In ΔBCA, BA = 17 cm, BF = 6 cm, CG = 9 cm. Find the perimeter of ΔBCA.

Triangle BCA with inscribed circle D. Segments BF and BH, CF and CG, and AG and GH are tangent to circle D.

32 cm
35 cm
47 cm
52 cm

answer is 52 cm

Step-by-step explanation:

BA = 17, BF = 6, and CG = 9

triangle sides BF and BH are congruent because they are tangent lines along the same curve

BH is equal to 6 since BF is equal to 6

this applies to CG and CF and AG and AH

CG is equal to 9

so CF is equal to 9

now we add 6 + 9 for the measurement of side CB

CB = 15

now we do the same for the other two sides

we already know BA = 17 so we subtract it from BH = 6 to find the measurement of AH

AH is equal to 11 and is tangent across the same curve as AG

AG = 11

now we add AG to CG

11 + 9 = 20

now we know  all the side lengths

CB + BA + AC = Perimeter

15 + 17 + 20 = 52

Perimeter of triangle BCA = 52

I don't know why I went through all the trouble if I know no one is going to see this :'(. I just thought the other guy gave the wrong reasons

i dont know nkw

step-by-step explanation:

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step-by-step explanation:

there is a formula i can give you:

it is r^2*n*sin(360/n) divided by 2 where r represents the "radius" and n is the number of sides.

plug them in:

(10sqrt(3))^2*6*sin(360/6) divided by 2

100*3*6*sin(60) divided by 2

900*sin(60)

900*sqrt(3)/2

450*sqrt(3)

approx is 779.4 cm^2