Question

can anyone help me with this question...
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Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

(i)

Since tangents to a circle from the same point have the same length:

AP = AR,    BP = BQ    and    CQ = CR.

So...

AB + CQ = AP + BP + CQ = AR + BQ + CR = AC + BQ

(ii)

A radius and a tangent at the same point meet at right angles.

So ΔBPO is a right angled triangle.

area(BPO) = (1/2) × base × height

               = (1/2) × BP × PO

               = (1/2) × BP × r

Similarly:

area(BQO) = (1/2) × BQ × r

area(CQO) = (1/2) × CQ × r

area(CRO) = (1/2) × CR × r

area(ARO) = (1/2) × AR × r

area(APO) = (1/2) × AP × r

Adding these six areas gives:

area(ABC) = (1/2) × ( BP + BQ + CQ + CR + AR + AP ) × r

               = (1/2) × (perimeter of ABC) × r

[ Notice there is a mistake in the book for part (ii) ... it says ΔABQ where it should say ΔABC. ]