Question

In $\Delta$ABC, $\angle B={90}^o$. Find the sides of the triangle if, AB$=$x cm, BC$=(4x+4)$ cm and AC $=(4x+5)$ cm.

Answer 1

We are given that in $△ABC,\angle ={90}^o$

tyhe sides of $△ABC$ is given by

$AB=xcm,BC=(4x+4)cm$ &

$AC=(4x+5)cm$

From the pythagoras theorem,

$A{C}^2=A{B}^2+B{C}^2$

So,

$(4x+5{)}^2-(x{)}^2+(4x+4{)}^2$

$16x^2+40x+25=x^2+16x^2+22z-16$

$\therefore 16x^2+40x+25=17x^2+32x+16$

$\therefore x^2-8x-9=0$

$\therefore (x-9)(x+1)=0$

$\therefore x=9$ or $x=-1$

but distance cannot be negative

So, $x=9$

Now, sides are $AB=x=9cm$

$BC=4\left(9\right)+4=40cm$

$AC=4\left(9\right)+5=41cm$