Question

In Fig. 10.80, the perimeter of ΔABC is
A. 30 cm
B. 60 cm
C. 45 cm
D. 15 cm

Answer 1

The Perimeter of the given triangle and be calculated by adding the length of all three sides of  the triangle.

Explanation:

So, what we have is al following:-

Given – ABC is a triangle.

BR = BP [ tangent to the circle from point B].

AR = AQ [ tangent to the circle from point A].

CP = CQ [ tangent to the circle from point C].

So, by this we have,  BR = BP = 6cm.

  AR = AQ = 4cm.

  CP = CQ = 5 cm.

But, AB = BR + RA

        BC = BP + PC

        AC = AQ + QC.

And Perimeter of the triangle ABC = AB + BC + CA

 So, Perimeter= (BR + RA) + (BP + PC) + (AQ + QC).

           =  (6 + 4) + (6 + 5) + (4 + 5)

           = ( 10) + (11) + (9)

           = 30 cm.

The perimeter of a triangle is equal to 30 cm. So, the correct answer is Option A.

Answer 2

$\Large{\underline{\underline{\bf{Solution :}}}}$

★ Taking A as tangent,

AQ = AR

AR = 4 cm

$\rule{150}{2}$

★ Taking C as center.

PC = QC

QC = 5 cm

$\rule{150}{2}$

★ Taking B as center

BR = BP

BP = 6 cm

$\rule{200}{2}$

$\small{\star{\boxed{\sf{Perimeter \: of \: triangle = Sum \: of \: all \: sides}}}}$

$\sf{→Perimeter = AR + BR + AQ + QC + BP + BC} \\ \\ \sf{→Perimeter = 4 + 4 + 5 + 5 + 6 + 6} \\ \\ \sf{→ Perimeter = 8 + 10 + 12} \\ \\ \sf{→Perimeter = 30 \: cm}$