Question

Find Find the area of a rectangle whose length and breadth are( 3a³-4a²+2a+5) × ( 5a+7a²-2a³) respectively.​

Answer 1

Answer

Given

Area =4a

2

+4a−3.

We know that

Area of rectangle = length × breadth

So, to find the possible expressions for the length and breadth we have to factorise the given expression.

Using the method of splitting the middle term,

4a+4a−3

=4a

2

+6a−2a−3

=2a(2a+3)−1(2a+3)

=(2a−1)(2a+3)

∴ length × breadth =(2a−1)(2a+3)

Hence, the possible expressions for the length and breadth of the rectangle are :

length =(2a−1) and breadth =(2a+3) or, length =(2a+3) and breadth =(2a−1).

Answer 2

Given:

• Length of the rectangle = ( 3a³-4a²+2a+5 ) units

• Breadth of the rectangle = ( 5a+7a²-2a³ ) units

To calculate:

• Area of the rectangle.

Calculation:

We know that,

${\underline {\boxed {\large {\rm { {Area}_{(Rectangle)} = Length \times Breadth } }}}}$

➛ Area = [( 3a³-4a²+2a+5 ) × ( 5a+7a²-2a³ )] sq. units

• Using distributive property.

➛ Area = [ 3a³ ( 5a+7a²-2a³ ) -4a²( 5a+7a²-2a³ ) + 2a( 5a+7a²-2a³ ) + 5( 5a+7a²-2a³ ) ] sq. units

• Now, perform the multiplication.

➛ Area = [ 15a⁴ + 21a⁵ - 6a⁶ - 20a³ - 28a⁴ + 8a⁵ + 10a² + 14a³ - 4a⁴ + 25a + 35a² - 10a³ ] sq. units

• Grouping all like terms.

➛ Area = [ - 6a⁶ + 21a⁵ + 8a⁵ + 15a⁴ - 28a⁴ - 4a⁴ - 20a³ + 14a³ - 10a³ + 10a² + 35a² + 25a ] sq. units

• Performing addition.

➛ Area = [ - 6a⁶ + 29a⁵ + 15a⁴ - 28a⁴ - 4a⁴ - 6a³ - 10a³ + 45a² + 25a ] sq. units

• Performing substraction.

➛ Area = [ - 6a⁶ + 29a⁵ - 13a⁴ - 4a⁴ - 16a³ + 45a² + 25a ] sq. units

• Again, performing substraction.

➛ Area = [ - 6a⁶ + 29a⁵ - 17a⁴ - 16a³ + 45a² + 25a ] sq. units

Hence, the area of the square is - 6a⁶ + 29a⁵ - 13a⁴ - 4a⁴ - 16a³ + 45a² + 25a sq. units .

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More about rectangles!!

• A rectangle is a type of parallelogram.

• Opposite sides of a rectangle are equal.

• It has all angle of 90°.

• Its diagonals bisect each other.

• Area of rectangle = Length × Breadth

• Perimeter of rectangle = 2 ( length + Breadth )