Question

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1. Subtract the sum of - x²+3xy+y² and 4x²-3y²+4xy from the sum of - 3x²-6y²+4xt and - x²-2xy-y²
2. The sides of a rectangle are (-3x+2y) and (2x+3y). Find the perimeter of the rectangle.
3. Find the difference between the perimeter of a square of side (2x+3y) cm and an equilateral triangle of side 4y cm. ​

Answer 1

Answer:

(1)

(- 3x²-6y²+4x+{- x²-2xy-y²})-

(-x²+3xy+y²+4x²-3y²+4xy)

(- 3x²-6y²+4x-x²-2xy-y²)-(-x²+3xy+y²+4x²-3y²+4xy)

(-4x²-7y²+4x-2xy)-(3x²+7xy-2y²)

-4x²-7y²+4x-2xy-3x²-7xy+2y²

=-7x²+5y²+4x-9xy {Answer}

(2)

Perimeter=2(l+b)

=2(-3x+2y+2x+3y)

=2(-x+5y)

=-2x+10y{Answer}

(3)

Perimeter of square =4×side

=4×(2x+3y)

=8x+12y

Perimeter of equilateral triangle=3×side

=3×4y

=12y

Difference=8x+12y-12y

=8x{Answer}

Answer 2

1. Subtract the sum of - x²+3xy+y² and 4x²-3y²+4xy from the sum of - 3x²-6y²+4xy and - x²-2xy-y²

-x²+3xy+y² + 4x²-3y²+4xy

=3x²+7xy-2y²

- 3x²-6y²+4xy + (- x²-2xy-y²)

=-4x²-7y²+2xy

-4x²-7y²+2xy-(3x²+7xy-2y²)

-4x²-7y²+2xy-5x²-7xy+2y²

= -7x²-5y²-5xy

2. The sides of a rectangle are (-3x+2y) and (2x+3y). Find the perimeter of the rectangle.

perimeter of rectangle=2(l+b)

so 2((-3x+2y)+(2x+3y))

2[-x+5y]

-2x+10y

3. Find the difference between the perimeter of a square of side (2x+3y) cm and an equilateral triangle of side 4y cm.

perimeter of square=4×side

4(2x+3y)=8x+12y

perimeter of equilateral Triangle=3×side

3×4y=12y

so difference=8x