Correct option is A
Given:
Side lengths: PQ=18 cm, QR=13 cm, RP=15 cm.
A circle is inscribed in \triangle PQR, touching sides PQ, QR, and RP at points D, E, and F respectively.
Concept Used:
In a triangle, the lengths of the tangents drawn from an external point to a circle are equal. Therefore:
PD=PF
QD=QE
RE=RF
Solution:
PQ = PD + QD = x + y = 18 ...........(1)
QR = QE + RE = y + z = 13.......(2)
RP = RF + FP = z + x = 15.........(3)
Add above three equations, we get
2x+2y+2z = 46
x+ y+z = 23.........(4)
From equation 2, Put the value of y +z in equation 4, Then:
x = 23 - 13 = 10
So, PD = x = 10 cm
QR = QE + RE = y + z = 13.......(2)
RP = RF + FP = z + x = 15.........(3)
Add above three equations, we get
2x+2y+2z = 46
x+ y+z = 23.........(4)
From equation 2, Put the value of y +z in equation 4, Then:
x = 23 - 13 = 10
So, PD = x = 10 cm
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