Question

please solve this one

Answer 1

Since the lengths of tangents from an external point to a circle are equal.
So, AP=AR i.e.,6cm
BP=BQ i.e.,8cm
RC=CQ i.e.,x cm
Now, AB=AP+PB
=6+8 =14cm
AC=AR+RC
=(6+x) cm
BC=BQ+QC
=(8+x) cm
Now, s=AB+BC+AC / 2
=14+8+x+6+x / 2
=28+2x / 2
=14+x
ar(∆ABC)=√s(s-a)(s-b)(s-c)
=√14+x(14+x-14)(14+x-6-x)(14+x-8-x)
=√x+14.x.8.6
=√x+14.48x
=√48x²+672x

After this, see the snap given above :)