{"id":4360,"date":"2024-03-11T13:08:00","date_gmt":"2024-03-11T17:08:00","guid":{"rendered":"http:\/\/mathfun4kids.com\/mlog\/?p=4360"},"modified":"2024-10-25T12:15:31","modified_gmt":"2024-10-25T16:15:31","slug":"mathcounts-2017-2018-handbook-problem-136","status":"publish","type":"post","link":"https:\/\/mathfun4kids.com\/mlog\/?p=4360","title":{"rendered":"Mathcounts 2017-2018 Handbook Problem 136"},"content":{"rendered":"\n

In right triangle ABC, with AB = 44 cm and BC = 33 cm, point D lies on side BC so that BD:DC = 2:1. If vertex A is folded onto point D to create quadrilateral BCEF, as shown, what is the area of triangle CDE? Click here<\/a> for the solution.\n<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n
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Solution 1: <\/strong>Obviously, $\\triangle{ABC}$ is a 3-4-5 triangle with $AC=55$.<\/p>\n\n\n\n

Because $BD:DC=2:1$, and $BC=33$, we have $BD=22$, $CD=11$.<\/p>\n\n\n\n

Let $BF=x$, $AF=DF=AB-BF=44-x$. Because $\\triangle{BDF}$ is a right triangle, according to Pythagorean Theorem<\/a>, we have <\/p>\n\n\n\n

$$DF^2=BD^2+BF^2\\ \\ \\ \\ i.e.\\ \\ (44-x)^2=22^2+x^2\\tag{1}$$<\/p>\n\n\n\n

Solve the above equation, we have $x=\\dfrac{33}{2}$. Therefore $BF=\\dfrac{33}{2}$, $AF=\\dfrac{55}{2}$.<\/p>\n\n\n\n

Draw line $AD$ intersecting $EF$ at $G$, and extend $EF$ and $CB$ intersecting at $H$, as shown below. We have $AG\\perp EF$ as $\\triangle{DEF}$ is a reflection of $\\triangle{AEF}$ along line $EF$. $\\triangle{AGF}$ is a right triangle.<\/p>\n\n\n\n

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

Therefore, $\\triangle{AGF}\\sim\\triangle{ABD}\\sim\\triangle{HBF}$. We have $$\\dfrac{HB}{BF}=\\dfrac{AB}{BD}=\\dfrac{44}{22}=2$$ Therefore $HB=2BF=33$, and $CH=HB + BC=66$.<\/p>\n\n\n\n

According to Menelaus’s Theorem<\/a>, we have $$\\dfrac{CH}{HB}\\cdot\\dfrac{BF}{FA}\\cdot\\dfrac{AE}{EC}=1$$i.e.<\/p>\n\n\n\n

$$dfrac{66}{33}\\cdot\\dfrac{\\dfrac{33}{2}}{\\dfrac{55}{2}}\\cdot\\dfrac{AE}{EC}=1$$<\/p>\n\n\n\n

Therefore $$\\dfrac{AE}{EC}=\\dfrac{5}{6}\\tag{2}$$<\/p>\n\n\n\n

Because $$AE+EC=AC=55\\tag{3}$$ Solving (2) and (3), we have: $AE=25$, $EC=30$.<\/p>\n\n\n\n

Because $\\triangle{DEF}$ is a reflection of $\\triangle{AEF}$, $DE=AE=25$. So the lengths of three sides of $\\triangle{CDE}$ are $a=11$, $b=25$, and $c=30$. According to Heron’s formula<\/a>, we have $$s=\\dfrac{a+b+c}{2}=33$$ and the area of $\\triangle{CDE}$ is<\/p>\n\n\n\n

$$S=\\sqrt{s(s-a)(s-b)(s-c)}=\\sqrt{33\\times 22\\times 8 \\times 3}$$ $$=11\\sqrt{3\\times 2\\times 3\\times 8}=11\\times 3\\times 4=\\boxed{132}$$<\/p>\n\n\n\n

Solution 2:<\/strong> Obviously, $\\triangle{ABC}$ is a 3-4-5 triangle with $AC=55$.<\/p>\n\n\n\n

Because $BD:DC=2:1$, and $BC=33$, we have $BD=22$, $CD=11$.<\/p>\n\n\n\n

Draw line $AD$ intersecting $EF$ at $G$, and draw line $DH$ so that $DH\\perp CA$ at $H$, as shown below:<\/p>\n\n\n\n

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

$$AD=\\sqrt{AB^2+BD^2}=\\sqrt{44^2+22^2}=22\\sqrt{5}$$<\/p>\n\n\n\n

Because $\\triangle{DEF}$ is a reflection of $\\triangle{AEF}$, $AG\\perp GE$. Therefore, $\\triangle{AGE}\\sim\\triangle{AHD}$ and $AG=\\dfrac{1}{2}AD=11\\sqrt{5}$.<\/p>\n\n\n\n

The area of $\\triangle{ACD}$ is $$[\\triangle{ACD}]=\\dfrac{1}{2}AB\\cdot CD=\\dfrac{1}{2}\\times 44\\times 11=242$$<\/p>\n\n\n\n

Because $[\\triangle{AHD}]=\\dfrac{1}{2}AC\\cdot DH$, we have $$242=\\dfrac{1}{2}\\times 55\\cdot DH$$<\/p>\n\n\n\n

Therefore $DH=\\dfrac{44}{5}$. <\/p>\n\n\n\n

Because $\\triangle{AHD}$ is a right triangle, we have $$AH=\\sqrt{AD^2-DH^2}=\\sqrt{(22\\sqrt{5})^2-(\\dfrac{44}{5})^2}=22\\sqrt{5-\\dfrac{4}{25}}=22\\times\\dfrac{11}{5}=\\dfrac{242}{5}$$<\/p>\n\n\n\n

Because $\\triangle{AGE}\\sim\\triangle{AHD}$, we have $$\\dfrac{GE}{AG}=\\dfrac{DH}{AH}=\\dfrac{\\dfrac{44}{5}}{\\dfrac{242}{5}}=\\dfrac{2}{11}$$ <\/p>\n\n\n\n

Therefore $$EG=\\dfrac{2}{11}AG=\\dfrac{2}{11}\\cdot 11\\sqrt{5}=2\\sqrt{5}$$<\/p>\n\n\n\n

Therefore the area of $\\triangle{ADE}$ is $$[\\triangle{ADE}]=\\dfrac{1}{2}AD\\cdot DH=\\dfrac{1}{2}\\cdot 22\\sqrt{5}\\cdot 2\\sqrt{5}=110$$<\/p>\n\n\n\n

The area of $\\triangle{CDE}$ is<\/p>\n\n\n\n

$$[\\triangle{CDE}]=[\\triangle{ACD}]-[\\triangle{ADE}]=242-110=\\boxed{132}$$<\/p>\n","protected":false},"excerpt":{"rendered":"

In right triangle ABC, with AB = 44 cm and BC = 33 cm, point D lies on side BC so that BD:DC = 2:1. If vertex A is folded onto point D to create quadrilateral BCEF, as shown, what … 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\/4360"}],"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=4360"}],"version-history":[{"count":29,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/4360\/revisions"}],"predecessor-version":[{"id":4564,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/4360\/revisions\/4564"}],"wp:attachment":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=4360"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=4360"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=4360"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}