## Area of a triangle

There are so many different approaches for finding the area of a triangle. A suitable result is applicable as per the availability of data.

**General formula**

**Heron's formula**

**Trigonometric application**

A, B and C → measure of three angles

**Application of circum radius (R)**

Area = abc/4r

a, b and c → lengths of three sides

R → circum radius

**Application of in-radius (r)**

Area = r.S

r → in-radius

S → semi perimeter

**Equilateral Triangle**

Area = √3 / 4 * a^{2}

a → length of each side.

Height of an equilateral triangle = √3/4 * a

**Isosceles triangle**

Area = b/4 * √(4a^{2} - b^{2})

a→ length of equal sides

b → length of third side

**Important theorems and useful results**

**Result 1:**

If AB/BC= m/n and AE/ED = p/q

then; Area of ΔABE/Area of ΔACD = pm/(p+q)(m+n)

Example:

In the given diagram ΔBDF is inscribed in ΔACE.

AB/BC= 12, CD/DE = 35 and AF/FE = 1/4

Find the ratio of the area of shaded region to the area of non-shaded region.

Solution:

Area of ΔABF/Area of ΔACE = 1*1/((1+2)(1+4)) = 1/15

Area of ΔBCD/Area of ΔACE= 2*3/((2+1)(3+5)) = 14

Area of ΔDEF/Area of ΔACE= 5*4/((5+3)(4+1)) = 12

Area of (ΔABF + ΔBCD + ΔDEF)/Area of ΔACE= 1/15+ 1/4 + 1/2= 49/60

i.e. Area of non-shaded region = 49/60 * Area of ΔACE

i.e. Area of shaded region = Area of ΔACE - Area of non-shaded region

Area of shaded region = [1- 49/60]Area of ΔACE = 11/60 * Area of ΔACE

Area of shaded region/Area of non-shaded region= 11/49

Hence, the required ratio = 11 : 49

**Result 2: Properties of right triangles**

ΔABC is a right angled triangle, AD is the altitude drawn from vertex A to hypotenuse BC.

ΔADB ΔCDA ΔCAB (similar triangles)

AD/CD= BD/AD

AD^{2} = BD * CD |

AB/BC= BD/AB

AB^{2} = BD * BC |

AC/CD= BC/AC

AC^{2} = BC * CD |

Examples:

- In the given diagram AB = 6 cm, AC = 8 cm. Find AD.

Solution:

AB = 6 cm and AC = 8 cm

BC = 10 cm

Area of ΔABC = 1/2 (AC*AB) = 1/2 (BC *AD)

AC*AB = BC *AD

8 * 6 = 10 * AD

AD = 4.8 cm.

- In right triangle ABC, AB : BC = 1: . Find AD : CD.

Solution:

Let AB = 1 and BC = √3

AC = √(1^{2} + √3^{2}) = 2

Let AD = a

CD = 2 - a

AB2 = AD * AC

1^{2} = a * 2

a = 1/2

i.e. AD = 1/2

CD = 2 - 1/2 = 3/2

AD : CD = 1/2 : 3/2

= 1 : 3

**Result 3: Internal angle bisector theorem**

In ΔABC, AD is the angle bisector of ∠A, then AD/AC = BD/CD and BD * AC - AB * CD = AD^{2}

**Result 4: External angle bisector theorem**

BE is the angle bisector of ∠CBD. then BC/AB = CE/AE

**Result 5: Apollonius theorem**

In triangle ABC, AD is the median from vertex A to side BC.

AB^{2} + AC^{2} = 2(AD^{2} + BD^{2})

Example:

Lengths of the diagonals of a parallelogram are 20cm and 24 cm. if one of its sides is 16 cm, find the length other non-parallel side.

Solution:

Diagonals of a parallelogram are bisecting each other. Therefore E is the midpoint of BD (and AE).

In ΔABD, AE is a median.

As per Apollonius theorem;

AB^{2} + AD^{2} = 2(AE^{2} + BE^{2})

16^{2} + x^{2} = 2(12^{2} + 10^{2})

x^{2} = 232

x = 2√58

Hence, the length of the other non-parallel side = 2√58 cm