Difference between revisions of "2021 JMPSC Accuracy Problems/Problem 6"
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==Solution== | ==Solution== | ||
| − | + | Notice that since <math>\overline{AC}</math> bisects a pair of opposite angles in quadrilateral <math>ABCD</math>, we can distinguish this quadrilateral as a kite. | |
| + | |||
| + | <math>\linebreak</math> | ||
| + | With this information, we have that <math>\overline{AD}=\overline{AB}=15</math> and <math>\overline{CD}=\overline{BC}=13</math>. | ||
| + | |||
| + | Therefore, the perimeter is <cmath>15+15+13+13=\boxed{56}</cmath> | ||
| + | <math>\square</math> | ||
| + | |||
| + | <math>\linebreak</math> | ||
| + | ~Apple321 | ||
| + | |||
| + | |||
| + | ==Solution 2== | ||
| + | |||
| + | We note that triangle <math>ABC</math> and <math>DAC</math> are congruent due to <math>ASA</math> congruency. Therefore, <math>AD + DC = 28</math> and the perimeter of the quadrilateral is <math>28+28 = \boxed{56}</math> | ||
| + | |||
| + | ~Grisham | ||
| + | |||
| + | ==See also== | ||
| + | #[[2021 JMPSC Accuracy Problems|Other 2021 JMPSC Accuracy Problems]] | ||
| + | #[[2021 JMPSC Accuracy Answer Key|2021 JMPSC Accuracy Answer Key]] | ||
| + | #[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]] | ||
| + | {{JMPSC Notice}} | ||
Latest revision as of 16:53, 11 July 2021
Contents
Problem
In quadrilateral
, diagonal
bisects both
and
. If
and
, find the perimeter of
.
Solution
Notice that since
bisects a pair of opposite angles in quadrilateral
, we can distinguish this quadrilateral as a kite.
With this information, we have that
and
.
Therefore, the perimeter is
~Apple321
Solution 2
We note that triangle
and
are congruent due to
congruency. Therefore,
and the perimeter of the quadrilateral is
~Grisham
See also
The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition.