Difference between revisions of "2021 JMPSC Sprint Problems/Problem 16"
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+ | |||
+ | == Solution 2 == | ||
+ | <cmath>[ACD] = \frac{24 \cdot 20}{2}=240</cmath> | ||
+ | <cmath>[ABC] = \frac{12 \cdot 16}{2}=96</cmath> | ||
+ | Therefore, <math>[ABCD] = 240-96=144</math> | ||
+ | |||
+ | - kante314 - | ||
+ | |||
+ | ==See also== | ||
+ | #[[2021 JMPSC Sprint Problems|Other 2021 JMPSC Sprint Problems]] | ||
+ | #[[2021 JMPSC Sprint Answer Key|2021 JMPSC Sprint Answer Key]] | ||
+ | #[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]] | ||
+ | {{JMPSC Notice}} |
Latest revision as of 09:39, 12 July 2021
Contents
Problem
is a concave quadrilateral with , , , and . Find the area of .
Solution
Notice that and by the Pythagorean Thereom. We then have that the area of triangle of is , and the area of triangle is , so the area of quadrilateral is .
~Mathdreams
Solution 2
Therefore,
- kante314 -
See also
The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition.