NCERT Class 9 Solutions: Heron's Formula (Chapter 12) Exercise 12.2 Part 1

  • Previous

Herons formula with semi-perimeter

Herens Formula in Two Parts

Herons formula with semi-perimeter

Q-1 A park, in the shape of a quadrilateral ABCD, has Equation . How much area does it occupy?

Solution:

Angle α Angle α: Angle between D, B, A Angle α Angle α: Angle between D, B, A Angle α Angle α: Angle between D, B, A Segment f Segment f: Segment [A, B] Segment g Segment g: Segment [A, C] Segment h Segment h: Segment [C, D] Segment i Segment i: Segment [D, B] Segment j Segment j: Segment [D, A] Point A A = (-1.2, 0.84) Point A A = (-1.2, 0.84) Point A A = (-1.2, 0.84) Point B B = (3.04, 0.84) Point B B = (3.04, 0.84) Point B B = (3.04, 0.84) Point C C = (-0.5, 4.24) Point C C = (-0.5, 4.24) Point C C = (-0.5, 4.24) Point D D = (3.04, 3.08) Point D D = (3.04, 3.08) Point D D = (3.04, 3.08) 12m text1 = "12m" 9m text2 = "9m" 8m text3 = "8m" 5m text4 = "5m"

Quadrilateral ABCD

Quadrilateral ABCD, ∠B=90°,CA=9m,AB=12m,BD=5mandCD=8

Given a quadrilateral ABCD with, Equation .

Construction: Join diagonal AD

To find the area of quadrilateral we can add the areas of two trianlges. In Equation , by applying Pythagoras theorem,

  • Equation

  • Equation

  • Equation

  • Equation

Also since triangle Equation is right triangle, Area of Equation Now, semi perimeter of Equation Equation

Using heron's formula, area of Equation

  • Equation

  • Equation

  • Equation

  • Equation

  • Equation (approx.)

Area of quadrilateral Equation

Q-2 Find the area of a quadrilateral ABCD in which Equation

Solution:

Quadrilateral poly1 Quadrilateral poly1: Polygon A, B, C, D Segment a Segment a: Segment [A, B] of Quadrilateral poly1 Segment b Segment b: Segment [B, C] of Quadrilateral poly1 Segment c Segment c: Segment [C, D] of Quadrilateral poly1 Segment d Segment d: Segment [D, A] of Quadrilateral poly1 Segment f Segment f: Segment [B, D] Segment g Segment g: Segment [A, C] Point A A = (-2.06, 4.18) Point A A = (-2.06, 4.18) Point A A = (-2.06, 4.18) Point B B = (2.58, 4.18) Point B B = (2.58, 4.18) Point B B = (2.58, 4.18) Point C C = (2.58, 0.14) Point C C = (2.58, 0.14) Point C C = (2.58, 0.14) Point D D = (-0.78, 0.12) Point D D = (-0.78, 0.12) Point D D = (-0.78, 0.12) 4cm text1 = "4cm" 4cm text2 = "4cm" 3cm text3 = "3cm" 5cm text4 = "5cm" 5cm text5 = "5cm"

Quadrilateral ABCD

Quadrilateral ABCD with CD=3cm,CB=4cm,CD=4cm,AD=5cmandBD=5cm.

Given a quadrilateral ABCD, Equation

We can calculate the area of the two triangles by using heroes’ formula on the two triangles. However we can make things simple by first proving that that Equation is a right triangle. Let’s see if it satisfies Pythagoras theorem,

  • Equation

  • Equation

  • Equation

  • Equation

Thus, Equation is a right angled at C. Area of Equation

  • Equation

Now, semi perimeter of Equation Equation

Using heron's formula, area of Equation

  • Equation

  • Equation

  • Equation

  • Equation

  • Equation (approx.)

Finally, area of quadrilateral

  • Equation

Explore Solutions for Mathematics

Sign In