Question

area of a rectangle is (3a^2+5ab+2b^2) one of its side is (a+b) what is the perimeter​

Answer 1

Given:-

• Area of rectangle is 3a² + 5ab + 2b².

• One of its side is a + b .

To Find:-

• Another side of rectangle.

• Perimeter of rectangle.

Solution:-

Here, Area of rectangle = Length × Breadth

Let the another side be $x$ units.

Now, $a + b \times x = 3a^{2} + 5ab + 3b^{2}$

   ⇒  $x = \dfrac{3a^{2} + 5ab + 2b^{2}}{a + b}$

   ⇒  $x = 3a + 2b$

So, Another side of rectangle will be 3a + 2b units.

Since, Perimeter of rectangle = $2 ( L + B )$

                                                  ⇒ $2 [(a + b) + (3a + 2b)]$

                                                  ⇒ $2 \times ( 4a + 3b )$

                                                  ⇒ $8a + 6b$

Hence, Perimeter of the rectangle is 8a + 6b units .

Some Important terms:-

• Area of square = $(Side)^{2}$

• Perimeter of square = 4 × Side

Answer 2

Concept used-:

~Here the concept of Area & perimetre is used. By using the area formula of rectangle we will find the length. Then after súbsitúte the value of length and breath in the Perimeter formula.

Formula used

$\to \bold\pink {Area\:of \:rectangle= length×breath}$

$\to\bold \purple{Perimeter\:of \: rectangle=2×(length×breath)}$

Solution

• Area of rectangle = 3a²+5ab+2b² sq.unit
• side (breath) = a+b unit

Let us assume the length as L

Area of rectangle= length× breath

$\sf{\implies3a²+5ab+2b² =L× (a+b)}\\ \\ \sf{ \implies L= \frac{3a²+5ab+2b²}{a+b}} \\ \\ \sf{ \implies \red{L = 3a+2b\:units }}$

therefore the length of the rectangle is 3a+2b units

Now let us find the perimeter.

perimeter of rectangle=2×( length+breath)

$\sf{ \implies perimeter \: of \: rectangle= 2×(3a+2b +(a+b))}$

$\sf{ \implies perimeter \: of \: rectangle= 2× (4a+3b)} \\ \\ \sf{ \implies \green{perimeter \: of \: rectangle = 8a +6b\:units}}$

Therefore the perimetre of the rectangle is 8a +6b units

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