**Why Heron’s Formula…?**

**When the triangle is a right angled triangle:**

** **

Area =

Example:

Now, Area of

**When the triangle is an equilateral triangle:**

Consider an equilateral triangle ABC with each side = 10cm

If we drop AM perpendicular to BC then BM = 5cm,

Now, AB^{2 }= AM^{2 }+ BM^{2 }(Pythagoras Theorem)

AM^{2} = 10^{2} – 5^{2}

Therefore, AM =

Now, Area for triangle ABC = 2×(Area of triangle ABM)

And,

Therefore,

^{2}

Now, Area for triangle ABC = 2×(^{2}

**Therefore, AreaofΔABC=253–√cm2**

**When the triangle is an isosceles triangle: **

Consider an isosceles triangle ABC with equal sides AB = AC = 13 cm and unequal side BC = 10cm

Drop AM perpendicular to BC. Therefore CM = 5cm

Now, AC^{2 }= AM^{2 }+ CM^{2 }(Pythagoras Theorem)

AM^{2} = 13^{2} – 5^{2}

Therefore AM = 12cm

Now, Area for triangle ABC = 2×(Area of triangle ACM)

And,

Therefore,

^{2}

Now, Area for triangle ABC = 2× 30 cm^{2}

**Therefore, Area of triangle ABC = 60cm ^{2}**

**When the triangle is a scalene triangle:**

Suppose that we know the lengths of sides of a scalene triangle and we have to find its area. Now, for applying the above formula we need to calculate its height. But in this case we have no idea about how to calculate the height of a scalene triangle and that is why it’s very difficult to find its area. Thus, Heron gave a formula to find the area of any triangle known as **Heron’s Formula**.

**Area of a Triangle ABC=s(s−a)(s−b)(s−c)−−−−−−−−−−−−−−−−−√**

where a, b and c are sides of the triangle.

**s = semi perimeter (half of perimeter of triangle)**

**Finding square roots:**

** (i) ****Find values of: 72895−−−−−√**

** **

**(ii)Find values of 745−−−√**

** **

**Q.1: The lengths of two sides of a triangle ABC are 10cm and 7cm and its perimeter is 28cm. Find the area of a triangle ABC.**

**Sol. **

Given perimeter = 28cm

Therefore, **s = perimeter2 = 14cm**

And a = 10cm, b =7cm and c = 11cm (Since, perimeter = a + b + c)

** On applying Heron’s Formula;**

Therefore, area of Triangle ABC

=

(or) =

=

**Therefore, area of triangle ABC = 146–√cm ^{2} **

**Q.2: If perimeter of a triangular plot is 250m and the lengths of the sides of a triangular plot are in the ratio of 4 : 7 : 9. Find its area.**

**Sol. **

Since, the lengths of the sides of a triangular plot are in the ratio of 4 : 7 : 9

Therefore, sides of triangular plot will be 4x, 7x and 9x

Given perimeter = 250m

Therefore, 4x + 7x + 9x = 250

Hence x = 12.5, that gives:

a = 4x = 50m, b = 7x = 87.5m and c = 9x = 112.5m

And **s = perimeter2= 125m**

**Now on applying Heron’s Formula;**

Therefore, area of plot ABC:

=

(or) =

=

=2096.3m^{2}

**Therefore, area of triangular plot ABC = ****2096.3****m ^{2} **

**Q.3: There is a triangular park in a locality having sides 110m, 90m and 40m. A gardener Fauzia has to plant grass inside. In how much area does she need to plant grass? Also Fauzia has to put a fence all around that triangular park. Find the cost of fencing that triangular park with barbed wire if the cost of barbed wire is Rs 25 per meter and also a space of 2m is to be left for a gate on any one side of that park.**

**Sol.**

Given a = 110m, b = 90m and c = 40m

Therefore, perimeter = 110 + 90 + 40 = 240m

And **s = perimeter2= 120m**

**Now applying Heron’s Formula;**

Therefore, area of park ABC:

=

(or) =

=

= 1697.07m^{2} (Approx.)

