Difference between revisions of "2018 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 4"
(→""90"") |
(→Solution) |
||
Line 4: | Line 4: | ||
− | == Solution== | + | == 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. | ||
Line 10: | Line 10: | ||
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 | + | 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 square units and 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 (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | ||
All UNM-PNM Problems and Solutions |