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Question
Prove that a parallelogram circumscribing a circle is a rhombus [4 MARKS]
Prove that a parallelogram circumscribing a circle is a rhombus [4 MARKS]
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Solution
Concept: 1 Mark
Application: 3 Marks
Since ABCD is a parallelogram,
AB = CD …(i)
BC = AD …(ii)
As the lengths of tangents drawn from an external point to a circle are equal
DR = DS
CR = CQ
BP = BQ
AP = AS
Adding all these equations, we obtain
DR + CR + BP + AP = DS + CQ + BQ + AS
(DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)
CD + AB = AD + BC
On putting the values of equations (i) and (ii) in this equation, we obtain
2AB = 2BC
AB = BC …(iii)
Comparing equations (i), (ii), and (iii), we obtain
AB = BC = CD = DA
Hence, ABCD is a rhombus.
Concept: 1 Mark
Application: 3 Marks
Since ABCD is a parallelogram,
AB = CD …(i)
BC = AD …(ii)
As the lengths of tangents drawn from an external point to a circle are equal
DR = DS
CR = CQ
BP = BQ
AP = AS
Adding all these equations, we obtain
DR + CR + BP + AP = DS + CQ + BQ + AS
(DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)
CD + AB = AD + BC
On putting the values of equations (i) and (ii) in this equation, we obtain
2AB = 2BC
AB = BC …(iii)
Comparing equations (i), (ii), and (iii), we obtain
AB = BC = CD = DA
Hence, ABCD is a rhombus.
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