**Therefore, area of triangular plot ABC = ****1697.07****m ^{2} (Approx.)**

Since, perimeter = 240m

Therefore for fencing length of wire needed = 240 – 2 = 238m

**So the total cost of fencing the triangular park = 238×25 = 5950Rs **

**Q.4: The side walls of a flyover are triangular in shape and are used for advertisements. Dimensions of side walls are: 120m, 140m and 50m. The cost of posting advertisements is 4500Rs per m ^{2} per year. How much rent an automobile company paid if the company hired one of its walls for 4 months?**

**Sol**.

Given a = 120m, b = 140m and c = 50m

Therefore, perimeter of each triangular wall = 120 + 140 + 50 = 310m

And s =

**Now applying Heron’s Formula;**

Therefore, area of each triangular wall:

=

(or) =

=

**Therefore, area of each triangular wall = 52531−−√m ^{2}**

Since, rent of posting advertisements is 4500 per m^{2} per year (12 months)

Therefore** rent for 52531−−√m ^{2} area = 52531−−√ × 4500**

=

Now, For 12 months’ rent =

**Hence, rent paid by company for hiring one of its walls for 4 months = 78750031−−√Rs.**

** **

**Q.5: If each side of a triangle is increased by six times then find the total percentage of increase in its area.**

**Sol. **

**Let a, b and c be the sides of the original triangle.** Then,

Perimeter of original triangle = a + b + c and s_{1} =

**And sides of the new triangle will be 6a, 6b and 6c.** Then,

Perimeter of new triangle = 6a + 6b + 6c and s_{2} = _{1 }(Since, 2s_{1 }= a + b + c)

Let A_{1} and A_{2} be the areas of the original and new triangle respectively. Then,

** Area of triangle A _{1}= s1(s1–a)(s1–b)(s1–c)−−−−−−−−−−−−−−−−−√**

And **Area of triangle A _{2}= s2(s2–a)(s2–b)(s2–c)−−−−−−−−−−−−−−−−−√
**

Since, **s _{2} = 6s_{1}**

Therefore, Area of triangle A_{2}=

Area of triangle A_{2}=

Therefore,** A _{2} = 12A_{1}**

Therefore, increase in the area of triangle = A_{2 }– A_{1 }

= 12A_{1 }– A_{1} = 11A_{1}

**Therefore, percentage increase in area = 11A1A1×100 = 1100%**

**Hence, area of the new triangle is increased by 1100%**

** **

**Q.6: If perimeter of a triangular plot is ****490**** and the lengths of the sides of a triangular plot are in the ratio of ****22**** : ****16**** : ****32****. Find its area.**

**Sol. **

Since, the lengths of the sides of a triangular plot are in the ratio of 22 : 16 : 32

Therefore, sides of triangular plot will be 22x, 16x and 32x

Given perimeter = 490m

Therefore, 22x + 16x + 32x = 490

Hence x =7 , that gives:

a =22x = 154m, b = 16x = 112m and c = 32x = 224m

Therefore perimeter = a + b + c = 490m

And s =

**Now on applying Heron’s Formula;**

Therefore, area of plot ABC:

=

(or) =

** = 7891.13m ^{2 }(Approx.)**

**Therefore, area of triangular plot ABC = ****7891.13 ****m ^{2} (Approx.)**

**Q.7: The length of**** equal ****sides of a****n isosceles**** triangle ABC ****is** **8****cm and its perimeter is 28cm. Find the area of a triangle ABC.**

**Sol. **

Given perimeter = 28cm

Therefore, **s = perimeter2 = 14cm**

And a = 8cm, b =8cm and c = 12m (Since, perimeter = a + b + c)

**On applying Heron’s Formula;**

Therefore, area of Triangle ABC

=

(or) =

=

=31.75cm^{2} **(Approx.)**

**Therefore, area of triangle ABC ****= 31.75cm ^{2}**

**(Approx.)**