Thread: area of triangle given angle bisector

1. area of triangle given angle bisector

BN=4 units
AN is an angle bisector
AB + AC = 30 units
m(ABC) = 90 degrees

What is the area of the triangle ABC?

I tried having three equations with three unknowns from phtagors plus the side ratios of angle bisector formulas but i am getting equations to 4th degree, which is not supposed to happen in this test, there must be a much simpler and nicer solution

the answer was given as 60

2. Re: area of triangle given angle bisector

Let's use letters to represent the different lengths:

$$\overline{AB} = w, \overline{AN} = x, \overline{NC}=y, \overline{AC} = z$$

Now, we have the following information:

By the Pythagorean Theorem:
$$w^2+4^2=x^2$$

By the Pythagorean Theorem:
$$(4+y)^2+w^2 = z^2$$

By the double angle formula for sine:
$$\dfrac{4+y}{z} = 2\dfrac{4}{x}\cdot \dfrac{w}{x}$$

Given:
$$w+z=30$$

That is four equations in four variables. So, this system may have a solution. You would need to solve it.

3. Re: area of triangle given angle bisector

$\displaystyle \frac{w}{4}=\frac{z}{y}=\frac{w+z}{y+4}=\frac{30}{ y+4}$

area $\displaystyle =\frac{1}{2}w(y+4)=\frac{1}{2}120=60$

4. Re: area of triangle given angle bisector

Changed the z to v (z is a confusing variable!)

https://www.wolframalpha.com/input/?...D(y%2B4)*x%5E2

Notice that triangleABN is close to being ye olde 3-4-5 right triangle.

5. Re: area of triangle given angle bisector

Originally Posted by Idea
$\displaystyle \frac{w}{4}=\frac{z}{y}=\frac{w+z}{y+4}=\frac{30}{ y+4}$

area $\displaystyle =\frac{1}{2}w(y+4)=\frac{1}{2}120=60$
great solution, so, w+z/y+4 also comes from a formula? i didnt know that one. if not , how , based on what, can you write that?

6. Re: area of triangle given angle bisector

Originally Posted by ketanco
great solution, so, w+z/y+4 also comes from a formula? i didnt know that one. if not , how , based on what, can you write that?
Did you mean to ask about $\displaystyle w + \dfrac{z}{y} + 4$ or $\displaystyle \dfrac{w + z}{y + 4}$?

Parenthesis are needed here! Also try (w + z)/(y + 4).

-Dan

7. Re: area of triangle given angle bisector

Originally Posted by topsquark
Did you mean to ask about $\displaystyle w + \dfrac{z}{y} + 4$ or $\displaystyle \dfrac{w + z}{y + 4}$?

Parenthesis are needed here! Also try (w + z)/(y + 4).

-Dan

again... where does (w+z)/(y+4) come from?

this is my question

is it part of the formula?

the formula i knew about angle bisector was: w/4=z/y.... i understand that.... i knew that..... but why is this ratio equal to : (w+z)/(y+4)? is this also part of the formula? i never seen this before... may be i missed.... if not part of the formula, how, from where did you write (w+z)/(y+4) being equal to w/4=z/y?

8. Re: area of triangle given angle bisector

it is a property of proportions and easy to prove that

if

$\displaystyle \frac{a}{b}=\frac{c}{d}$

then

$\displaystyle \frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}$

9. Re: area of triangle given angle bisector

ahh ok)))) thanks

10. Re: area of triangle given angle bisector

Originally Posted by DenisB
Changed the z to v (z is a confusing variable!)

https://www.wolframalpha.com/input/?...D(y%2B4)*x%5E2

Notice that triangleABN is close to being ye olde 3-4-5 right triangle.
Hello,

Area of the triangle is not 60. it is $\frac12*26.2655*3.50207=45.9918097$. Why the difference in the both answers(given by you and idea)?

11. Re: area of triangle given angle bisector

Originally Posted by ketanco
BN=4 units
AN is an angle bisector
AB + AC = 30 units
m(ABC) = 90 degrees

What is the area of the triangle ABC?

I tried having three equations with three unknowns from phtagors plus the side ratios of angle bisector formulas but i am getting equations to 4th degree, which is not supposed to happen in this test, there must be a much simpler and nicer solution

the answer was given as 60

Hello, Your question contains wrong information. Please check your question. The information given in the question can not be used to create an valid triangle. Please look at https://www.calculator.net/triangle-...8.797&vz=&vb=&

12. Re: area of triangle given angle bisector

Originally Posted by Vinod
Hello, Your question contains wrong information. Please check your question. The information given in the question can not be used to create an valid triangle. Please look at https://www.calculator.net/triangle-...8.797&vz=&vb=&
Here is a triangle

$AB \approx 13.7262$

$BC \approx 8.74243$

$CA \approx 16.2738$

13. Re: area of triangle given angle bisector

Originally Posted by Idea
Here is a triangle

$AB \approx 13.7262$

$BC \approx 8.74243$

$CA \approx 16.2738$
Hello,
Can you tell me $m\angle bac$ which is bisected? and $m\angle anc$and $m\angle anb$.You also tell me length of angle bisector $\overline{an}$

14. Re: area of triangle given angle bisector

angle $\displaystyle BAC \approx 32.5$ degrees

15. Re: area of triangle given angle bisector

Originally Posted by Vinod
Hello,
Can you tell me $m\angle bac$ which is bisected? and $m\angle anc$and $m\angle anb$.You also tell me length of angle bisector $\overline{an}$
Hello,