Question

The perimeter of a rectangle is 42 metres and it's diagonal is15 metres.what are the length s of it's sides?​

Answer 1

Answer:

What are the sides of a rectangle if its perimeter is 42 meters and diagonal is 15 meters?

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The perimeter of a rectangle is 2L+2W=P . The diagonals can be represented as L2+W2=D2 . Plugging in, we have:

2L+2W=42⟹L+W=21

L2+W2=152=225

We can square the first equation and subtract the second:

L+W=21

(L+W)2=212

L2+2LW+W2=441

L2+2LW+W2−(L2+W2)=441−225

2LW=216

It doesn’t tell us much, but we can now more easily substitute and solve for L .

W=21−L & LW=108

L(21−L)=108

21L−L2=108

L2−21L+108=0

We can use the quadratic formula

x=−b±b2−4ac√2a

L=21±441−432√2=21±9√2=21±32

L=21+32=242=12

[math]L=\frac{21 - 3}{[/math]

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Let’s say a and b are the length and breadth of the rectangle, and given that

perimeter = 42 m => 2(a+b) = 42 => (a+b) = 21 …..{eq.1}

diagonal = 15 m => sqrt(a^2+b^2) = 15 => (a^2+b^2) = 225 …..{eq.2}

Using the algebraic identity below:

2(a^2+b^2) = (a+b)^2 + (a-b)^2

We can find out that (a-b) = 3 …..{eq.3}

Solving eq.1 and eq.3

We get a = 12 m and b = 9m

Note : Interchanging length and width doesn’t make any different, but shape of the rectangle.