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3 years ago

Prove: segment SD = segment PD

Mathematics

1 answer:

Elza [17]3 years ago

7 0

Answer:

Step-by-step explanation:

SD is congruent to PD and both triangles share the same side with each other.

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On a coordinate plane, triangles P Q R and S T U are shown. Triangle P Q R has points (4, 4), (negative 2, 0), (negative 2, 4).

mamaluj [8]

Answer:

The answer is below

Step-by-step explanation:

Two triangles are said to be similar if their corresponding angles are equal and the corresponding sides are in proportion.

The distance between two points on the coordinate plane is given as:

$Distance=\sqrt{(x_2-x_1}^2+(y_2-y_1)^2 \\\\Therefore\ in\ triangle\ PQR:\\\\|QR|=\sqrt{(-2-(-2))^2+(4-0)^2}=4\\\\|PQ|=\sqrt{(-2-4)^2+(0-4)^2}=\sqrt{52}=2\sqrt{13} \\\\|PR|=\sqrt{(-2-4)^2+(4-4)^2}=6$

In triangle STU:

$|ST|=\sqrt{(-1-2)^2+(-2-(-4))^2}=\sqrt{13}\\\\|SU|=\sqrt{(-1-2)^2+(-4-(-4))^2} =3\\\\|TU|=\sqrt{(-1-(-1))^2+(-4-(-2))^2}=2$

|QR| / |TU| = 4/2 = 2

|PR| / |SU| = 6/3 = 2

|PQ| / |ST| = 2√13 / √13 = 2

Hence:

|QR| / |TU| = |PR| / |SU| = |PQ| / |ST|

Therefore, △PQR and △STU are similar triangles since the ratio of their sides are in the same proportion.

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