Question

4. Diagonal of a rectangle is
$\sqrt{89}$
cm and
area is 40 sq.cm . Find the length and
breadth of the rectangle​

Answer 1

Given:-

Length of one diagonal of rectangle (d) = $\sqrt{89}$

Area of rectangle = 40cm²

To find:-

Length and breadth of the rectangle.

Solution:-

First let's derive a relation of area and diagonal to find length + Breadth for our ease.

First, we will find Length + Breadth.

$\rm Let \ Length = l, \ Breadth = b, Diagonal = d$

We know that,

$\rm \dashrightarrow d=\sqrt{l^2+b^2} \quad Or \quad d^2 = l^2+b^2$

Add 2lb to both the sides of the equation.

$\rm \dashrightarrow d^2 + 2lb = l^2+b^2+2lb$

We know that, $\rm (a+b)^2=a^2+b^2+2ab$

Using this identity,

$\rm \dashrightarrow d^2 + 2lb = (l+b)^2$

$\rm \boxed{\bf l+b = \sqrt{ d^2+2lb}}$

____________________________

We have $\rm d = \sqrt{89}cm \ and \ Area \ of \ rectangle \ or \ lb = 40cm$

Using the relation above, let's find out length + Breadth.

$\rm\dashrightarrow l+b = \sqrt{ d^2+2lb}$

Substituting the values, we get,

$\rm \dashrightarrow l+b = \sqrt{ (\sqrt{89})^2+2(40)}$

$\rm \dashrightarrow l+b = \sqrt{ 89+80}$

$\rm \dashrightarrow l+b = \sqrt{ 169}$

$\rm \dashrightarrow l+b = 13$

Since, we have

$\rm d^2 = l^2+b^2$

$\rm \dashrightarrow \sqrt{89} = l^2+b^2$

$lb = 40$

And

$\rm l+b=13$

Therefore, we can find $\rm l-b$ and form simultaneous equation with $\rm l+b$and get the values of $l$ and b

We know that,

(a-b)² = a²+b²-2ab

Using this identity,

$\rm\dashrightarrow (l-b)^2 = l^2+b^2-2lb$

$\rm\dashrightarrow (l-b)^2 = d^2 -2(40) \qquad \qquad ..[since, l^2+b^2=d^2]$

$\rm\dashrightarrow (l-b)^2 = (\sqrt{89})^2 -80\qquad \qquad ..[since, d= \sqrt{89}]$

$\rm\dashrightarrow (l-b)^2 = 89 -80$

$\rm\dashrightarrow (l-b)^2 = 9$

$\rm\dashrightarrow l-b =\sqrt{9}$

$\rm\dashrightarrow l-b = 3$

Now we have $\rm l + b = 13 \ and \ l-b=3$

Solving the simultaneous equations.

Now add the equations.

$\rm l + b = 13 \\ \rm + \underline{l-b=3} \\\rm 2l = 16 \\ \rm \dashrightarrow l = \dfrac{16}{2} = 8$

$\boxed{\bf l=8}$

Substituting the value of $l$ in l - b = 3

$\rm l - b = 3 \\ \rm 8-b = 3 \\ \rm b = 8-3 \\ \boxed{ \bf b = 5}$

Hence, Length is 8cm and Breadth of the rectangle is 5cm.