Let ABC be a triangle such that AB=13 BC=14 and CA=15. D is a point on BC such that AD Bisects
Answers
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 [tex]x[/tex] ; hence, DC will be [tex]14-x[/tex]
and AD be [tex]y[/tex]
To, find the area of triangle ADC
Area of triangle ADC = [tex]\frac{1}{2} (DC)(AD)[/tex]
= [tex]\frac{1}{2}(14-x)(y)[/tex]
We will have to determine [tex]x[/tex] and [tex]y[/tex]
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
[tex]Area =\sqrt{s(s-a)(s-b)(s-c)}[/tex]
Where [tex]s = \frac{a+b+c}{2}[/tex]
For the given triangle ABC
Let [tex]a[/tex] = AB, [tex]b[/tex] = BC, and [tex]c[/tex] = CA
Hence, [tex]a = 13, b= 14,[/tex] and [tex]c = 15[/tex]
∴ [tex]s = \frac{13+14+15}{2} \\s= \frac{42}{2}\\s = 21[/tex]
Then,
Area of triangle ABC = [tex]\sqrt{(21)(21-13)(21-14)(21-15)}[/tex]
Area of triangle ABC = [tex]\sqrt{(21)(8)(7)(6)}[/tex] = [tex]\sqrt{7056}[/tex]
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 = [tex]\frac{1}{2} (BD)(AD)[/tex]
Area of triangle ADB = [tex]\frac{1}{2}(x)(y)[/tex]
Hence,
Area of triangle ABC = [tex]\frac{1}{2}(x)(y)[/tex] + [tex]\frac{1}{2}(14-x)(y)[/tex]
84 = [tex]\frac{1}{2}(x)(y)[/tex] + [tex]\frac{1}{2}(14-x)(y)[/tex]
∴ [tex]84 = \frac{1}{2}(xy) + 7y - \frac{1}{2}(xy)[/tex]
[tex]84 = 7y\\y = \frac{84}{7}[/tex]
∴ [tex]y = 12[/tex]
Hence, [tex]y =[/tex] 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/ = [tex]\sqrt{25}[/tex]
/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 = [tex]\frac{1}{2} (DC)(AD)[/tex]
Area of triangle ADC = [tex]\frac{1}{2} (9)(12)[/tex]
Area of triangle ADC = 9 × 6
Area of triangle ADC = 54 square unit
Hence, Area of triangle ADC is 54 square unit.
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Answers
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Answers
Answer:
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Answers
Answer:
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