Herens Formula in Two Parts

Herons formula with semi-perimeter

Q-1 A park, in the shape of a quadrilateral ABCD, has ∠ B = 90 ° , C A = 9 m , A B = 12 m , B D = 5 m a n d C D = 8 m . How much area does it occupy?

Solution:

Angle Î± Angle Î±: Angle between D, B, A Angle Î± Angle Î±: Angle between D, B, A Angle Î± Angle Î±: 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, ∠ B = 90 ° , C A = 9 m , A B = 12 m , B D = 5 m a n d C D = 8 m .

To find the area of quadrilateral we can add the areas of two trianlges. In Δ A B D , by applying Pythagoras theorem,

A D 2 = A B 2 + B D 2

A D 2 = 12 2 + 5 2

A D 2 = 169

A D = 13 m

Also since triangle Δ A B D is right triangle, Area of Δ A B D = 1 2 × 12 × 5 = 30 m 2 Now, semi perimeter of Δ C A D 8 + 9 + 13 2 m = 30 2 m = 15 m

Using heron's formula, area of Δ C A D

s ( s − a ) ( s − b ) ( s − c )

15 ( 15 − 13 ) ( 15 − 9 ) ( 15 − 8 ) m 2

15 × 2 × 6 × 7 m 2

6 35 m 2

35.49 m 2 (approx.)

Area of quadrilateral A B C D = A r e a o f Δ A B D + A r e a o f Δ C A D = 30 m 2 + 35.49 m 2 = 65.49 m 2

Q-2 Find the area of a quadrilateral ABCD in which C D = 3 c m , C B = 4 c m , C D = 4 c m , A D = 5 c m a n d B D = 5 c m .

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, C D = 3 c m , C B = 4 c m , C D = 4 c m , A D = 5 c m a n d B D = 5 c m .

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 Δ D C B is a right triangle. Let’s see if it satisfies Pythagoras theorem,

B D 2 = D C 2 + C B 2

5 2 = 3 2 + 4 2

25 = 9 + 16

25 = 25

Thus, Δ D C B is a right angled at C. Area of Δ C B A

Now, semi perimeter of Δ D B A = 5 + 5 + 4 2 c m = 14 2 c m = 7 m

Using heron's formula, area of Δ D C A

= s ( s − a ) ( s − b ) ( s − c )

= 7 ( 7 − 5 ) ( 7 − 5 ) ( 7 − 4 ) c m 2

= 7 × 2 × 2 × 3 c m 2

= 2 21 c m 2

≈ 9.17 c m 2 (approx.)

Finally, area of quadrilateral