Two side of the triangle are 4cm and 6cm the third side must be less than..?
Answers
Answer:
We know that the sum of two sides should be less than the third side.
So, let the length of the third side be x cm.
4 + x > 6 = x > 2
x < 4 + 6 = x < 10
Hence, x=3,4,5,6,7,8,9
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Answer:
Answer:
The triangle inequality says that the sum of the two smaller sides must exceed the largest side to have a real triangle. So here that means the remaining side is between 2 and 10, exclusive.
There’s an algebraic form to the triangle inequality that’s not well known and pretty interesting. If we have a triangle with sides a,b and c, which we can assume are all positive, the triangle inequality is the rather complicated
a+b>c and a+c>b and b+c>a
Of course the perimeter is positive as well. So in a real triangle we have four positive quantities,
a+b+c>0
−a+b+c>0
a−b+c>0
a+b−c>0
We get a real triangle precisely when the product of all four is positive, A>0 where
A=(a+b+c)(−a+b+c)(a−b+c)(a+b−c)
There are a number of different ways to multiply this out; the identity (x+y)(x−y)=x2−y2 is helpful. Let’s single out a .
A=((b+c)+a)((b+c)−a)(a−(b−c))(a+(b−c))
=((b+c)2−a2)(a2−(b−c)2)
=(b2+2bc+c2−a2)(a2−(b2−2bc+c2))
=(2bc+(b2+c2−a2))(2bc−(b2+c2−a2))
A=4b2c2−(b2+c2−a2)2
That’s the algebraic triangle inequality. A>0 means a real triangle, with a positive area. A=0 is the degenerate triangle, vertices three collinear points. A<0 is an impossible triangle, a,b,c are collectively not possible distances between three points, they strictly fail the triangle inequality.
Note that A only depends on the squares of the sides. It would work correctly if we allowed negative sides (signed lengths are sometimes called displacements).
A is symmetrical, remaining the same if we swap any pair of a,b or c . We know this because we started from a symmetrical expression to derive it.
There’s a more symmetrical-looking form involving only squared sides that you might want to derive yourself before reading on.
Let’s abbreviate A=a2,B=b2,C=c2 and write
A=4BC−(B+C−A)2
A=4BC−(B2+C2+A2−2(AB+AC−BC))
A=4BC−(B2+C2+A2−2AB−2AC+2BC)
A=−(A2+B2+C2)+2(AB+AC+BC)
Subtracting the identity
(A+B+C)2=(A2+B2+C2)+2(AB+AC+BC)
A−(A+B+C)2=−2(A2+B2+C2)
A=(A+B+C)2−2(A2+B2+C2)
or in terms of the side lengths,
A=(a2+b2+c2)2−2(a4+b4+c4)
That’s a symmetrical looking form for A, very pretty but usually the asymmetrical (looking) form is easier to use.
Archimedes’ Theorem says A is sixteen times the area of the triangle with sides a,b and c . A is called the quadrea of the triangle.
Let’s summarize. A triangle with sides a,b,c and squared sides A=a2,B=b2,C=c2, area Δ and quadrea A satisfies
A=16Δ2=4BC−(B+C−A)2
=(A+B+C)2−2(A2+B2+C2)
=(a+b+c)(−a+b+c)(a−b+c)(a+b−c)
A>0 indicates a real triangle, A=0 a degenerate triangle (vertices three collinear points), A<0 an impossible triangle.
Using the appropriate one of these formulas to calculate area almost always works out better than using Heron’s formula.
Step-by-step explanation:
