Question

What’s the answer to this please ?

Answer 1

Answer:

Yes

Step-by-step explanation:

The formula of a distance between two points:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

ΔABC:

A(2, 5), B(5, 5), C(5, 9)

$AB=\sqrt{(5-2)^2+(5-5)^2}=\sqrt{3^2+0^2}=\sqrt{9+0}=\sqrt9=3\\\\AC=\sqrt{(5-2)^2+(9-5)^2}=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5\\\\BC=\sqrt{(5-5)^2+(9-5)^2}=\sqrt{0^2+4^2}=\sqrt{0+16}=\sqrt{16}=4$

ΔDEF:

D(-7, 8), E(-4, 8), F(-4, 4)

$DE=\sqrt{-4-(-7))^2+(8-8)^2}=\sqrt{3^2+0^2}=\sqrt{9+0}=\sqrt9=3\\\\DF=\sqrt{(-4-(-7))^2+(4-8)^2}=\sqrt{3^2+(-4)^2}=\sqrt{9+16}=\sqrt{25}=5\\\\EF=\sqrt{(-4-(-4))^2+(4-8)^2}=\sqrt{0^2+(-4)^2}=\sqrt0+16}=\sqrt{16}=4$

AB ≅ DE

AC ≅ DF

BC ≅ EF

Therefore ΔABC ≅ ΔDEF