Question

In a rectangle $ABCD$, if $AB+BC=23,AC+BD=34$, then find the area of the rectangle.

Answer 1

Let $AB=CD=y$

and $BC=AD=x$ because opposite side of rectangle are equal

Now,

$BD=AC=\sqrt{x^2+y^2}$

Given,

$AB+BC=23$

$y+x=23$

$x=23-y$ ____(i)

Now,

$AC+BD=34$

But

$BD=AC=\sqrt{x^2+y^2}$

$\therefore \sqrt{x^2+y^2}+\sqrt{x^2+y^2}=34$

$2\sqrt{x^2+y^2}=34$

$\sqrt{x^2+y^2}=17$

Squaring both side we get

$x^2+y^2=289$

also, $x=23-y$

$\therefore \left(23-y^2\right)+y^2=289$

$529-46y+y^2+y^2=289$

$2y^2-46y+529-289=0$

$2y^2-46y+240=0$

$2\left(y^2-46y+120\right)=0$

$y^2-23y+120=0$

$y^2-15y+8y+120=0$

$y(y-15)-8(y-15)=0$

$(y-8)(y-15)=0$

$y-8=0$ and $y-15=0$

$y=8$ and $y=15$

Putting the value of $y=8$ in $eqn\left(i\right)$ we get

$x=23-8$

$x=15$

$AB=CD=y=8$

$BC=AD=x=15$

Area of rectangle$=15\times 8$

$=120m^2$

Which is the required answer.