Question

Find the dimensions of a rectangle whose perimeter is 28 meters and whose area is 40square meters


​

Answer 1

Answer:

HOPE THIS HELPS

PLEASE LIKE AND RATE

AND MARK IT AS BRAINLIEST

AND PLEASE FOLLOW ME

I WILL CLEAR ALL YOUR DOUBTS

Answer 2

Step-by-step explanation:

HEYA!!!!

Here is your answer :

Given,

Perimeter of rectangle = 28mts

2 ( l + b ) = 28

2 (l + b ) = 28

l + b = 28/2

l + b = 14

l = 14 - b ---------(1)

Now area of rectangle = 40

l × b = 40

from eq 1

( 14-b ) × b = 40

\begin{gathered}(14 - b) \times b= 40 \\ \\ (14 - \ {b^{2} } ) = 40 \\ \\ {b }^{2} - 14b + 40 = 0 \\ \\ after \: splitting \: th e \: zeros\: \\ \\ \\ {b}^{2} - 4b - 10b + 40 = 0 \\ \\ b(b - 4) - 10(b - 4) = 0 \\ \\ b - 4 = 0 \: \: \: \: \: b - 10 = 0 \\ \\ b = 4 \: \: \: \: \: \: b = 10\end{gathered}

(14−b)×b=40

(14− b

2

)=40

b

2

−14b+40=0

b

2

−4b−10b+40=0

b(b−4)−10(b−4)=0

b−4=0b−10=0

b=4b=10