Answer:
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 ![y](https://tex.z-dn.net/?f=y)
To, find the area of triangle ADC
Area of triangle ADC = ![\frac{1}{2} (DC)(AD)](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20%28DC%29%28AD%29)
= ![\frac{1}{2}(14-x)(y)](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%2814-x%29%28y%29)
We will have to determine
and ![y](https://tex.z-dn.net/?f=y)
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
![Area =\sqrt{s(s-a)(s-b)(s-c)}](https://tex.z-dn.net/?f=Area%20%3D%5Csqrt%7Bs%28s-a%29%28s-b%29%28s-c%29%7D)
Where ![s = \frac{a+b+c}{2}](https://tex.z-dn.net/?f=s%20%3D%20%5Cfrac%7Ba%2Bb%2Bc%7D%7B2%7D)
For the given triangle ABC
Let
= AB,
= BC, and
= CA
Hence,
and ![c = 15](https://tex.z-dn.net/?f=c%20%3D%2015)
∴ ![s = \frac{13+14+15}{2} \\s= \frac{42}{2}\\s = 21](https://tex.z-dn.net/?f=s%20%3D%20%5Cfrac%7B13%2B14%2B15%7D%7B2%7D%20%5C%5Cs%3D%20%5Cfrac%7B42%7D%7B2%7D%5C%5Cs%20%3D%2021)
Then,
Area of triangle ABC = ![\sqrt{(21)(21-13)(21-14)(21-15)}](https://tex.z-dn.net/?f=%5Csqrt%7B%2821%29%2821-13%29%2821-14%29%2821-15%29%7D)
Area of triangle ABC =
= ![\sqrt{7056}](https://tex.z-dn.net/?f=%5Csqrt%7B7056%7D)
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 = ![\frac{1}{2} (BD)(AD)](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20%28BD%29%28AD%29)
Area of triangle ADB = ![\frac{1}{2}(x)(y)](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%28x%29%28y%29)
Hence,
Area of triangle ABC =
+ ![\frac{1}{2}(14-x)(y)](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%2814-x%29%28y%29)
84 =
+ ![\frac{1}{2}(14-x)(y)](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%2814-x%29%28y%29)
∴ ![84 = \frac{1}{2}(xy) + 7y - \frac{1}{2}(xy)](https://tex.z-dn.net/?f=84%20%3D%20%5Cfrac%7B1%7D%7B2%7D%28xy%29%20%2B%207y%20-%20%5Cfrac%7B1%7D%7B2%7D%28xy%29)
![84 = 7y\\y = \frac{84}{7}](https://tex.z-dn.net/?f=84%20%3D%207y%5C%5Cy%20%3D%20%5Cfrac%7B84%7D%7B7%7D)
∴ ![y = 12](https://tex.z-dn.net/?f=y%20%3D%2012)
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/ = ![\sqrt{25}](https://tex.z-dn.net/?f=%5Csqrt%7B25%7D)
/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 = ![\frac{1}{2} (DC)(AD)](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20%28DC%29%28AD%29)
Area of triangle ADC = ![\frac{1}{2} (9)(12)](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20%289%29%2812%29)
Area of triangle ADC = 9 × 6
Area of triangle ADC = 54 square unit
Hence, Area of triangle ADC is 54 square unit.