CBSE Class 10 Answered
Since, OT is perpendicular bisector of PQ, therefore
PR = RQ… (1)
ButPQ = 8 km (given)
PR + RQ = 8
PR + PR= 8[using (1)]
PR = 4
RQ = PR = 4 km... (2) [using (1)]
In right triangle ORP, we have:
OP2 = OR2 + PR2
OR2 = OP2 - PR2
OR2 = 25 - 16 = 9
OR = 3 km
Since, TP is tangent to circle with centre O and OP is its radius, therefore,
OP TP
[ The tangent at any point of a circle is perpendicular to the radius through the point of contact]
OPT = 90o
In right triangle OPT, we have:
OT2 = PT2 + OP2
(TR + OR)2 = PT2 + 25
(TR + 3)2 = PT2 + 25... (4)
In right triangle PRT, we have:
PT2 = TR2 + PR2
PT2 = TR2 + 16... (5) [using (2)]
From(4) and (5), we have:
(TR + 3)2 = (TR2 + 16) + 25
TR2 + 9 + 6TR = TR2 + 41
6TR = 32
TR = ... (6)
Now, from (5) and (6), we get
We know tangents drawn from an external point to a circle are equal in length.
So, TP = TQ
Total length of the roads TP and TQ = km
Total cost of roads = = Rs. 1,60000
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