Application Of Heron’s Formula

Area of parallelogram ABCD = Area of triangle ABD + Area of triangle BCD
= 2* Area of triangle ABD:
s = (3+4+6)/2 = 7.5 cm
Area = 2√ 7.5(7.5-3)(7.5-4)(7.5-6) = 26.62 sq. cm.

s = 30/2 = 15
Third side = 30 – 12 – 12 =6
Area = √ 15(15-12)(15-12)(15-6) = 9√15 sq.cm

Area of square ABCD = Area of triangle ABC + Area of triangle ACD = 2 * Area of triangle ABC
s = (5 + 5 + 5 √ 2 )/2 = (10+5 √ 2)/2
Area = 2 √ ((10+5 √ 2)/2 ((10+5 √ 2)/2 – 5)((10+5 √ 2)/2 – 5)((10+5 √ 2)/2 – 5 √ 2)) = 25 sq. cm.

Area of Quadrilateral ABCD = Area of triangle ABC + Area of triangle ACD
Area of triangle ABC:
s = (6+6.5+8)/2 = 10.25
Area = √ 10.25(10.25-6)(10.25-6.5)(10.25-8) = 19.17 sq. m.
Area of triangle ACD:
s = (7+5+8)/2 = 10
Area = √10(10-7)(10-5)(10-8) = 17.32 sq. m.
Area of Quadrilateral ABCD = 36.5 sq. m.

Area of Trapezoid ABCD = Area of triangle ABC + Area of triangle ACD
Area of ACD
s = (20+21+29)/2 = 35
Area = √ 35(35-20)(35-21)(35-29) = 210 sq. units
Area of ABC
s = (34+42+20)/2 = 48
Area = √ 48(48-20)(48-42)(48-34) = 336 sq. units
Area = 546 sq. units

Let the sides of the Rhombus be x.
Area of Rhombus ABCD = Area of triangle ABC + Area of triangle ACD = 2 * Area of triangle ABC
s =( x+x+4)/2 = x+2
Area = 2 √ x+2 (x+2-4)(x+2-x)(x+2-x) = 16 sq. cm.
x = 2 √ 5 cm

Let the sides of the triangle be 5x, 12x, 13x
given 5x+12x+13x =150
x = 5
s = (25+60+65)/2 = 75
Area of triangle ABC = √ 75(75-25)(75-60)(75-65) = 750

Side = 136/4 = 34
Area of Rhombus ABCD = Area of triangle ABC + Area of triangle ACD = 2 * Area of triangle ABC
s = (34+34+60)/2 = 64
Area = 2 √ 64(64-34)(64-34)(64-60) = 960 sq. cm

Sides of the equilateral triangle = 23.5
s = 23.5* 3 / 2 = 35.25
Area = √ 35.25(35.25-23.5)^3 = 239 sq. inches

Area of rectangle ABCD = Area of triangle ABC + Area of triangle ACD = 2 * Area of triangle ABC
s = (9+12+15)/2 = 18
Area = 2 √ 18(18-15)(18-9)(18-12) = 108 sq. units
Cost = 108 * 10.5 = $ 1134