An Easter Conundrum

Thread Starter

studiot

Joined Nov 9, 2007
4,998
Most folks have no trouble doing a 3-4-5 triangle, however there are other interesting ones some even without right angles.

None of the sides in the obtuse triangle in the sketch are whole numbers, yet each forms a whole number when squared. So the squares on each side have whole number areas.

What is the area of the triangle itself?

Is it also a whole number?
 

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jasperthecat

Joined Mar 26, 2009
20
From your diagram

the lengths of the sides are Sqrt(41), Sqrt(97) and 14

using

area = sqrt(s*(s-a)*(s-b)*(s-c))

where a,b,c are sides and s = semi perimeter

I'd be surprised if the area is a whole number
 

Ratch

Joined Mar 20, 2007
1,070
To the Ineffable All,

A simpler way is to drop a perpendicular line from the top vertice to the base. Then solving x^2+y^2 = 41, (14-x)^2+y^2 = 97 , where x is the distance from the left vertice to the perpendicular line and y is the altitude of the triangle. Solving, we easily get y=4 . So (1/2)*(14)*4 = 28

Ratch
 

zgozvrm

Joined Oct 24, 2009
115
None of the sides in the obtuse triangle in the sketch are whole numbers, yet each forms a whole number when squared.
Whoops!

If you meant for the sides of the triangle to NOT be whole numbers in order to make the problem more difficult, you failed ... you have the square on the bottom of the triangle as having an area of 196 sq. m. The square root of 196 is 14 which IS a whole number!
 
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