Which statements are true about triangle qrs? check all that apply.

the side opposite ∠q is rs.
the side opposite ∠q is rq.
the hypotenuse is qr.
the side adjacent to ∠r is sq.
the side adjacent to ∠q is qs.

More Details:

A) True. Side RS is as far away from Q as you can get. The letter Q is nowhere in the sequence RS.

B) False. The letter Q is found in the sequence RQ so this side is either adjacent or the hypotenuse (assuming we know nothing about angle S). It turns out RQ is the hypotenuse since angle S is the 90 degree angle. The hypotenuse is always opposite the 90 degree angle.

C) True. Refer to choice B above. QR is the same as RQ. The order of the letters does not matter when you label a segment.

D) False. Adjacent sides must include the reference angle. Angle R is the reference angle in this case, so R must be in the sequence. The adjacent side is RS or SR.

E) True. The letter Q is found in QS. This angle is touching or close to the segment QS. Therefore QS is adjacent to angle Q.

For reference angle Q,

* The opposite side is RS or SR

* The adjacent side is QS or SQ

For reference angle R

* The opposite side is QS or SQ

* The adjacent side is RS or SR

The hypotenuse is always the side opposite the largest angle, which in this case is the 90 degree angle. The hypotenuse will be the same no matter what reference angle you use. You do not use the 90 degree angle as the reference angle.

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step-by-step explanation:

tell you what. i'll find the length of the medians and you can fill in the blanks where they belong.

br

b = (2a,0)

r = (a/2,b/2) remember these things are medians. they go to the 1/2 way point of the line opposite the vertex they face.

br^2 =   (2a - a/2)^2 + (0 - b/2)^2

br^2 = (3/2 a) ^2 + b^2 / 4

br^2 = 9/4 a^2 + b^2 / 4   we need to find some relationship between a and b.

let's try ab = bc  

ab = 2a

bc = (2a - a)^2 + (b - 0)^2

bc = sqrt(a^2 + b^2)

ab = bc

2a = sqrt(a^2 + b^2) square both sides.

4a^2 = a^2 + b^2 subtract a^2 from both sides.

sqrt(3a^2) = sqrt(b^2)

sqrt(3)a = b

let's put b into br^2

br^2 = 9/4 a^2 + 3b^2 / 4

br^2 = 12 a^2 / 4

br^2 = 3a^2 

br = sqrt(3) * a

cp

c = (a,b) ; p = (a,0) this is another application of the distance formula, and it is a good one.

cp^2 = (a - a)^2 + (b - 0)^2

cp^2 = 0 + b^2

cp^2 = b^2 which you can see from the diagram. 

cp = sqrt(b^2)

cp = b but b = sqrt(3) * a 

cp = sqrt(3)*a

aq

a = (0,0)

q = (3/2) a, b/2

aq^2 = (3a/2 -0)^2 + (b/2 - 0)^2

aq^2 = 9a^2/4 + b^2/4

but b = sqrt(3) * a 

aq^2 = 9a^2 / 4 + (sqrt(3)*a)^2 /4

aq^2 = 9 a^2/ 4 + 3 a^2 / 4

aq^2 = 12 a^2/4

aq^2 = 3 a^2

aq = (sqrt  3) * a which agrees with the other two.