Let ABC be a triangle such that AB=13 BC=14 and CA=15. D is a point on BC such that AD Bisects

Area of triangle ADC  is 54 square unit

Step-by-step explanation:

Here is the complete question:

Let ABC be a triangle such that AB=13, BC=14, and CA=15. D is a point on BC such that AD bisects angle A. Find the area of triangle ADC .

Step-by-step explanation:

Please see the attachment below for an illustrative diagram

Considering the diagram,

BC = BD + DC = 14

Let BD be ; hence, DC will be

and AD be

To, find the area of triangle ADC

Area of triangle ADC  =

=

We will have to determine and

First we will find the area of triangle ABC

The area of triangle ABC can be determined using the Heron's formula.

Given a triangle with a,b, and c

Where

For the given triangle ABC

Let = AB, = BC, and = CA

Hence, and

∴

Then,

Area of triangle ABC =

Area of triangle ABC = =

Area of triangle ABC = 84 square unit

Now, considering the diagram

Area of triangle ABC = Area of triangle ADB + Area of triangle ADC

Area of triangle ADB =

Area of triangle ADB =

Hence,

Area of triangle ABC =   +

84 =   +

∴

∴

Hence, AD = 12

Now, we can find BD

Considering triangle ADB,

From Pythagorean theorem,

/AB/² = /AD/² + /BD/²

∴13² = 12² + /BD/²

/BD/² = 169 - 144

/BD/ =

/BD/ = 5

But, BD + DC = 14

Then, DC = 14 - BD = 14 - 5

BD = 9

Now, we can find the area of triangle ADC

Area of triangle ADC  =

Area of triangle ADC  =

Area of triangle ADC  = 9 × 6

Area of triangle ADC  = 54 square unit

Hence, Area of triangle ADC  is 54 square unit.

11) obtuse

13)right

15) acute