Question

The area of rectangle is $\sf \: 5(y + 3)( {y}^{2} - 16)$. It's one side is $\sf5( {y}^{2} - y - 12)$. What is its other side?

Answer this correctly!!

All the best :) ​

Answer 1

Answer:

15(y+3)(y2−16)÷5(y2−y−12)

y2−16=(y)2−(4)2

=(y+4)(y−4)

y2−y−12=y2−4y+3y−12

=y(y−4)+3(y−4)

=(y−4)(y+3)

Now,

5(y2−y−12)15(y+3)(y2−16)

=5(y−4)(y+3)15×(y+3)(y+4)(y−4)

=3(y+4)

Answer 2

★ $\rm{Answer}$

➪ y + 4 is the answer

Given,

→Area of rectangle=$\sf \: 5(y + 3)( {y}^{2} - 16)$

→ One side of rect =$\sf5( {y}^{2} - y - 12)$

∴ Area of rectangle = Length × Breadth

$\rm \: ➟ 5(y + 3)( {y}^{2} - 16) = 5( {y}^{2} - y - 12) \times breadth$

Now finding the 2nd side :-

$\rm = \frac{y + 3( {(y)}^{2} - {(4)}^{2} )}{y - 4y + 3y - 12}$

$= \rm \: \frac{(y + 3)(y - 4)(y + 4)}{y(y.4) + 3(y - 4)}$

$\rm \: \bold=y + 4$

|I{•------» Breadth = y + 4 «------•}I|

$\Huge\fbox{{More \: Information }}$

Rectangle formulas

• Area = l ⋅ w

• Perimeter = 2(l + w)

__________________★★★

More questions like this :-

1)) The perimeter of a rectangle is 42 cm. If its width is 3 more than twice its length, then find its length and with.

Solution:

Let x be the length of the rectangle.

Then, the width is (2x + 3)

Perimeter of the rectangle = 42 cm

• 2(l + w) = 42

Divide each side by 2.

• l + w = 21

Substitute x for l and (2x + 3) for w.

• x + (2x + 3) = 21

• x + 2x + 3 = 21

• 3x + 3 = 21

Subtract 3 from each side.

• 3x = 18

Divide each side by 3.

• x = 6

Therefore, the length is 6 cm.

And the width is :-

• 2x + 3 = 2(6) + 3

• 2x + 3 = 12 + 3

• 2x + 3 = 15

So, the length and width of the rectangle are 6 cm and 15 cm respectively.