Heron's Formula

Let a, b and c are the sides of triangle and s is the semi-perimeter, then its area is given by:

A = where [Heron’s Formula]

= = 235

A =

A =

Let a, b and c are the sides of triangle and s is the semi-perimeter, then its area is given by:

A = where [Heron’s Formula]

= = 18

A =

A = = 54 cm²

Let the third side of the triangle is

Since the perimeter of a triangle is given by:

cm

Let a, b and c are the sides of triangle and s is the semi-perimeter, then its area is given by:

A = where [Heron’s Formula]

= = 21

A =

A = = cm²

Let a, b and c are the sides of triangle and s is the semi-perimeter, then its area is given by:

A = where [Heron’s Formula]

= = 21

A =

A = = cm²

Area of triangle =

84 =

Altitude = 12 cm

Sides of the triangle are in ratio: 25: 17: 12

Since the perimeter of a triangle is given by:

Therefore sides of the triangle are:

When a, b and c are the sides of the triangle and s is the semiperimeter, then its area is given by:

A = where [Heron’s Formula]

= = 270

A =

A =

= m²

Sides of triangle are in ratio: 3 : 5 : 7

Since the perimeter of a triangle is given by:

Therefore sides of the triangle are:

When a, b and c are the sides of triangle and s is the semi-perimeter, then its area is given by:

A = where [Heron’s Formula]

= = 150

A =

A =

Let the third side of the triangle is

Since the perimeter of a triangle is given by:

dm

Let a, b and c are the sides of triangle and s is the semi-perimeter, then its area is given by:

A = where [Heron’s Formula]

= = 120

A =

A = = dm²

Area of triangle =

1680 =

Altitude = 67.2 dm

Let a, b and c are the sides of triangle and s is the semi-perimeter, then its area is given by:

A = where [Heron’s Formula]

= = 75

A =

A = = cm²

Altitude on side 35 cm:

Area of triangle =

939.15 =

Altitude = 53.66 cm

Altitude on side 54 cm:

Area of triangle =

939.15 =

Altitude = 34.78 cm

Altitude on side 61 cm:

Area of triangle =

939.15 =

Altitude = 30.79 cm

Therefore smallest Altitude is: 30.79 cm

Sides of triangle are in ratio: 3 : 4 : 5

Since the perimeter of a triangle is given by:

Therefore sides of the triangle are:

When a, b and c are the sides of triangle and s is the semi-perimeter, then its area is given by:

A = where [Heron’s Formula]

= = 72

A =

A = = cm²

Area of triangle =

864 =

Altitude = 28.8 cm

Let the equal sides of isosceles triangle is

Base of the triangle =

Since the perimeter of a triangle is given by:

(2

cm

Therefore sides of triangle are:

Let a, b and c are the sides of triangle and s is the semi-perimeter, then its area is given by:

A = where [Heron’s Formula]

= = 21

A =

A = = cm²

Area of triangle =

71.43 =

Altitude = 7.93 cm

In side (AB)² = (AD)² + (BD)²

(AB)² = (12)² + (16)²

AB =

AB = = 20 cm

In

Let a, b and c are the sides of triangle and s is the semi-perimeter, then its area is given by:

A = where [Heron’s Formula]

= = 24

A₁ =

A₁ = = cm²

In

Let a, b and c are the sides of triangle and s is the semi-perimeter, then its area is given by:

A₂ = where

= = 60

A₂ =

A₂ = = cm²

Area of shaded region is A₂-A₁ = 480 – 96 = 384 cm²

Let consider a quadrilateral ABCD

In ∆ABC;

AB = a = 3 cm, BC = b = 4cm, AC = c = 5 cm

Let a, b and c are the sides of triangle and s is

the semi-perimeter, then its area is given by:

A = where [Heron’s Formula]

= = 6

A₁ =

A₁ = = cm²

In ∆ADC;

DA = a = 5 cm, CD = 4 = 4cm, AC = c = 5 cm

Let a, b and c are the sides of triangle and s is

the semi-perimeter, then its area is given by:

