Question

In the given fig. AP = 4 cm, BQ = 6 cm and AC = 9 cm. Find the semi perimeter of ∆ ABC.

Answer 1

GIVEN:
AP = 4  cm, BQ = 6 cm  and AC = 9 cm.
Lengths of the tangent from an exterior point to a circle are equal.

AR = AP  = 4 cm [ From A].........(1)
BQ = BP  = 6 cm [From B]..........(2)
CR = CQ =   5 cm [From C].........(3)

[AC = AR + RC , 9 = 4 cm + RC, RC = 9 - 4 = 5 cm]

Adding equations 1, 2 & 3.
AR + BQ + CR = AP + BP + CQ

Perimeter of ∆ABC = AB + BC + AC

Perimeter of ∆ABC = (AP +PB) + ( CQ + BQ) + (AR + RC )
Perimeter of ∆ABC = (AP + AR) + (PB + BQ )+ (CQ + RC)
Perimeter of ∆ABC = (AP + AP) + (PB + PB)+ (CQ + CQ)

[FROM EQUATION 1, 2 AND 3]

Perimeter of ∆ABC = 2AP + 2PB +2CQ
Perimeter of ∆ABC = 2(AP + PB +CQ)
AP + PB +CQ = ½(Perimeter of ∆ABC)
4 + 6 +5 = ½(Perimeter of ∆ABC)

[FROM EQUATION 1, 2 AND 3]
15 = ½(Perimeter of ∆ABC)

Hence, semi perimeter of ∆ABC is 15 cm.

HOPE THIS WILL HELP YOU...

Answer 2

s= a+b+ c/2
where a,b and c are lengths of the sides
AC=9cm
∆PQB is an equilateral triangle
therefore ,PB= 6cm
AB= AP+PB
AB=10cm
∆ARP is an equilateral triangle
therefore AR=AP
AR= 4cm
now, RC=AC-AR
RC=5cm
now, similarly
QC=5cm
hence
BC=11cm
semi perimeter =AB+BC+AC/2
=10+11+9/2
=30/2
=15