Answer
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Hint: AR and AQ are tangents to the circle from A. Similarly BP and BQ are tangent from point B. Thus these values are equal. Now from the figure get an equation for AQ and AR. Add them to get perimeter of $\Delta ABC$
Complete step-by-step answer:
From the figure you can understand $\Delta ABC$ , AB and AC are produced to Q and R. Thus it becomes AQ and AR. Now the circle touches $\Delta ABC$ at point P, where P is a point on the side BC of the given triangle ABC.
From the figure you say that AR and AQ are tangents from the point A. Thus AR will be equal to AQ.
i.e. AR = AQ, which is the length of the tangents from A to the circle. Hence as AR = AQ, we can also say that BQ = BP i.e. they are length of tangent from the point B to the circle.
Similarly CR = CP = length of the tangent from point C onto the circle. Hence we found out that
AR = AQ = 5cm
BQ = BP and CR = CP
From the figure we can say that AQ = AB + BQ as BQ = BP, it becomes
AQ = AB + BP…………………..(i)
Similarly from figure AR = AC + CR,
CR = CP
AR = AC + CP…………….(ii)
Now let add both (i) and (ii)
AQ + AR = AB + AC + BP + CP………………….(iii)
The perimeter of $\Delta ABC=AB+BC+CA$ .
Here BC = BP + BP and AQ = AR
Thus equation (iii) becomes
AQ + AQ = AB + AC + BC.
Perimeter of $\Delta ABC=5+5=10cm$
Hence we got the perimeter of $\Delta ABC$ as 10cm.
Therefore option (a) is correct.
Note: If a line is tangent to a circle, it is perpendicular to the radius drawn to the point of tangency. The tangent segment to a circle from the same external point is congruent.
Complete step-by-step answer:
From the figure you can understand $\Delta ABC$ , AB and AC are produced to Q and R. Thus it becomes AQ and AR. Now the circle touches $\Delta ABC$ at point P, where P is a point on the side BC of the given triangle ABC.
From the figure you say that AR and AQ are tangents from the point A. Thus AR will be equal to AQ.
i.e. AR = AQ, which is the length of the tangents from A to the circle. Hence as AR = AQ, we can also say that BQ = BP i.e. they are length of tangent from the point B to the circle.
Similarly CR = CP = length of the tangent from point C onto the circle. Hence we found out that
AR = AQ = 5cm
BQ = BP and CR = CP
From the figure we can say that AQ = AB + BQ as BQ = BP, it becomes
AQ = AB + BP…………………..(i)
Similarly from figure AR = AC + CR,
CR = CP
AR = AC + CP…………….(ii)
Now let add both (i) and (ii)
AQ + AR = AB + AC + BP + CP………………….(iii)
The perimeter of $\Delta ABC=AB+BC+CA$ .
Here BC = BP + BP and AQ = AR
Thus equation (iii) becomes
AQ + AQ = AB + AC + BC.
Perimeter of $\Delta ABC=5+5=10cm$
Hence we got the perimeter of $\Delta ABC$ as 10cm.
Therefore option (a) is correct.
Note: If a line is tangent to a circle, it is perpendicular to the radius drawn to the point of tangency. The tangent segment to a circle from the same external point is congruent.
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