A = where

= = 7

A₂ =

A₂ = = cm²

Therefore area of quadrilateral ABCD = A₁ + A₂ = 6+9.16 = 15.16 cm²

Let consider a quadrilateral ABCD

In ∆ADC;

AC = = 25 cm

In ∆ABC

AB = a = 26 cm, BC = b = 27 cm, AC = c = 25 cm

Let a, b and c are the sides of triangle and s is

the semi-perimeter, then its area is given by:

A = where

= = 39

A₁ =

A₁ = = cm²

In ∆ADC;

DA = a = 24 cm, CD = b = 7cm, AC = c = 25 cm

Let a, b and c are the sides of triangle and s is

the semi-perimeter, then its area is given by:

A = where [Heron’s Formula]

= = 28

A₂ =

A₂ = = cm²

Therefore area of quadrilateral ABCD = A₁ + A₂ = 291.85+84 = 375.85 cm²

Let consider a quadrilateral ABCD

In ∆ABC;

AC = = 13 m

AB = a = 5 m, BC = b = 12 m, AC = c = 13 m

Let a, b and c are the sides of triangle and s is

the semi-perimeter, then its area is given by:

A = where [Heron’s Formula]

= = 15

A₁ =

A₁ = = m²

In ∆ADC;

DA = a = 15 m, CD = b = 14 m, AC = c = 13 m

Let a, b and c are the sides of triangle and s is

the semi-perimeter, then its area is given by:

A = where

= = 21

A₂ =

A₂ = = m²

Therefore area of quadrilateral ABCD = A₁ + A₂ = 30+84 = 114 m²

Let consider a quadrilateral ABCD

In ∆BCD;

BD = = 13 m

BC = a = 12 m, CD = b = 5 m, BD = c = 13 m

Let a, b and c are the sides of triangle and s is

the semi-perimeter, then its area is given by:

A = where [Heron’s Formula]

= = 15

A₁ =

A₁ = = m²

In ∆ABD;

AB = a = 9 m, AD = b = 8 m, BD = c = 13 m

Let a, b and c are the sides of triangle and s is

the semi-perimeter, then its area is given by:

A = where

= = 15

A₂ =

A₂ = = m²

Therefore area of quadrilateral ABCD = A₁ + A₂ = 30+35.50 = 65.50 m²

Let ABCD is a trapezium in which AB = 77 cm, BC = 26 cm, CD = 60 cm, DA = 25 cm

Draw CE// AD

Now, ACDE is a parallelograme

BE = AB-DC = 77-60 = 17 cm

In ∆BEC, Let a, b and c are the sides of triangle and

And s be the semi-perimeter, then its area

A = where

= = 34

A=

A = = m²

Therefore area of ∆BCE =

204 × 2 = 17× Altitude

Altitude = 24 cm

Area of trapezium ABCD = =

⇒

Therefore area of trapezium ABCD = 1644 cm²

Let ABCD be the rhombus of perimeter 80 m and diagonal AC = 24 m

We have,

AB + BC + CD + DA = 80

4 AB = 80 [AB=BC=CD=DA sides of Rhombus]

AB = 20 m

In ΔABC, we have

AB = a = 20 cm, BC = b = 20 cm, AC = c = 24 cm

Let a, b and c are the sides of triangle and s is

the semi-perimeter, then its area is given by:

A = where [Heron’s Formula]

= = 32

A=

A = = cm²

Hence, area of rhombus ABCD = 2 ×192 m²

= 384 m²

Since the sides of a rhombus are equal there fore each side = = = 8 m

Let a, b and c are the sides of triangle and s is

the semi-perimeter, then its area is given by:

A = where [Heron’s Formula]

= = 13

A=

A = = cm²

Hence, area of rhombus ABCD = 2 ×31.22 m²

= 62.44 m²

Total painting area of rhombus = 62.44 × 2 = 124.88 m²

Cost of painting of rhombus on both sides = 124.88 × 5 = Rs 624.50

Let ABCD is a quadrilateral in which AD = 24 cm and ∆BCD is an equilateral.

