Question

The area of a rectangle is a^2+4a-21. Find the width of the rectangle if its length is (a+7)?

Answer 1

Answer & Step-by-step explanation:

Area of a triangle = width x length

$a^{2} +4a-21 = W * ( a + 7)$

Factorise $a^{2} +4a-21$

- 21 can be (3 x -7) or (-3 x 7) or (1 x -21) or (-1 x 21)

But the two values should add upto + 4 the middle vale in the expression.

(-3 x 7) gives -21 and aslo the additon of the two values gives (-3 + 7)  + 4

so we get (a - 3)(a + 7) => $a^{2} +4a-21$

back to the question

$a^{2} +4a-21 = W * ( a + 7)$ ; $a^{2} +4a-21$ this can now be written as (a - 3)(a + 7)

hence

(a - 3)(a + 7) = W x (a + 7)

W = $\frac{(a - 3)(a +7)}{(a + 7)}$

cancel off (a + 7) with (a +7)

and we finally get the width of the rectangle

W = (a - 3)

Verification

take a as 2 or any number u like

2^2 + 4 x 2 - 21 = (2 + 7) x (2 - 3)

-9 = 9 x -1

-9 = -9

and yes thats correct...

Answer 2

SOLUTION

GIVEN

The area of a rectangle is a² + 4a - 21.

The length is (a + 7)

TO DETERMINE

The width of the rectangle

CONCEPT TO BE IMPLEMENTED

Area of a rectangle

= Length × Width

EVALUATION

Here it is given that the area of a rectangle

$\sf = {a}^{2} + 4a - 21$

$\sf = {a}^{2} + (7 - 3)a - 21$

$\sf = {a}^{2} + 7a - 3a - 21$

$\sf = a(a + 7) - 3(a + 7)$

$\sf = (a + 7) (a - 3)$

Now it is given that length is (a + 7)

Hence the required width of the rectangle

$\displaystyle \sf{ = \frac{ Area}{ Length }}$

$\displaystyle \sf{ = \frac{ (a + 7)(a - 3)}{(a + 7) }}$

$\displaystyle \sf{ = (a - 3)}$

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