# An Easter Conundrum

Discussion in 'Math' started by studiot, Apr 8, 2009.

1. ### studiot Thread Starter AAC Fanatic!

Nov 9, 2007
5,005
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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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2. ### jpanhalt Expert

Jan 18, 2008
6,041
1,054
28 using Heron's formula.

John

3. ### jasperthecat Member

Mar 26, 2009
20
0

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

4. ### jasperthecat Member

Mar 26, 2009
20
0
My surprise for the day

5. ### Ratch New Member

Mar 20, 2007
1,068
4
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

6. ### hgmjr Moderator

Jan 28, 2005
9,030
218
The answer I obtained was 28.

hgmjr

Last edited: Apr 8, 2009
7. ### zgozvrm Active Member

Oct 24, 2009
115
3
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!