Question

The length of common chord of two intersecting circles is $30cm$. If the diameters of these two circles are $50$ cm and $34$ cm, Calculate the distance between their centres.

Answer 1

length of common chord $=30cm=CD$

$radiusofcircl{e}_1=\frac{diameters}{2}=\frac{5D}{2}=25cm=AC$

$radiusofcircl{e}_2=\frac{diameters}{2}=\frac{34}{2}=17cm=BC$

$OC=\frac{CD}{2}=\frac{30}{2}=15cm$

In triangle $ACD$

$(AD{)}^2+(OC{)}^2=(AC{)}^2$

$(AO{)}^2=(AC{)}^2-(OC{)}^2$

$=(25{)}^2-(15{)}^2=625-225=400$

$\Rightarrow AO=20$

In a triangle $BCO$

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

$\Rightarrow (BO{)}^2=(CB{)}^2-(OC{)}^2$

$=(17{)}^2-(15{)}^2=52\times 2=64$

$\Rightarrow BO=8cm$

$\therefore$ distance between centres $AB=OB+OA$

$=8+20$

$AB=28cm$

$dostancebetweenthecentreis28cm$