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

Answer 1

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.