Question

In the given figure, $\angle B={90}^0$, $XY//BC$, $AB=12$ cm, $AY=8$ cm and $AX:XB=1:2$. Find the sum of the lengths of $AC$ and $BC$.( Rounded off to the nearest integer)

Answer 1

Since, If a line parallel to one side of a triangle intersects the other two sides of the triangle, then the line divides these two sides proportionally.

$\therefore \frac{AX}{XB}=\frac{AY}{YC}=\frac{1}{2}$

given, $AY=8$

$\Rightarrow YC=16$

$\Rightarrow AC=AY+YC=8+16=24$

Now by pythagoras theorem:

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

$\Rightarrow {12}^2+B{C}^2={24}^2$

$\Rightarrow B{C}^2=576-144=432$

$\Rightarrow BC=12\sqrt{3}=20.78$

$\therefore AC=24cm$and $BC=12\sqrt{3}=20.78$ $\approx$ $21cm$

Thus after rounding off the value of $BC$ to the nearest integer, we have

$AC+BC=24+21=45$