Question

A1. The length and breadth of a rectangle are (4x2 + 3x – 5) units and (2x2 + x – 3) units,
respectively. Find the perimeter of the rectangle.
A2. If the lengths of the sides of a triangle are (x + 1), (x + 2) and (x + 3), find the perimeter​

Answer 1

Answer:

The area and perimeter in the form of algebraic expression are 3x^2+12x+93x

2

+12x+9 and 6x+126x+12 respectively.

Step-by-step explanation:

Length of rectangle = 3x+3

Breadth of rectangle = x+3

Area of rectangle =Length \times BreadthLength×Breadth

Area of rectangle =(3x+3)(x+3)=3x^2+9x+3x+9=3x^2+12x+9(3x+3)(x+3)=3x

2

+9x+3x+9=3x

2

+12x+9

Perimeter of rectangle =2(l+b)=2(2x+3+x+3)=2(3x+6)=6x+122(l+b)=2(2x+3+x+3)=2(3x+6)=6x+12

Hence The area and perimeter in the form of algebraic expression are 3x^2+12x+93x

2

+12x+9 and 6x+126x+12 respectively.

#Learn more:

The length and breadth of rectangular field are (2x+3y) unit and (3x-y) unit respectively find the area and perimeter of the field in the form of algebraic expression

Answer 2

GIVEN :-

• Length of Rectangle = (5x² + 8x - 5).
• Breadth of Rectangle = (2x² + 3x - 3).

TO FIND :-

• The perimeter of Rectangle.

SOLUTION :-

$\\ : \implies \displaystyle \sf \: perimeter = 2(length + breadth)$

$: \implies \displaystyle \sf \: perimeter =2(length) + 2(breadth)$

$: \implies \displaystyle \sf \: perimeter =2(5x ^{2} + 8x - 5) + 2(2x ^{2} + 3x - 3)$

$: \implies \displaystyle \sf \: perimeter =(10x ^{2} + 16x - 10) + (4x ^{2} + 6x - 6)$

$: \implies \displaystyle \sf \: perimeter =10x ^{2} + 16x - 10 + 4x ^{2} + 6x - 6$

$: \implies \displaystyle \sf \: perimeter =10x ^{2} + 4x ^{2} + 16x + 6x - 10 - 6$

$: \implies \underline{ \boxed{\displaystyle \sf \: perimeter =14x ^{2} + 22x - 16 \: units}}$