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Question
In the given figure, AB and DE are perpendicular to BC.
1) Prove that ΔABC ∼ ΔDEC
2) If AB = 6 cm; DE = 4 cm and AC = 15 cm. Calculate CD.
3) Find the ratio of area of ΔABC: area of ΔDEC
Solution
1) In ΔABC and ΔDEC,
∠ABC = ∠DEC = 90°`
∠ACB = ∠DCE (Common)
∴ ΔABC ~ ΔDCE (AA criterion)
2) Since ΔABC ∼ ΔDEC,
`"AB"/"DE" = "AC"/"CD"`
`=> 6/4 = 15/"CD" => 6 xx CD = 60 => CD = 60/6 = 10 cm`
3) It is known that the ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
`=> (Ar(ABC))/(Ar(DEC)) = "AB"^2/"DE"^2 = 6^2/4^2 = 36/16 = 9/4`
`=> ar(ABC) : ar(DEC) = 9 : 4`
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