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Difference between revisions of "2021 JMPSC Sprint Problems/Problem 16"

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==Solution==
 
==Solution==
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Notice that <math>[ABCD] = [ADC] - [ABC]</math> and <math>AC = \sqrt{12^2 + 16^2} = 20</math> by the Pythagorean Thereom. We then have that the area of triangle of <math>ADC</math> is <math>\frac{20 \cdot \sqrt{26^2 - 10^2}}{2} = 240</math>, and the area of triangle <math>ABC</math> is <math>\frac{12 \cdot 16}{2} = 96</math>, so the area of quadrilateral <math>ABCD</math> is <math>240 - 96 = 144</math>.
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~Mathdreams
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== Solution 2 ==
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<cmath>[ACD] = \frac{24 \cdot 20}{2}=240</cmath>
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<cmath>[ABC] = \frac{12 \cdot 16}{2}=96</cmath>
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Therefore, <math>[ABCD] = 240-96=144</math>
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- kante314 -
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==See also==
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#[[2021 JMPSC Sprint Problems|Other 2021 JMPSC Sprint Problems]]
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#[[2021 JMPSC Sprint Answer Key|2021 JMPSC Sprint Answer Key]]
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#[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]
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{{JMPSC Notice}}

Latest revision as of 09:39, 12 July 2021

Problem

$ABCD$ is a concave quadrilateral with $AB = 12$, $BC = 16$, $AD = CD = 26$, and $\angle ABC=90^\circ$. Find the area of $ABCD$.

Sprint16.jpg

Solution

Notice that $[ABCD] = [ADC] - [ABC]$ and $AC = \sqrt{12^2 + 16^2} = 20$ by the Pythagorean Thereom. We then have that the area of triangle of $ADC$ is $\frac{20 \cdot \sqrt{26^2 - 10^2}}{2} = 240$, and the area of triangle $ABC$ is $\frac{12 \cdot 16}{2} = 96$, so the area of quadrilateral $ABCD$ is $240 - 96 = 144$.

~Mathdreams

Solution 2

\[[ACD] = \frac{24 \cdot 20}{2}=240\] \[[ABC] = \frac{12 \cdot 16}{2}=96\] Therefore, $[ABCD] = 240-96=144$

- kante314 -

See also

  1. Other 2021 JMPSC Sprint Problems
  2. 2021 JMPSC Sprint Answer Key
  3. All JMPSC Problems and Solutions

The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition. JMPSC.png

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