Given: δabc is a right triangle. prove: a2 + b2 = c2 right triangle bca with sides of length a, b, and c. perpendicular cd forms right triangles bdc and cda. cd measures h units, bd measures y units, da measures x units. the following two-column proof with missing justifications proves the pythagorean theorem using similar triangles: statement justification draw an altitude from point c to line segment ab let segment bc = a segment ca = b segment ab = c segment cd = h segment db = y segment ad = x y + x = c c over a equals a over y and c over b equals b over x a2 = cy; b2 = cx a2 + b2 = cy + b2 a2 + b2 = cy + cx a2 + b2 = c(y + x) a2 + b2 = c(c) a2 + b2 = c2 which is not a justification for the proof? addition property of equality pythagorean theorem pieces of right triangles similarity theorem cross product property

Pythagorean Theorem is not the justification for the proof.

Step-by-step explanation:

We can get the right answer with help of the below proof of Pythagoras theorem,

Since, Here ABC is the triangle in which CD is the altitude from the point C.

Where,   ( shown in figure)

Here, segment BC = a, segment CA = b, segment AB = c, segment CD = h

segment DB = y , segment AD = x  

y + x = c( c over a equals a over y and c over b equals b over x)

Since, ⇒ a/c=y/a and ⇒  b/x=c/b ( Pieces of Right Triangles Similarity Theorem)

⇒ a^2=cy and b^2=cx  (  Cross Product Property)

(Addition Property of Equality)

b: |12 – x| ≤ 0.25

step-by-step explanation:      

let x be the possible lengths in inches, of the ribbons jaime cut.

we have been given that jaime is cutting strips of ribbon for a project. he tries to cut each piece to be 12 inches long, but sometimes he cuts it too long or too short by as much as 0.25 inch.  

we can represent this information in an inequality as:

by the definition of absolute value function we will get,

upon combining our inequalities we will get,

therefore, the inequality   describes all the possible lengths (x), in inches, of the ribbons jaime cut.

step-by-step explanation: