![]() ![]() ![]() Lowest possible positive value for x is 24 and the highest is 40 As an example from earlier, if p=98 and A=420 then the Lower positive root but with the sign before 2π/3 (the third, negative, root is the same as the Solving cubic equations is not particularĪttractive but for those who care, the highest possible positive There are exactly two isosceles triangles Third of the perimeter the area is then p 2 /√432. Given perimeter is the equilateral triangle with each side a Leaving y and z equal to each other: in other words theĪssociated triangles are isosceles. The subinterval and only there the square root will vanish, X which will allow y and z to be calculated. Its maximum possible value for positive x when x=p/3 in whichĬlearly any horizontal line between zero and p 4/432Īs mentioned earlier, there will be a subinterval of for With turning points at x=0 and x=p/3, so A 2 achieves So we needįor given positive p, the left hand side is aĬubic in x: with roots at x=p/2 and at x=0 (twice), positive for X 2/4 - 4A 2/(p(p-2x)) ) being square-rootedĮarlier to be non-negative when it is zero we will have y=z and The axes of the ellipse and so congruent.įor all this to make sense, we need the number ( Triangles with this height they are reflections of each other in The base and the perpendicular height, so given the base and theĪrea, the height is fixed. The area of a triangle is half of the product of Major axis being the difference between the perimeter and the Vertices of the given edge as its foci and the length of its Positions for the third vertex lie on an ellipse with the Imagine trying to identifyĪll triangles with a given edge and a known perimeter. So we can also conclude that for given A and p, xĪnd area and with one side the same are congruent. Y = (p-x)/2 +/- √( x 2/4 - 4A 2/(p(p-2x))Īnd z is the same but with the opposite sign for We have perimeter p = 2s, and clearly s-z = x+y-s, The easiest way to show this is to use Heron's formula for theĪrea of a triangle with semi-perimeter s and with sides x, y and Square of the perimeter must be more than the area times the The perimeter is large enough for there to be at least one non-equilateral Here "sensible cases" means that for a given area, Triangles with a given perimeter and area. Somos's list throws up some more examples MohammedĪassila has shown there are an infinite number of Heronianĭuplicates (though it might be difficult to access the article).īut I am more interested in the general case, where the sides are Sides and areas) with sides 17,25,28 and 20,21,29: they both Consider the Heronian triangles ( integer Probably did not help on this question, so here are some thoughts They found my page on calculating the area of a triangle, which Somebody was searching for triangles with the same area and ![]() Substituting the height and base into the formula for area gives us: Area of triangle = 1/2 x 6 x 4 = 12.Triangles with the same area and perimeter Triangles with the same area and perimeter by Henry Bottomley By Pythagoras' Theorem, 5 2 = 3 2 + height 2, therefore height 2 = 16 and so height = 4. To find the height, we can split our isosceles into two identical right angled triangles whose hypotenuse (the side opposite to the right angle) has length 5 and the base has length 1/2 x 6 = 3. We know that the base = 6, but don't know the height yet. Area of a triangle = 1/2 x base x height. Therefore, the base has length x + 4 = 2 + 4 = 6, and the other sides have length x + 3 = 2 + 3 = 5.Now that we know what the actual lengths of each side are, we need to calculate the area. We are told that the perimeter is equal to 16, so setting the equation equal to 16 gives 3x + 10 = 16, meaning that 3x = 6 and so x = 2. The perimeter of a shape is the sum of the length of all of its sides, so the perimeter of this isosceles is x + 4 + 2(x + 3) = x + 4 + 2x + 6 = 3x +10. ![]()
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