Question

pls answer fast.
pls give a good explanation ​

Answer 1

Here,

BP = BR and CR = CQ

Perimeter of ΔABC = AB + BR + RC + CA

= AB + BP + QC + CA

= AP + QA (AP = QA)

= 2AP

ΔAPO

AO²=AP²+PO²

13² = AP² + 5²

AP² = 144

AP = 12

The perimeter of the ΔABC

2 x 12 = 24 cm is the answer



Here we have,

OA = 13cm

Radius = OP = 5cm

Since AP is a tangent to the circle with center O and OP is its radius, OP ⊥ AP

Now, In ΔOPA

∠OPA = 90°

AP^{2} = OA^{2} - OP^{2} {using pythagoras theorem}

= 13^{2} - 5^{2}

= 169 - 25

AP^{2} = 144

AP = \sqrt{144}

⇒ AP = 12

Now,

AP = \frac{1}{2} \times Perimeter of ΔABC

⇒ 12 = \frac{1}{2} \times Perimeter of ΔABC

⇒ Perimeter of ΔABC = 24cm