In the given figure, PR, RT and PT are tangents to the circle at Q,
S and U respectively. If PR = (RT + 3) cm, PR = (PT + 1) cm and
perimeter of A PRT is 26 cm, then QR + RT equals to:
(A) 9 cm
(B) 7 cm
(C) 13 cm
(D) 11 cm
Answers
If PR = (RT + 3) cm, PR = (PT + 1) cm and perimeter of A PRT is 26 cm, then QR + RT equals to option (D): 11 cm
Step-by-step explanation:
Step 1:
It is given that,
PR, RT and PT are tangents to the circle at Q, S and U.
Since two tangents drawn from an external point to a circle are always equal in length.
So, we get
PQ = PU, TU = TS and RS = RQ
Let’s assume,
PQ = PU = “a”
TU = TS = “b”
RS = RQ = “c”
Therefore,
PR = PQ + RQ = a + c …… (i)
RT = RS + TS = c + b ….. (ii)
PT = PU + TU = a + b ….. (iii)
Step 2:
Also given,
PR = (RT + 3) cm
PR = (PT + 1) cm
Perimeter of ∆ PRT = 26 cm
We know,
The perimeter of ∆ PRT = PR + RT + PT
⇒ 26 = PR + PR – 3 + PR – 1
⇒ 3PR = 26 + 1 + 3
⇒ PR = 30/3
⇒ PR = 10 cm ….. (iv)
∴ RT = PR – 3 = 10 – 3 = 7 cm ….. (v)
And, PT = PR – 1 = 10 – 1 = 9 cm ….. (vi)
Step 3:
Substituting (iv), (v) & (vi) in (i), (ii) & (iii) and then adding the three equations so received, we get
a + c = 10
c + b = 7
a + b = 9
-------------------------
2 (a + b + c) = 26
--------------------------
⇒ a + b + c = 13
∴ QR = c = a+b+c – (a+b) = 13 – 9 = 4 cm ….. (vii)
Thus, from (v) & (vii), we get
The value of QR + RT is,
= 4 + 7
= 11 cm
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