Question

(b) In ASTU, I(ST) = 7 cm,
(TU) = 4 cm, I(SU) = 5 cm
(c) In APOR, I(PQ) = 6 cm,
I(QR) = 3.8 cm, 7(PR) = 4.5 cm​

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.

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