Question

0. In A PQR, RS is the bisector of ZR. If pq = 6cm, QR = 8cm
RP = 4cm then PS is equal to
a) 2cm b) 4cm c) 3cm d) 6cm
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Answer 1

Answer:

I don't know answer for this question

Answer 2

Step-by-step explanation:

It's quite simple :-

In ∆PQR,

PQ = 6cm

QR = 8cm

RP = 4cm

Let Sides PQ,QR and RP be a,b and c respectively.

Then, S of ∆PQR = a+b+c/2

= 6+8+4/2

= 18/2

= 9cm

Now, Using Heron's formula, we get :-

Area of ∆PQR = √S(S-a)(S-b)(S-c)

= √9(9-6)(9-8)(9-4)

= √9 × 3 × 1 × 5

= √27 × 5

= √135

= √3 × 3 × 15

= 3√15 cm^2

= 11.62 cm^2

Hope it helps :)