{"id":1316,"date":"2021-01-01T18:39:52","date_gmt":"2021-01-01T18:39:52","guid":{"rendered":"http:\/\/mathfun4kids.com\/mlog\/?p=1316"},"modified":"2024-10-25T02:56:58","modified_gmt":"2024-10-25T06:56:58","slug":"mathcounts-exercise-22","status":"publish","type":"post","link":"https:\/\/mathfun4kids.com\/mlog\/?p=1316","title":{"rendered":"MATHCOUNTS Exercise – 22"},"content":{"rendered":"\n

\"\"Point $E$ and $F$ are inside the square $ABCD$, with $DE=10$, $EF=6$, $BF=4$, and $\\angle{DEF}=\\angle{BFE}=90^\\circ$. Find the area of the square $ABCD$. Click here for the solutions.<\/a>



<\/p>\n\n\n\n

\n\n\n\n

\"\"Solution 1<\/strong>: Draw line $DB$, intersecting $EF$ at $G$. We have $\\triangle{DEG}\\sim\\triangle{BFG}$.<\/p>\n\n\n\n

$\\therefore \\dfrac{FG}{EG}=\\dfrac{BF}{DE}=\\dfrac{4}{10}$<\/p>\n\n\n\n

$\\because FG+EG=EF=6$<\/p>\n\n\n\n

$\\therefore FG=\\dfrac{12}{7}$, and $EG=\\dfrac{30}{7}$<\/p>\n\n\n\n

$DG=\\sqrt{DE^2+EG^2}=\\sqrt{10^2+(\\dfrac{30}{7})^2}=\\dfrac{10}{7}\\sqrt{58}$<\/p>\n\n\n\n

$BG=\\sqrt{BF^2+FG^2}=\\sqrt{4^2+(\\dfrac{12}{7})^2}=\\dfrac{4}{7}\\sqrt{58}$<\/p>\n\n\n\n

$\\therefore BD=DG+BG=\\dfrac{10}{7}\\sqrt{58}+\\dfrac{4}{7}\\sqrt{58}=2\\sqrt{58}$<\/p>\n\n\n\n

$\\therefore$ Area of $ABCD=\\dfrac{BD^2}{2}=\\dfrac{(2\\sqrt{58})^2}{2}=\\boxed{116}$<\/p>\n\n\n\n

\"\"Solution 2<\/strong>: Draw line $DB$. Point $G$ is on $DE$ so that $DG=6$. Extend $BF$ to $H$ so that $BH=6$. Draw line $GH$ intersecting $DB$ at $I$. We have $\\triangle{DGI}\\cong\\triangle{BHI}$. <\/p>\n\n\n\n

$\\therefore GI=HI=\\dfrac{1}{2}GH=\\dfrac{1}{2}{EF}=3$<\/p>\n\n\n\n

$\\therefore DI^2=BI^2=DG^2+GI^2=6^2+3^2=58$<\/p>\n\n\n\n

$\\therefore$ Area of $ABCD=2\\cdot DI^2=2\\cdot 58=\\boxed{116}$<\/p>\n\n\n\n

\"\"Solution 3<\/strong>: Let point $G$ on $DE$ so that $GE=6$. Extend $BF$ to $H$ so that $FH=6$. Draw line $AH$ and $CE$. <\/p>\n\n\n\n

$\\therefore GE=EF=FH=GH=6$, and $EFGH$ is a square.<\/p>\n\n\n\n

$\\because B$, $F$, and $H$ are co-liner, $D$, $G,$ and $E$ are co-liner<\/p>\n\n\n\n

$\\therefore A$, $H$, and $G$ are co-linear, $C$, $E$, and $F$ are co-liner<\/p>\n\n\n\n

$\\therefore \\triangle{ABH}\\cong\\triangle{BCF}\\cong\\triangle{CDE}\\cong\\triangle{DAG}$<\/p>\n\n\n\n

