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Thread: area of triangle given angle bisector

  1. #1
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    area of triangle given angle bisector

    Please see attached
    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

    area of triangle given angle bisector-01-2.jpg
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    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.
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    Re: area of triangle given angle bisector

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

    area $\displaystyle =\frac{1}{2}w(y+4)=\frac{1}{2}120=60$
    Thanks from SlipEternal, ketanco and Vinod
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    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.
    Last edited by DenisB; Nov 16th 2018 at 08:50 AM.
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    Re: area of triangle given angle bisector

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

    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?
    Last edited by ketanco; Nov 16th 2018 at 11:42 AM.
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    Re: area of triangle given angle bisector

    Quote Originally Posted by ketanco View Post
    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
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    Re: area of triangle given angle bisector

    Quote Originally Posted by topsquark View Post
    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
    i meant to ask about exactly as i wrote

    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?
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    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

    $\displaystyle \frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}$
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    Re: area of triangle given angle bisector

    ahh ok)))) thanks
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    Re: area of triangle given angle bisector

    Quote Originally Posted by DenisB View Post
    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)?
    Last edited by Vinod; Nov 20th 2018 at 04:51 AM.
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    Senior Member Vinod's Avatar
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    Re: area of triangle given angle bisector

    Quote Originally Posted by ketanco View Post
    Please see attached
    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

    Click image for larger version. 

Name:	01-2.jpg 
Views:	1 
Size:	20.4 KB 
ID:	39128
    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=&
    Last edited by Vinod; Nov 20th 2018 at 08:40 AM.
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    Re: area of triangle given angle bisector

    Quote Originally Posted by Vinod View Post
    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

    $AB \approx 13.7262$

    $BC \approx 8.74243$

    $CA \approx 16.2738$
    Thanks from Vinod
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    Senior Member Vinod's Avatar
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    Re: area of triangle given angle bisector

    Quote Originally Posted by Idea View Post
    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 $\overline{an}$
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    Re: area of triangle given angle bisector

    angle $\displaystyle BAC \approx 32.5$ degrees
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  15. #15
    Senior Member Vinod's Avatar
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    Re: area of triangle given angle bisector

    Quote Originally Posted by Vinod View Post
    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,
    Your answer is correct. Then why did WolframAlpha computaional intelligence give wrong answer?
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