SQ is the midsegment of triangle NOP. what is the length of OP? (SQ = 5x-2 & OP = 8x+6)

Answer: question no. 1 -a. yes, the sides are in the ratio 2 : 5.question no. 2 -b. 59°question no. 3 -c. angles x and y are remote interior angles.question no. 4 -d. 90°, 45° and 45°question no. 5 -b. mn and qrstep-by-step explanation: question no. 1 -rs : xu = 3 : 7.5 = 2 : 5rt : x? = 6.4 : 16 = 2 : 5st : u? = 6 : 15 = 2 : 5as the ratios of all sides are equivalent, the triangles must be congruent and the sides must be proportional.therefore, the answer is: a. yes, the sides are in the ratio 2 : 5.question no. 2 -in the second diagram, angle b is the only letter not shown. therefore, the number must be equivalent to angle b.if we substitute the value of angle b as being 22° in the first diagram, we can now use the angle sum of a triangle to obtain the value of angle a.angle sum of a triangle = 180°therefore: let angle a = aa + 99 + 22 = 180a = 180 - 99 - 22a = 59°therefore, the answer is: b. 59°question no. 3 -angles x and y don't share the vertex of the triangle with the exterior angle w. remote interior angles don't share a vertex of a triangle with an exterior angle. this means that angles x and y are remote interior angles.therefore, the answer is: c. angles x and y are remote interior angles.question no. 4 -the sum of all interior angles of a triangle must be equivalent to 180°.90°, 45° and 45° are the only set of angle measures that are equivalent to 180°.therefore, the answer is: d. 90°, 45° and 45°question no. 5 -the only pair of sides stated in the options which are corresponding are mn and qr.therefore, the answer is: b. mn and qr.

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

ac ≅ ab (given)< cax ≅ < bax   (given)ax ≅ ax (common side)

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

ac ≅ ab (given)< cay ≅ < bay (since < cax ≅ < bax is given and xy is the extension of line ax)ay ≅ ay (common side)

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)