In right angled ∆BAD applying pythagorous theorem:

(BD)² = (AB)² + (AD)²

(26)² = (AB)² + (24)²

AB = 10 cm

Area of right angled ∆BAD =

⇒ = 120 cm²

Now in equilateral ∆BCD

Let a, b and c are the sides of triangle and s is

the semi-perimeter, then its area is given by:

A = where [Heron’s Formula]

= = 39

A=

A = = cm²

Hence, area of quad ABCD = 120+292.72 =412.72 cm²

In ∆ABD;

AB = a = 42 cm, BD = b = 20 cm, DA = c = 34 cm

Let a, b and c are the sides of triangle and s is

the semi-perimeter, then its area is given by:

A = where

= = 48

A₁ =

A₁ = = m²

In ∆BCD;

BC = a = 21 cm, DC = b = 29 cm, BD = c = 20 cm

Let a, b and c are the sides of triangle and s is

the semi-perimeter, then its area is given by:

A = where

= = 35

A₂ =

A₂ = = m²

Therefore area of quadrilateral ABCD = A₁ + A₂ = 336+210 = 546 cm²

In right ∆ACB using pythagorous theorem:

(AB)² = (AC)² + (BC)²

(17)² = (15)² + (BC)²

BC = 8 cm

Perimeter of quad ABCD = AB+BC+CD+DA = 17+8+12+9 = 46 cm

Area of right angled ∆ACB =

⇒ = 60 cm²

Now in equilateral ∆ACD

Let a, b and c are the sides of triangle and s is

the semi-perimeter, then its area is given by:

A = where [Heron’s Formula]

= = 18

A=

A = = cm²

Hence, area of quad ABCD =60+54 =114 cm²

Now in ∆ABC

AB = a = 34 cm, BC = b = 20 cm, AC = c = 42 cm

Let a, b and c are the sides of triangle and s is

the semi-perimeter, then its area is given by:

A = where

= = 48

A=

A = = cm²

Hence, area of Parallelograme ABCD =2× 336 =672 cm²

Let a, b and c are the sides of blade and s is

the semi-perimeter, then its area is given by:

A = where [Heron’s Formula]

= = 5.5

A=

A = = cm²

Hence, total area of both the blades = 2 × 2.49 = 4.98 cm²

Let a, b and c are the sides of triangular strips and s is

the semi-perimeter, then its area is given by:

A = where [Heron’s Formula]

= = 32

A=

A = = cm²

Hence, total area of 5 Nos of triangular strips of one type = 5 × 168 = 840 cm²

Let a, b and c are the sides of triangle and s is

the semi-perimeter, then its area is given by:

A = where [Heron’s Formula]

= = 21

A=

A = = cm²

Therefore area of ∆ =

84 × 2 = 14× Altitude

Altitude = 12 cm

Base = 5 cm, Altitude = 4 cm

Area of ∆ =

Area of ∆ = = 10 cm²

Let a, b and c are the sides of triangle and s is the semi-perimeter, then its area is given by:

A = where [Heron’s Formula]

= = 40

A =

A = = 240 cm²

Base = 30 cm, Altitude = 30 cm

Area of ∆ =

Area of ∆ = = 450 cm²

Let a, b and c are the sides of triangle and s is

the semi-perimeter, then its area is given by:

A = where [Heron’s Formula]

= = 6

A=

A = = cm²

In ∆ABC, AB = ………………….Given

Since , and are the sides of an isosceles triangle and s is

the semi-perimeter, then its area is given by:

A = where [Heron’s Formula]

=

A=

A = = cm²

Let a, b and c are the sides of triangle and s is the semi-perimeter, then its area is given by:

A = where [Heron’s Formula]

= = 15

A =

A = = 12√5 cm²

Area of an equilateral triangle = = = 4√3 cm²

Let a, b and c are the sides of triangle and s is the semi-perimeter, then its area is given by:

A = where [Heron’s Formula]

