Show that $△ A B C$ , where A (-2, 0), B (2, 0), C (0, 2) and $△ P Q R$ , where P (-4, 0), Q (4, 0) R (0, 4) are similar.

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Solution

In ABC
AB =
BC =
AC =

In PQR
PQ =
QR =
RP =

$fraction numerator P Q over denominator A B end fraction equals 8 over 4 equals 2$
$fraction numerator Q R over denominator B C end fraction equals fraction numerator 4 square root of 2 over denominator 2 square root of 2 end fraction equals 2$
$fraction numerator R P over denominator A C end fraction equals fraction numerator 4 square root of 2 over denominator 2 square root of 2 end fraction equals 2$

$fraction numerator P Q over denominator A B end fraction equals fraction numerator Q R over denominator B C end fraction equals fraction numerator R P over denominator A C end fraction$

Hence PQR and ABC are similar.

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Similar questions

△ A B C

A (- 2, 0), B (2, 0), C (0, 2)

△ P Q R

P (- 4, 0), Q (4, 0), R (0, 4)

Show that $△ A B C$ , where A(-2, 0), B(0, 2), and C(2, 0) and $△ P Q R$ where P(-4, 0), Q(4, 0), R(0, 4) are similar triangles.

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