2021 JMPSC Sprint Problems/Problem 16: Difference between revisions

Revision as of 10:58, 11 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 $ABCD$ .

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

@@ Line 6: / Line 6: @@
 ==Solution==
+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>.
+~Mathdreams

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2021 JMPSC Sprint Problems/Problem 16: Difference between revisions

Revision as of 10:58, 11 July 2021

Problem

Solution