Question

In Fig. 3, the radius of incircle of ABC of area 84 cm2

is 4 cm and the lengths of the

segments AP and BP into which side AB is divided by the point of contact are 6 cm and

8 cm. Find the lengths of the sides AC and BC.

Answer 1

In ∆ BPO and ∆ BQO,
OB = OB (Common sides)
OP =OQ. (Radius )
angle BPO = angle BQO. (both 90°)

then ,

∆ BPO is congruent to∆ BQO. (RHS)

BP = BQ. (CPCT)

BQ = 8 cm
Similarly
∆BQO is congruent to ∆ CQO,
BQ = QC
QC = 8cm
BC = QC + BQ
BC = 8 + 8 = 16cm

Similarly,
∆ CQO and ∆ COR are congruent,
CQ = CR. (CPCT)
CR = 8cm.

In ∆ APO and ∆ ARO,
AO = AO. (Common)
angle P = angle R. (each 90°)
PO = RO. ( Radius)

then ,
∆ APO is congruent to ∆ ARO
AP = AR. (CPCT)
AR = 6 cm
And,
AC = AR + RC = 6 + 8 = 14 cm
AC = 14 cm

Answer 2

Answer:

AC=13 cm ; AB=15 cm

Step-by-step explanation:

                           AC=(x+6) cm ,AB=(x+8) cm

                            Semi perimeter=x+14

                            Area of ΔABC =√s(s-a)(s-b)(s-c)

                                                    =√$48x^{2}$+672x

                                 ar ΔABC=ar(ΔOBC)+ar(ΔBOC)+ar(ΔAOC)

                                   √$48x^{2} +672x$=$16(x+14)^{2}$

                                                   x=7                  

                                          AC=13cm ; AB=15cm