude is the same for both pairs of sides. With a diagonal d and sides a, b the area is 2 × the Heron area of the triangle (a, b, d).
Rhombus: area = ½ d1 d2, side = ½ √(d1² + d2²), perimeter = 2 √(d1² + d2²); given a side a and a diagonal d1, the other diagonal is √(4a² − d1²).

Worked Examples

Example 1: In △PQR, PQ is found from right-angled △PSQ with PS = 12 and SQ = 16, so PQ = 20; with QR = 48 and PR = 52, s = 60 and ar(△PQR) = √(60 × 40 × 12 × 8) = 480 cm². Subtract ar(△PSQ) = 96 to get the shaded area 384 cm².
Example 2: A quadrilateral with sides 9, 40, 28, 15 m and a right angle between the first two: the diagonal is 41 m. Area = ½ × 9 × 40 + √(42 × 1 × 14 × 27) = 180 + 126 = 306 m².
Example 3: A trapezium has parallel sides 25 m and 10 m and slant sides 14 m and 13 m. Draw a line parallel to the 13 m side to cut off a parallelogram and a triangle with sides 14, 13, 15; the triangle has area 84 m² and height 56/5 m, so the parallelogram has area 112 m² and the total is 196 m².
Example 4: The area of a regular hexagon of side a is 6 × (√3/4)a². If it is 24√3 cm², then a = 4 cm and the perimeter is 24 cm.