The vertices of δabc are a(1,–2), b(1,1), and c(5,–2).

Ответ:

duartealondra45

13.07.2020

Option B.

Step-by-step explanation:

The vertices of ΔABC are A(1,–2), B(1,1), and C(5,–2).

Distance formula:

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

Using distance formula we get

$AB=\sqrt{(1-1)^2+(1-(-2))^2}=\sqrt{3^2}=3$

$BC=\sqrt{(5-1)^2+(-2-1)^2}=\sqrt{4^2+3^2}=\sqrt{25}=5$

$AC=\sqrt{(5-1)^2+(-2-(-2))^2}=\sqrt{4^2}=4$

The length of sides are 3,4 and 5.

We need to find the side lengths of a triangle that is similar but not congruent to ΔABC.

It means the sides of new triangle are 3k, 4k and 5k, where k is a positive number and not equal to 1.

For k=3, the sides of new triangle are

3k=3(3)=9

4k=4(3)=12

5k=5(3)=15

The side lengths of a triangle that is similar but not congruent to ΔABC are 9, 12, and 15 units .

Therefore, the correct option is B.

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