Question

The sides AB , BC , and AC of a △ABC are tangent to a circle at points P, Q, and R respectively. Find AP, PB, BQ, QC, CR, and RA if AB = 10 cm, BC = 12 cm, and CA = 5 cm. Really need an answer

Answer 1

Answer:

AP = RA = 1.5 cm

PB = BQ = 8.5 cm

QC = CR = 3.5 cm

Step-by-step explanation:

The distance from the vertex to the two nearest tangent points is the same. If we say the distance AP = RA = x, then PB = BQ = 10-x, and QC = CR = 5 -x.

Since we know

BQ +QC = 12

We can substitute the above expressions involving x to find what x is.

(10 -x) +(5 -x) = 12

15 -2x = 12

x = (15 -12)/2 = 1.5

This tells us ...

AP = RA = 1.5 cm

PB = BQ = 8.5 cm

QC = CR = 3.5 cm

Answer 2

Answer:

AP = RA   = 1.5cm

BP=BQ=8.5 cm

QC=CR=3.5cm

Given :

AB = 10 cm, BC = 12 cm, and CA = 5 cm.

Step-by-step explanation:

The diagram is attached below. We know that the tangent lines  from a point are equal

Let AP = RA   = x  

BP=BQ=y

QC=CR=z

We know AB=10

AP+BP=10

x+y=10

BQ+QC=12

y+z=12

CR+RA=5

z+x=5

WE know that AB+BC+CA=10+12+5=27

$AP+BP+BQ+QC+CR+RA=27\\x+x+y+y+z+z=27\\2(x+y+z)= 27\\x+y+z=13.5$

we know that x+y=10, Replace it in above equation

$x+y+z=13.5\\Replace x+y with 10\\10+z=13.5\\z=3.5$

$x+y+z=13.5\\y+z=12\\x+12=13,5\\x=1.5\\\\x+y+z=13.5\\z+x=5\\5+y=13.5\\y=8.5$

Learn more : brainly.in/question/17213757