Question

The perimeter of a triangle is (x square y +10) units. One of the sides is of length (x square y-4) units and another side is (3-2x square y) units. Find the length of the third side.​

Answer 1

Solution:

Given:

⇒ Perimeter of triangle = (x²y + 10)

⇒ First side of triangle (A) = (x²y - 4)

⇒ Second side of triangle (B) = (3 - 2x²y)

To Find:

⇒ Length of third side (C).

Formula used:

⇒ Perimeter of triangle = A + B + C

Now, put all the values in the formula. We get,

⇒ Perimeter of triangle = A + B + C

⇒ (x²y + 10) = (x²y - 4) + (3 - 2x²y) + C

⇒ (x²y + 10) - (x²y - 4) - (3 - 2x²y) = C

⇒ x²y + 10 - x²y + 4 - 3 + 2x²y = C

⇒ 10 + 1 + 2x²y = C

⇒ C = 11 + 2x²y

Answer 2

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Third side of the triangle = 11+2x²y

$\bold{\underline{\underline{\huge{\sf{StEp\:by\:stEp\:explanation:}}}}}$

GIVEN :

• Perimeter of a triangle = x²y + 10
• Length of sides of the triangle :

1. First side = x²y - 4
2. Second side = 3 - 2x²

TO FIND :

• Length of the third side

SOLUTION :

Since the given figure is a triangle, we know that it has three sides.

We have the measurement for the two sides of the triangle. We have the perimeter of the triangle.

So we will use the formula for perimeter of a triangle and calculate the length of the third side as well.

FORMULA :

$\bold{\large{\boxed{\sf{\red{Perimeter\:of\:triangle\:=\:Side\:1\:+\:Side\:2\:+\:Side\:3}}}}}$

We have,

• Perimeter = x²y + 10
• Side 1 (a) = x²y - 4
• Side 2 (b) = 3 - 2x²y

We need to calculate side 3 (c)

Block in the values,

$\rightarrow$ $\bold{x^2y+10=x^2y-4+3-2x^2y+c}$

$\rightarrow$ $\bold{10=-4+3-2x^2y+c}$

$\rightarrow$ $\bold{10=-1-2x^2y+c}$

$\rightarrow$ $\bold{10+1=-2x^2y+c}$

$\rightarrow$ $\bold{11=-2x^2y+c}$

$\rightarrow$ $\bold{11+2x^2y=c}$

•°• Side 3 (c) = 11 + 2x²y