= = 375

A =

A = = 18750 cm²

Area of an equilateral triangle = = = cm²

Let a, b and c are the sides of triangle and s is the semi-perimeter, then its area is given by:

A = where [Heron’s Formula]

= =120

A =

A = = cm²

Area of triangle =

1680 =

Altitude = 30 cm

Sides of triangle are in ratio: 3 : 4 : 5

Since the perimeter of a triangle is given by:

Therefore sides of the triangle are:

When a, b and c are the sides of triangle and s is the semi-perimeter, then its area is given by:

A = where [Heron’s Formula]

= = 72

A =

A = = cm²

Let a, b and c are the sides of triangle and s is the semi-perimeter, then its area is given by:

A = where [Heron’s Formula]

= = 66

A =

A = = m²

Area of triangle =

330 =

Altitude = 60 m

Let each side of equilateral triangle is a cm

Using pythagorous theorem in right ∆ADB

(AB)² = (AD)² + (BD)²

()² = ()² + ()²

= cm

Area of an equilateral triangle = = = = = cm²

Let a, b and c are the sides of triangle and s is the semi-perimeter, then its area is given by:

A = where [Heron’s Formula]

= = 21

A =

A = = cm²

Area of triangle =

=

Altitude = cm

When each side of triangle = a

When a, b and c are the sides of triangle and s is the semi-perimeter, then its area is given by:

A = where [Heron’s Formula]

A’ = where [Heron’s Formula]

When each side of triangle = 2a

= = 2

A’ =

A’ = = 4= 4A

In right isosceles ∆ABC

Area of triangle =

8 =

(AC)² = (AB)² + (BC)²

Perimeter of ∆ABC = AB + BC + CA = 4 + 4 + 4√2 = 8 + 4√2 cm

Let the sides of triangle are a, b, c

When a, b and c are the sides of triangle and s is the semi-perimeter, then its area is given by:

A = where [Heron’s Formula]

A’ = where [Heron’s Formula]

When each side of triangle is doubled

= = 2

A’ =

A’ = = 4= 4A

Increase in area = 4A-A = 3A

Percentage increase in area = = = 300%

Perimeter of an equilateral triangle with side 9 cm = 9× 3 = 27 cm

Let the sides of ∆ABC are: AB = (Since the sides of ∆ABC are consecutive intergers)

Perimeter of ∆ABC = Perimeter of equilateral triangle

When each side of triangle = a

Area of an equilateral triangle = = A

When each side of triangle = 3a

Area of an equilateral triangle = = = 9A

Increase in area = 9A-A = 8A

Percentage increase in area = = = 800%

In ∆ABC

AD:DC = 3:2

Area of ∆ABD = =

Area of ∆BDC = =

Area of ∆ABC = area of ∆ABD+ area of ∆BDC = 40 cm²

Therefore area of ∆ABD =

Therefore area of ∆BDC =

Using Pythagorous theorem:

()² = ² + (Altitude)²

(Altitude)² = 169 – 25 = 144

Altitude = 12 cm

Area of triangle =

Median of an equilateral =

In an equilateral triangle median is an altitude.

Let the side of triangle be

(AB)² = (AD)² + (BD)²

()² = ()² + ()² C

= cm

Area of an equilateral triangle = = = = = cm²

Area of an equilateral triangle =

Let side of equilateral triangle =

Area of an equilateral triangle = =

⇒

⇒

Let the sides of triangle are a, b, c

When a, b and c are the sides of triangle and s is the semi-perimeter, then its area is given by:

A = where [Heron’s Formula]

A’ = where [Heron’s Formula]

When each side of triangle is doubled

= = 2

A’ =

A’ = = 4= 4A

Increase in area = 4A-A = 3A

Percentage increase in area = = = 300%

Let each side of square = and each side of an equilateral triangle =

Perimeter of square = perimeter of an equilateral triangle

…………(1)

Diagonal of square = 12√2 cm

Therefore using Pythagorous theorem:

Therefore substituting value of y in equation (1) we get:

Area of an equilateral triangle = =