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

Answer 1

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 $x$ ; hence, DC will be $14-x$

and AD be $y$

To, find the area of triangle ADC

Area of triangle ADC  = $\frac{1}{2} (DC)(AD)$

= $\frac{1}{2}(14-x)(y)$

We will have to determine $x$ and $y$

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

$Area =\sqrt{s(s-a)(s-b)(s-c)}$

Where $s = \frac{a+b+c}{2}$

For the given triangle ABC

Let $a$ = AB, $b$ = BC, and $c$ = CA

Hence, $a = 13, b= 14,$ and $c = 15$

∴ $s = \frac{13+14+15}{2} \\s= \frac{42}{2}\\s = 21$

Then,

Area of triangle ABC = $\sqrt{(21)(21-13)(21-14)(21-15)}$

Area of triangle ABC = $\sqrt{(21)(8)(7)(6)}$ = $\sqrt{7056}$

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 = $\frac{1}{2} (BD)(AD)$

Area of triangle ADB = $\frac{1}{2}(x)(y)$

Hence,

Area of triangle ABC =  $\frac{1}{2}(x)(y)$ + $\frac{1}{2}(14-x)(y)$

84 =   $\frac{1}{2}(x)(y)$ + $\frac{1}{2}(14-x)(y)$

∴ $84 = \frac{1}{2}(xy) + 7y - \frac{1}{2}(xy)$

$84 = 7y\\y = \frac{84}{7}$

∴ $y = 12$

Hence, $y =$ 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/ = $\sqrt{25}$

/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  = $\frac{1}{2} (DC)(AD)$

Area of triangle ADC  = $\frac{1}{2} (9)(12)$

Area of triangle ADC  = 9 × 6

Area of triangle ADC  = 54 square unit

Hence, Area of triangle ADC  is 54 square unit.