The diagonals of a cyclic quadrilateral ABCD
intersect at P and the area of the triangle APB
is 24 square cm. If AB is 8 cm and CD is 5 cm,
then what is the area of the triangle CPD ?
Answers
Area of ΔCPD = 75/8 cm² = 9.375 cm²
Step-by-step explanation:
ABCD is cyclic Quadrilateral
Angle formed by chord AD in same segment
=> ∠DCA = ∠ABD
=> ∠DCP = ∠ABP (as P is intersection of AC & BD)
Similarly
Angle formed by chord BC in same segment
=> ∠CDB = ∠BAC
=> ∠CDP = ∠BAP (as P is interscetion of AC & BD)
∠APB = ∠DPC ( opposite angles)
=> in Δ APB & ΔCPD
∠ABP = ∠DCP
∠BAP = ∠CDP
∠APB = ∠DPC
Δ APB ≈ ΔCPD
Ration of Area of Similar triangle = (ratio of sides)²
=> Area of Δ APB / Area of ΔCPD = ( AB/CD)²
=> 24/Area of ΔCPD = (8/5)²
=> Area of ΔCPD = 24 * 5²/8²
=> Area of ΔCPD = 75/8 cm² = 9.375 cm²
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Answer:
75/8 cm²
Step-by-step explanation:
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