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

______Heyy Buddy ❤_______

_____Here's your Answer _________

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

Therefore,

∆ 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:

we know that tangents drawn from an external point are equal in length

Step-by-step explanation:

hence,ap=ar

and pb=bq

so AR=6cm

and

BQ=8cm

we know that radius is perpendicular to the tangent

so BQ=BC

BC=8cm

hence length of BC is 16 cm

and length of AC IS

AR+RC                             (RC=QC)

So length of AC is 14 cm