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Question
∆ABC with vertices A(–2, 0), B(2, 0) and C(0, 2) is similar to ∆DEF with vertices D(–4, 0), E(4, 0) and F(0, 4).
Options
True
False
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Solution
This statement is True.
Explanation:
Using distance formula,
d = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`
We can find,
AB = `sqrt((2 + 2)^2 + 0) = sqrt(16)` = 4
BC = `sqrt((0 - 2)^2 + (2 - 0)^2) = sqrt(8) = 2sqrt(2)`
CA = `sqrt((-2 - 0)^2 + (0 - 2)^2) = sqrt(8) = 2sqrt(2)`
DE = `sqrt((4 + 4)^2 + 0) = sqrt(64)` = 8
EF = `sqrt((0 - 4)^2 + (4 - 0)^2) = sqrt(32) = 4sqrt(2)`
FD = `sqrt((-4 - 0)^2 + (0 - 4)^2) = sqrt(32) = 4sqrt(2)`
∴ `("AB")/("DE") = ("BC")/("EF") = ("CA")/("FD") = 1/2`
⇒ ΔABC ∼ ΔDEF
Hence, triangle ABC and DEF are similar.
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