$\\therefore$ Area of $ABCD=EFHG+4\\cdot\\triangle{CDE}=EF^2+4\\cdot\\dfrac{DE\\cdot CE}{2}=6^2+4\\cdot\\dfrac{4\\cdot 10}{2}=\\boxed{116}$<\/p>\n\n\n\n

\"\"Solution 4<\/strong>: Extend line $DE$ to $G$ so that $EG=4$. Draw line $BD$ and $BG$. <\/p>\n\n\n\n

$\\because EG=BF=4$ and $\\angle{DEF}=\\angle{BFE}=90^\\circ$<\/p>\n\n\n\n

$\\therefore BGEF$ is a rectangle<\/p>\n\n\n\n

$\\therefore \\triangle{BDG}$ is a right triangle<\/p>\n\n\n\n

$\\therefore BD^2=BG^2+DG^2=EF^2+(DE+EG)^2=6^2+(10+4)^2=232$<\/p>\n\n\n\n

$\\therefore$ Area of $ABCD=\\dfrac{1}{2}\\cdot BD^2=\\dfrac{1}{2}\\cdot 232=\\boxed{116}$<\/p>\n\n\n\n

Generalization<\/strong>: Point $E$ and $F$ are inside square $ABCD$, with $DE=a$, $EF=b$, $BF=c$, and $\\angle{DEF}=\\angle{BFE}=90^\\circ$, then the area of the square $ABCD$ is $$ABCD=\\boxed{\\dfrac{(a+c)^2+b^2}{2}}$$<\/p>\n\n\n\n

Extra<\/strong>: Andrew started at the entry point $D$, which is one corner of a square shaped maze $ABCD$, with the exit point $B$ directly across the diagonal of the square. The following is his track record:<\/p>\n\n\n\n

  1. He entered the maze at point $D$ and walked $20$ feet first along the path.<\/li>
  2. He changed his direction $90^\\circ$ clock-wise and walked another $20$ feet.<\/li>
  3. He changed his direction $90^\\circ$ counter-clock-wise and walked another $20$ feet<\/li>
  4. He repeated step 2 and step 3 two more times.<\/li>
  5. He repeated step 3 once again, then step 2 twice.<\/li>
  6. He changed his direction $90^\\circ$ counter-clock-wise and walked $40$ feet to reach the exit point $B$.<\/li><\/ol>\n\n\n\n

    Based on the above track record, calculate the area of the maze. Click here for the solution.<\/a><\/p>\n\n\n\n

    \n\n\n\n

    \"\"Solution<\/em>:<\/strong> Based on the track record, the net distance Andrew walked parallel to the initial path direction $DE$ is $140$ feet, and the net distance perpendicular to the initial path direction $DE$ is $60$ feet, as shown the diagram in the right. The length of the diagonal $BD$ of the square $ABCD$ is:<\/p>\n\n\n\n

    $BD=\\sqrt{PD^2+PB^2}=\\sqrt{140^2+60^2}=20\\sqrt{58}$<\/p>\n\n\n\n

    Therefore, the area of $ABCD$ is:<\/p>\n\n\n\n

    $ABCD=\\dfrac{BD^2}{2}=\\dfrac{(20\\sqrt{58})^2}{2}=\\boxed{11600}$<\/p>\n\n\n\n<\/div><\/div>\n\n\n\n

    <\/p>\n","protected":false},"excerpt":{"rendered":"

    Point $E$ and $F$ are inside the square $ABCD$, with $DE=10$, $EF=6$, $BF=4$, and $\\angle{DEF}=\\angle{BFE}=90^\\circ$. Find the area of the square $ABCD$. Click here for the solutions. Solution 1: Draw line $DB$, intersecting $EF$ at $G$. We have $\\triangle{DEG}\\sim\\triangle{BFG}$. $\\therefore … 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":[9,11],"tags":[],"_links":{"self":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/1316"}],"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=1316"}],"version-history":[{"count":58,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/1316\/revisions"}],"predecessor-version":[{"id":1381,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/1316\/revisions\/1381"}],"wp:attachment":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=1316"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=1316"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=1316"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}