Question

Answer this question

Answer 1

Answer:

the answer must be 4√6cm OK

Answer 2

Answer:

$\dfrac{4\sqrt{6}}{3}$ cm

A is correct

Step-by-step explanation:

Length of pair of tangents from external point is equal.

So, AP=AR, BP=BQ, RC=QC

AP = AR = 4 cm

BP = BQ = 6 cm

AR+RC=AC

4+RC=12

RC = 8 = QC

Side of triangles,

• AB (c) = AP + PB = 4 + 6 = 10 cm
• BC (a) = BQ+QC = 6 + 8 = 14 cm
• AC (b) = AR+RC = 4 + 8 = 12 cm

Area of triangle $=\sqrt{s(s-a)(s-b)(s-c)}$

$s=\dfrac{a+b+c}{2}=\dfrac{10+14+12}{2}=18$

Let radius of circle be r cm

$sr=\sqrt{s(s-a)(s-b)(s-c)}$

$18r=\sqrt{18\cdot 8\cdot 4\cdot 6$

$r=\dfrac{4\dqrt{6}}{3}$ cm

Hence, The radius of circle is $\dfrac{4\sqrt{6}}{3}$ cm