On a coordinate plane, triangle A B C has points (negative 2, 7), (negative 2, 3), and (negative 6, 3) and triangle D E F has points (negative 2, negative 10), (negative 2, negative 2), and (6, negative 2). Given that StartFraction A B Over D E EndFraction = StartFraction B C Over E F EndFraction = one-half, complete the statements to show that △ABC ~ △DEF by the SAS similarity theorem. Horizontal and vertical lines are . So, angles are right angles by definition of perpendicular lines. All right angles are . Therefore, △ABC ~ △DEF by the SAS similarity theorem.

Question:

On a coordinate plane, △ABC has points (-2, 7), (-2, 3), and (-6, 3) and △DEF has points (-2,-10), (-2, -2), and (6, -2).

Given that

Complete the statements to show that △ABC ~ △DEF by the SAS similarity theorem.

Horizontal and vertical lines are __(1)__. So, angles are right angles by definition of perpendicular lines. All right angles are __(2)__

Therefore, △ABC ~ △DEF by the SAS similarity theorem.

(1) perpendicular

(2) congruent

Step-by-step explanation:

Required

Fill in the gaps

implies that:

In other words:

AB ~ DE ---- Side (S)

and

BC ~ EF ---- Side (S)

Now to fill in the gap with:

(1) perpendicular

(2) congruent

Further explanation

(1) Vertical lines and horizontal lines meet at a right angle (i.e. 90 degrees). Any two lines that meet at a right angle are perpendicular

(2) Angles with equal measures are congruent. Because right angles have a congruent angles of 90 degrees, then they are congruent.

(sin 42)/31. = sinx)/37

sinx = 37 (sin42)/31

= 37× 0.66/31 = 0.787

x= 51.9

other angle opposite to y = 180-51.9-42

= 180- 93.9 =86.1

so

sin42)/31 = sin 86.1)/y

0.021 = 0.997 /y

y= 0.997/0.021 =47.47