In the given figure, if RP: PK = 3:2, then
find the following ratios.
1. A(TRP): A(TPK)
11. A(TRK): A(TPK)
A(TRP): A(TRK)
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ΔTRP and ΔTPK have their bases on the same line and have a common vertex.
∴ The heights of these triangles are equal.
∴ Their areas are proportional to their corresponding bases.
RP:PK=3:2
(Given)
∴ PK RP = 2 3 ....... (i)
∴ A(ΔTPK) A(ΔTRP) = PK RP = 2 3
[From (i)] ∴A(ΔTRP):A(ΔTPK)=3:2
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