{"id":4472,"date":"2024-10-15T23:31:40","date_gmt":"2024-10-16T03:31:40","guid":{"rendered":"http:\/\/mathfun4kids.com\/mlog\/?p=4472"},"modified":"2024-10-25T15:14:49","modified_gmt":"2024-10-25T19:14:49","slug":"geometry-challenge-area-of-irregular-pentagon","status":"publish","type":"post","link":"https:\/\/mathfun4kids.com\/mlog\/?p=4472","title":{"rendered":"Geometry Challenge – Area of Irregular Pentagon"},"content":{"rendered":"\n

As shown in the diagram below, a star sign consists of five straight lines. It produces five triangles and a pentagon. If areas of five triangles are 3, 10, 7, 15, and 8 square unit respectively. Find the area of the pentagon.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\nClick here<\/a> for the solution.\n
\n\n\n\n

Solution:<\/strong> Draw line AB, BC, CD, DE, EA, and label the area of $5$ newly formed triangles as $a$, $b$, $c$, $d$, and $e$ respectively. Additionally, label the area of the pentagon $FGHIJ$ as $f$, we have the following diagram:<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Based on the common ratios of areas for triangles sharing the same bases, we have the following:<\/p>\n\n\n\n

$$\\dfrac{Area\\ \\triangle{CDG}}{Area\\ \\triangle{DEG}}=\\dfrac{Area\\ \\triangle{CBG}}{Area\\ \\triangle{BEG}}$$<\/p>\n\n\n\n

i.e. $$\\dfrac{a}{b+15}=\\dfrac{e+7}{f+18}\\tag{1}$$<\/p>\n\n\n\n

Similarly, we have the following additional equations:<\/p>\n\n\n\n

$$\\dfrac{b}{a+15}=\\dfrac{c+8}{f+10}\\tag{2}$$<\/p>\n\n\n\n

$$\\dfrac{c}{b+8}=\\dfrac{d+3}{f+25}\\tag{3}$$<\/p>\n\n\n\n

$$\\dfrac{d}{c+3}=\\dfrac{e+10}{f+15}\\tag{4}$$<\/p>\n\n\n\n

$$\\dfrac{e}{d+10}=\\dfrac{a+7}{f+18}\\tag{5}$$<\/p>\n\n\n\n

Since there are $6$ variables in the above 5 equations, we can add co-linear constraint $C$, $F$, and $G$.<\/p>\n\n\n\n

By using Barycentric Coordinate System<\/a> for triangles, we have the following non-normalized coordinates for $C$, $F$, and $G$, relative to $\\triangle{ABD}$:<\/p>\n\n\n\n

$$C=(d+e+10,-(a+e+7),a+f+25)\\ \\ \\ \\ \\ \\ F=(c+8,b,0)\\ \\ \\ \\ \\ \\ G=(e+7,0,a)$$<\/p>\n\n\n\n

Because $C$, $F$, and $G$ are co-linear, with Barycentric Coordinate System<\/a>, we have:<\/p>\n\n\n\n

$$\\begin{vmatrix} d+e+10 & -(a+e+7) & a+f+25\\\\ c+8 & b & 0 \\\\ e+7 & 0 & a \\end{vmatrix}=0\\tag{6}$$<\/p>\n\n\n\n

Expanding the above 6 equations, we have:<\/p>\n\n\n\n

\n$$\\begin{cases} \n\\ \\ a(f+18)=(e+7)(b+15)\\\\\n\\ \\ b(f+10)=(a+15)(c+8)\\\\\n\\ \\ c(f+25)=(b+8)(d+3)\\\\\n\\ \\ d(f+15)=(c+3)(e+10)\\\\\n\\ \\ e(f+18)=(d+10)(a+7)\\\\\n\\ \\ ab(d+e+10)+a(c+8)(a+e+7)-b(e+7)(a+f+25)=0\n\\end{cases}$$<\/p>\n\n\n\n

By using a symbolic Computer Algebra System, such as Wolfram Alpha<\/a> or Wolfree Alpha<\/a>, with an input such as:<\/p>\n\n\n\n

a>0,b>0,c>0,d>0,e>0,f>0,a(f+18)=(b+15)(e+7), b(f+10)=(c+8)(a+15), \nc(f+25)=(d+3)(b+8), d(f+15)=(e+10)(c+3), e(f+18)=(a+7)(d+10),\nab(d+e+10)+a(c+8)(a+e+7)-b(e+7)(a+f+25)=0<\/code><\/pre>\n\n\n\n
We have the following unique solution:<\/p>\n\n\n\n
$$a=21,\\ \\ \\ \\ \\ b=20,\\ \\ \\ \\ \\ c=7,\\ \\ \\ \\ \\ d=\\dfrac{15}{2},\\ \\ \\ \\ \\ e=14,\\ \\ \\ \\ \\ f=17$$<\/p>\n\n\n\n
Therefore the area of the pentagon is $\\boxed{17}$ square unit.<\/p>\n\n\n\n
Note: Please refer to Barycentric Coordinates in Olympiad Geometry<\/a> or Barycentric Coordinates for the Impatient<\/a> for additional information related to Barycentric Coordinates.<\/p>\n\n\n\n<\/div>\n","protected":false},"excerpt":{"rendered":"
As shown in the diagram below, a star sign consists of five straight lines. It produces five triangles and a pentagon. If areas of five triangles are 3, 10, 7, 15, and 8 square unit respectively. Find the area of … Continue reading →<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"inline_featured_image":false},"categories":[13,9],"tags":[],"_links":{"self":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/4472"}],"collection":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=4472"}],"version-history":[{"count":27,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/4472\/revisions"}],"predecessor-version":[{"id":4584,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/4472\/revisions\/4584"}],"wp:attachment":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=4472"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=4472"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=4472"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}