Beth is planning a playground and has decided to place the swings in such a way that they are the same distance from the jungle gym and the monkey bars. If Beth places the swings at point D, how could she prove that point D is equidistant from the jungle gym and monkey bars?

If segment AD ≅ segment CD, then point D is equidistant from points A and B because congruent parts of congruent triangles are congruent.

If segment AD ≅ segment CD, then point D is equidistant from points A and B because a point on a perpendicular bisector is equidistant from the endpoints of the segment it intersects.

If m∠ACD = 90° then point D is equidistant from points A and B because congruent parts of congruent triangles are congruent.

If m∠ACD = 90° then point D is equidistant from points A and B because a point on a perpendicular bisector is equidistant from the endpoints of the segment it intersects.

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step-by-step explanation:

d≈7.07in

step-by-step explanation:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

answer:

f(5/2) = 10/3

i hope this you!

the correct option is 2.

step-by-step explanation:

according to the given information triangle abc is a right angle triangle with right angle at a. segment ab is 5 and segment ac is 7. point d is on segment bc and angles adb and adc are right angles.

it is given that triangles abd, cad, and cba are similar.

two triangles are similar if their corresponding sides are proportional.

in triangle cba, using pythagoras theorem

                      )

triangle abd and cba are similar,

                            (2)

triangle cad and cba are similar,

                      )

using (1), (2) and (3), we get

hence proved.

therefore option 2 is correct.