Question

what area of a rectangle is(x^2+10x+21)sq.cm,and its length is (x+7) cm. find its breadth and perimeter...
please solve this question then give answer​

Answer 1

Step-by-step explanation:

breadth X+3

perimeter 4x +20

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Answer 2

Answer :-

The breadth of the rectangle is " x + 3cm " and perimeter is 4x + 20cm²

Step-by-step explanation :-

To find the breadth, we need to divide the area with the length, As we know that,

$\bf{Area = Length × Breadth}$

By putting the values,

$\sf{ \: x^2+10x+21 = (x + 7) \times (Breadth)}$

$\sf{ Breadth = \: \dfrac{{x}^{2}+10x+21}{x + 7}}$

$\sf{ \leadsto \: \: \dfrac{(x + 3)(x + 7)}{x + 7}}$

$\sf{ \leadsto \boxed{ \sf{ \: x + 3}}} \: \: \: \blue{\star}$

Now, let's find the perimeter,

We know that,

$\bf{Perimeter = 2( Length + Breadth )}$

By putting the values,

$\sf{\leadsto \: 2(x + 7 + x + 3)}$

$\sf{\leadsto \: (2)(x) + (2)(7) + (2)(x) + (2)(3)}$

$\sf{\leadsto \: 2x + 14 + 2x + 6}$

$\sf{\leadsto \: \boxed{ \sf{4x + 20}} \: \: \: \blue{\star}}$

Therefore, the perimeter of the rectangle is 4x+20 cm²