Question

the perimeter of a triangle is (x²y+10) units . one of the side is of length (x²y-4) units and another side is of length (3-2x²y) units.Find the length of the third side.
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Answer 1

EXPLANATION.

Perimeter of a triangle = (x²y + 10) units.

One of the side is of length = (x²y - 4) units.

Another sides is of length = (3 - 2x²y) units.

As we know that,

Perimeter of triangle = Sum of all sides.

Using this formula in the equation, we get.

Perimeter of triangle = x + y + z.

⇒ (x²y + 10) = (x²y - 4) + (3 - 2x²y) + z.

⇒ x²y + 10 = x²y - 4 + 3 - 2x²y + z.

⇒ x²y - x²y + 2x²y + 10 + 4 - 3 = z.

⇒ 2x²y + 11 = z.

Length of the third side = 2x²y + 11.

Answer 2

Answer:

Given :-

• The perimeter of a triangle is (x²y + 10).
• One side of a length of triangle is (x²y - 4) units.
• Another side of a length of a triangle is (3 - 2x²y) units.

To Find :-

• What is the length of the third side of a triangle.

Formula Used :-

$\clubsuit$ Perimeter of a triangle Formula :

$\footnotesize \bigstar \: \: \sf\boxed{\bold{\pink{Perimeter_{(Triangle)} =\: Sum\: of\: all\: sides}}}\: \: \: \bigstar$

Solution :-

Let,

$\mapsto \bf Third\: Side_{(Triangle)} =\: c$

Given :

• Perimeter of triangle = (x²y + 10)
• First Side of triangle (a) = (x²y - 4) units
• Second Side of triangle (b) = (3 - 2x²y) units.

According to the question by using the formula we get,

$\implies \bf Perimeter_{(Triangle)} =\: a + b + c$

$\implies \sf (x^2y + 10) =\: (x^2y - 4) + (3 - 2x^2y) + c$

$\implies \sf x^2y + 10 =\: x^2y - 4 + 3 - 2x^2y + c$

$\implies \sf x^2y + 10 - x^2y + 4 - 3 + 2x^2y =\: c$

$\implies \sf {\cancel{x^2y}} {\cancel{- x^2y}} + 2x^2y + 10 + 4 - 3 =\: c$

$\implies \sf 2x^2y + 14 - 3 =\: c$

$\implies \sf 2x^2y + 11 =\: c$

$\implies \sf\bold{\purple{c =\: 2x^2y + 11}}$

Hence, the third side of a triangle is :

$\dashrightarrow \sf Third\: Side_{(Triangle)} =\: c$

$\dashrightarrow \sf\bold{\red{Third\: Side_{(Triangle)} =\: 2x^2y + 11}}$

$\therefore$ The length of third side of a triangle is 2x²y + 11 .