Construct a trapezoid ABCD such that AB is parallel to CD, having AB = 8 cm, BC = 5cm, and ABC = 60o.

let's solve your equation step-by-step.

4x+2=2x+12

step 1: subtract 2x from both sides.

4x+2−2x=2x+12−2x

2x+2=12

step 2: subtract 2 from both sides.

2x+2−2=12−2

2x=10

step 3: divide both sides by 2.

2x

2

=

10

2

x=5

step-by-step explanation:

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

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

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

in δacy and δaby,

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)

3 qrs + 3/2 xy hope this : /

step-by-step explanation:

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