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Question-6553




Question Number 6553 by Rasheed Soomro last updated on 01/Jul/16
Commented by Rasheed Soomro last updated on 01/Jul/16
•∣AB∣=c, ∣BC∣=a, ∣CA∣=b  •AE^(↔)  , BF^(↔)  , CD^(↔)   are angular bisectors      of ∠A, ∠B, ∠C  respectively.    ★ ∣DE∣=f=?  , ∣EF∣=d=? , ∣FD∣=e=?  ★ Area of polygon LMNOPQ = ?
$$\bullet\mid{AB}\mid={c},\:\mid{BC}\mid={a},\:\mid{CA}\mid={b} \\ $$$$\bullet\overset{\leftrightarrow} {{AE}}\:,\:\overset{\leftrightarrow} {{BF}}\:,\:\overset{\leftrightarrow} {{CD}}\:\:{are}\:\boldsymbol{{angular}}\:\boldsymbol{{bisectors}} \\ $$$$\:\:\:\:{of}\:\angle{A},\:\angle{B},\:\angle{C}\:\:{respectively}. \\ $$$$ \\ $$$$\bigstar\:\mid{DE}\mid={f}=?\:\:,\:\mid{EF}\mid={d}=?\:,\:\mid{FD}\mid={e}=? \\ $$$$\bigstar\:{Area}\:{of}\:{polygon}\:{LMNOPQ}\:=\:? \\ $$
Commented by Rasheed Soomro last updated on 01/Jul/16
Series of questions of general triangle  (similar to above) is completed with  this last question.The list is given below:  •Triangle with median lines  •Triangle with altitude  lines  •Triangle with angular bisector lines    Thanks to forum-friends for cooperation!
$${Series}\:{of}\:{questions}\:{of}\:{general}\:{triangle} \\ $$$$\left({similar}\:{to}\:{above}\right)\:{is}\:{completed}\:{with} \\ $$$${this}\:{last}\:{question}.{The}\:{list}\:{is}\:{given}\:{below}: \\ $$$$\bullet{Triangle}\:{with}\:\boldsymbol{{median}}\:{lines} \\ $$$$\bullet{Triangle}\:{with}\:\boldsymbol{{altitude}}\:\:{lines} \\ $$$$\bullet{Triangle}\:{with}\:\boldsymbol{{angular}}\:\boldsymbol{{bisector}}\:{lines} \\ $$$$ \\ $$$$\boldsymbol{\mathrm{Thanks}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{forum}}-\boldsymbol{\mathrm{friends}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{cooperation}}! \\ $$
Commented by Yozzii last updated on 04/Jul/16
For the part concerning the area of the  polygon, it can be found by (1) finding the  total area of the ′generated′ triangle  that lies outside the original triangle  and then (2) subtract from the area of  the generated triangle the result of (1).  By the vector analysis I shared on   one of your posts, you′d need to find  the vertices of each of the external   triangles, calculate the sides of each  triangle and apply Heron′s formula (or use  any other method for finding the area of a triangle   given its side lengths)  for each to find the total answer for  part (1). This method I′ve described works  but it is rather long to type here given the   multitude of assumptions and conditions  that would need to be delineated in a  formal answer.
$${For}\:{the}\:{part}\:{concerning}\:{the}\:{area}\:{of}\:{the} \\ $$$${polygon},\:{it}\:{can}\:{be}\:{found}\:{by}\:\left(\mathrm{1}\right)\:{finding}\:{the} \\ $$$${total}\:{area}\:{of}\:{the}\:'{generated}'\:{triangle} \\ $$$${that}\:{lies}\:{outside}\:{the}\:{original}\:{triangle} \\ $$$${and}\:{then}\:\left(\mathrm{2}\right)\:{subtract}\:{from}\:{the}\:{area}\:{of} \\ $$$${the}\:{generated}\:{triangle}\:{the}\:{result}\:{of}\:\left(\mathrm{1}\right). \\ $$$${By}\:{the}\:{vector}\:{analysis}\:{I}\:{shared}\:{on}\: \\ $$$${one}\:{of}\:{your}\:{posts},\:{you}'{d}\:{need}\:{to}\:{find} \\ $$$${the}\:{vertices}\:{of}\:{each}\:{of}\:{the}\:{external}\: \\ $$$${triangles},\:{calculate}\:{the}\:{sides}\:{of}\:{each} \\ $$$${triangle}\:{and}\:{apply}\:{Heron}'{s}\:{formula}\:\left({or}\:{use}\right. \\ $$$${any}\:{other}\:{method}\:{for}\:{finding}\:{the}\:{area}\:{of}\:{a}\:{triangle}\: \\ $$$$\left.{given}\:{its}\:{side}\:{lengths}\right) \\ $$$${for}\:{each}\:{to}\:{find}\:{the}\:{total}\:{answer}\:{for} \\ $$$${part}\:\left(\mathrm{1}\right).\:{This}\:{method}\:{I}'{ve}\:{described}\:{works} \\ $$$${but}\:{it}\:{is}\:{rather}\:{long}\:{to}\:{type}\:{here}\:{given}\:{the}\: \\ $$$${multitude}\:{of}\:{assumptions}\:{and}\:{conditions} \\ $$$${that}\:{would}\:{need}\:{to}\:{be}\:{delineated}\:{in}\:{a} \\ $$$${formal}\:{answer}. \\ $$$$ \\ $$
Commented by Yozzii last updated on 04/Jul/16
Commented by Yozzii last updated on 04/Jul/16
Here, there are two external triangles.  Your diagram has three external triangles.  This diagram is based on angle bisectors.
$${Here},\:{there}\:{are}\:{two}\:{external}\:{triangles}. \\ $$$${Your}\:{diagram}\:{has}\:{three}\:{external}\:{triangles}. \\ $$$${This}\:{diagram}\:{is}\:{based}\:{on}\:{angle}\:{bisectors}. \\ $$
Commented by Yozzii last updated on 04/Jul/16
Commented by Yozzii last updated on 04/Jul/16
Here, there is one external triangle.
$${Here},\:{there}\:{is}\:{one}\:{external}\:{triangle}. \\ $$
Commented by Rasheed Soomro last updated on 04/Jul/16
THankS for good analysis  of the problem  and guiding to a way of solution.
$$\mathcal{TH}{ank}\mathcal{S}\:{for}\:\boldsymbol{{good}}\:\boldsymbol{{analysis}}\:\:{of}\:{the}\:{problem} \\ $$$${and}\:{guiding}\:{to}\:{a}\:{way}\:{of}\:{solution}. \\ $$$$ \\ $$
Commented by Rasheed Soomro last updated on 04/Jul/16
•The problems can be extended to any  n-sided polygon in a variety of ways!    •Also they can be extended to a general  polygon.
$$\bullet{The}\:{problems}\:{can}\:{be}\:{extended}\:{to}\:{any} \\ $$$${n}-{sided}\:{polygon}\:{in}\:{a}\:{variety}\:{of}\:{ways}! \\ $$$$ \\ $$$$\bullet{Also}\:{they}\:{can}\:{be}\:{extended}\:{to}\:{a}\:{general} \\ $$$${polygon}. \\ $$
Commented by Yozzii last updated on 04/Jul/16
Yes yes.
$${Yes}\:{yes}. \\ $$

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