Difference between revisions of "2018 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 4"

(""90"")
(Solution)
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== Solution==
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== Solution (Involves Hit and Trial)==
 
Let any one side be x and other side be y.
 
Let any one side be x and other side be y.
  
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let [(y)^2-(x)^2]^(1/2) be z.
 
let [(y)^2-(x)^2]^(1/2) be z.
 +
 
So x*z=39*95^(1/2)
 
So x*z=39*95^(1/2)
  
Here x = 13 satisfies with y = 32
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Here x = 13 satisfies with y = 32 { By Hit And Trial Method }
 +
 
 
so Perimeter is 2(13+32)=90
 
so Perimeter is 2(13+32)=90
  

Revision as of 01:26, 8 August 2019

Problem

Suppose ABCD is a parallelogram with area $39\sqrt{95}$ square units and $\angle{DAC}$ is a right angle. If the lengths of all the sides of ABCD are integers, what is the perimeter of ABCD?


Solution (Involves Hit and Trial)

Let any one side be x and other side be y.

Then one diagonal is [(y)^2-(x)^2]^(1/2)

let [(y)^2-(x)^2]^(1/2) be z.

So x*z=39*95^(1/2)

Here x = 13 satisfies with y = 32 { By Hit And Trial Method }

so Perimeter is 2(13+32)=90

See also

2018 UNM-PNM Contest II (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6 7 8 9 10
All UNM-PNM Problems and Solutions